Convert Ajax submit form then get HTML response to Pure Javascript - javascript

//THIS AJAX CODE WORKING GREAT.
$(document).ready(function(e) {
$("#ajaxupload").on('submit',(function(e) {
e.preventDefault();
$.ajax({
url: "http://example.com/upload",
type: "POST",
data: new FormData(this),
mimeType:"multipart/form-data",
contentType: false,
cache: false,
processData:false,
success: function(data){
//if success. Response is HTML. data = html. insert to my .result-wrapper.
$(".result-wrapper").prepend(data);
},
error: function(){
console.log('there\'s error!')
}
});
}));
});
How to convert this Ajax Jquery to pure Javascript? I've try to find solution around Stackoverflow then try to implement codes by answer was mark as accepted still error, my backend Controller not detect the data input value..
With ajax codes above.
1. Sumbit and get response without refresh the page.
Response is HTML.
No need to set data. because data already set in my HTML form.
Form have multiple input file upload and multiple input name. So length of input not a static number, depend files.
How to implement it with javascript? Submit without refresh, get response, then possible to submit data filled in HTML dinamically?
I've try do that with code below.
document.getElementById('ajaxupload').addEventListener('submit', function(e) {
e.preventDefault();
//e.submit();
var xhr = new XMLHttpRequest();
xhr.open('POST', 'http://example.com/upload/', true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
xhr.onload = function () {
// do something to response
console.log(this.responseText);
};
//xhr.send();
xhr.send(document.getElementById('ajaxupload').innerHTML); //my form id = ajaxupload
});
HTML:
<form action="http://example.com/upload" id="ajaxupload" enctype="multipart/form-data" method="post" accept-charset="utf-8">
<input id="insert-file" name="usr_files[]" type="file" multiple="">
<!--other input will generate inside this form on change `insert-file` .. depend on how many length file selected. ex if 2 files selected:
<input class="custom-file-name" name="usr_files[text][0]" type="text" value="My custom file name" required/>
<input class="custom-file-name" name="usr_files[text][1]" type="text" value="My custom file name no.2" required/> -->
</form>

Use the same formdata object in the native code
document.getElementById('ajaxupload').addEventListener('submit', function(e) {
e.preventDefault();
var formData = new FormData(this);
var xhr = new XMLHttpRequest();
xhr.open('POST', 'http://example.com/upload/');
xhr.onreadystatechange = function() {
if ( xhr.readyState === 4 && xhr.status === 200 ) {
console.log( xhr.responseText );
}
}
xhr.send( formData ); //my form id = ajaxupload
});

function ajaxGet(url, cb, token){
      var ajaxReq = new XMLHttpRequest();
      ajaxReq.addEventListener('load', function(){
        if(ajaxReq.status === 200) cb(null, {responseText: ajaxReq.responseText, rawAjaxRequest: ajaxReq});
        else cb({statusCode: ajaxReq.status, rawAjaxRequest: ajaxReq}, null);
      });
      ajaxReq.addEventListener('error', function(data){
        console.dir(data);
        var err = new Error('A fatal error occurred during ajaxGet, see console for more information');
        err.name = 'XMLHttpRequestError';
        cb(err, null);
      });
      ajaxReq.open('GET', url, true);
      if(token){
        ajaxReq.setRequestHeader('Authorization', token);
      }
      ajaxReq.send();
    },

Related

JavaScript Multiple Forms [duplicate]

If I have a form like this,
<form action="/Car/Edit/17" id="myForm" method="post" name="myForm"> ... </form>
how can I submit it without redirecting to another view by JavaScript/jQuery?
I read plenty of answers from Stack Overflow, but all of them redirect me to the view returned by the POST function.
You can achieve that by redirecting the form's action to an invisible <iframe>. It doesn't require any JavaScript or any other type of scripts.
<iframe name="dummyframe" id="dummyframe" style="display: none;"></iframe>
<form action="submitscript.php" target="dummyframe">
<!-- Form body here -->
</form>
In order to achieve what you want, you need to use jQuery Ajax as below:
$('#myForm').submit(function(e){
e.preventDefault();
$.ajax({
url: '/Car/Edit/17/',
type: 'post',
data:$('#myForm').serialize(),
success:function(){
// Whatever you want to do after the form is successfully submitted
}
});
});
Also try this one:
function SubForm(e){
e.preventDefault();
var url = $(this).closest('form').attr('action'),
data = $(this).closest('form').serialize();
$.ajax({
url: url,
type: 'post',
data: data,
success: function(){
// Whatever you want to do after the form is successfully submitted
}
});
}
Final solution
This worked flawlessly. I call this function from Html.ActionLink(...)
function SubForm (){
$.ajax({
url: '/Person/Edit/#Model.Id/',
type: 'post',
data: $('#myForm').serialize(),
success: function(){
alert("worked");
}
});
}
Since all current answers use jQuery or tricks with iframe, figured there is no harm to add method with just plain JavaScript:
function formSubmit(event) {
var url = "/post/url/here";
var request = new XMLHttpRequest();
request.open('POST', url, true);
request.onload = function() { // request successful
// we can use server response to our request now
console.log(request.responseText);
};
request.onerror = function() {
// request failed
};
request.send(new FormData(event.target)); // create FormData from form that triggered event
event.preventDefault();
}
// and you can attach form submit event like this for example
function attachFormSubmitEvent(formId){
document.getElementById(formId).addEventListener("submit", formSubmit);
}
Place a hidden iFrame at the bottom of your page and target it in your form:
<iframe name="hiddenFrame" width="0" height="0" border="0" style="display: none;"></iframe>
<form action="/Car/Edit/17" id="myForm" method="post" name="myForm" target="hiddenFrame"> ... </form>
Quick and easy. Keep in mind that while the target attribute is still widely supported (and supported in HTML5), it was deprecated in HTML 4.01.
So you really should be using Ajax to future-proof.
Okay, I'm not going to tell you a magical way of doing it because there isn't.
If you have an action attribute set for a form element, it will redirect.
If you don't want it to redirect simply don't set any action and set onsubmit="someFunction();"
In your someFunction() you do whatever you want, (with AJAX or not) and in the ending, you add return false; to tell the browser not to submit the form...
One-liner solution as of 2020, if your data is not meant to be sent as multipart/form-data or application/x-www-form-urlencoded:
<form onsubmit='return false'>
<!-- ... -->
</form>
You need Ajax to make it happen. Something like this:
$(document).ready(function(){
$("#myform").on('submit', function(){
var name = $("#name").val();
var email = $("#email").val();
var password = $("#password").val();
var contact = $("#contact").val();
var dataString = 'name1=' + name + '&email1=' + email + '&password1=' + password + '&contact1=' + contact;
if(name=='' || email=='' || password=='' || contact=='')
{
alert("Please fill in all fields");
}
else
{
// Ajax code to submit form.
$.ajax({
type: "POST",
url: "ajaxsubmit.php",
data: dataString,
cache: false,
success: function(result){
alert(result);
}
});
}
return false;
});
});
See jQuery's post function.
I would create a button, and set an onClickListener ($('#button').on('click', function(){});), and send the data in the function.
Also, see the preventDefault function, of jQuery!
The desired effect can also be achieved by moving the submit button outside of the form as described here:
Prevent page reload and redirect on form submit ajax/jquery
Like this:
<form id="getPatientsForm">
Enter URL for patient server
<br/><br/>
<input name="forwardToUrl" type="hidden" value="/WEB-INF/jsp/patient/patientList.jsp" />
<input name="patientRootUrl" size="100"></input>
<br/><br/>
</form>
<button onclick="javascript:postGetPatientsForm();">Connect to Server</button>
Using this snippet, you can submit the form and avoid redirection. Instead you can pass the success function as argument and do whatever you want.
function submitForm(form, successFn){
if (form.getAttribute("id") != '' || form.getAttribute("id") != null){
var id = form.getAttribute("id");
} else {
console.log("Form id attribute was not set; the form cannot be serialized");
}
$.ajax({
type: form.method,
url: form.action,
data: $(id).serializeArray(),
dataType: "json",
success: successFn,
//error: errorFn(data)
});
}
And then just do:
var formElement = document.getElementById("yourForm");
submitForm(formElement, function() {
console.log("Form submitted");
});
Fire and forget vanilla js + svelte
function handleSubmit(e) {
const request = new Request(`/products/${item.ItemCode}?_method=PUT`, {
method: 'POST',
body: new FormData(e.target),
});
fetch(request)
}
Used in Svelte:
<form method="post" on:submit|preventDefault={handleSubmit}>
If you control the back end, then use something like response.redirect instead of response.send.
You can create custom HTML pages for this or just redirect to something you already have.
In Express.js:
const handler = (req, res) => {
const { body } = req
handleResponse(body)
.then(data => {
console.log(data)
res.redirect('https://yoursite.com/ok.html')
})
.catch(err => {
console.log(err)
res.redirect('https://yoursite.com/err.html')
})
}
...
app.post('/endpoint', handler)

ajax submit wrong data to PHP

This is html file that submit ajax by button click to PHP script(on IIS).
But PHP script received wrong formatted data (there are brackets added [] and no parameter 'section' transmitted
What can be wrong
It would be good to have solution both: in JQuery and pure javascript
------------------- HTML
<!DOCTYPE html>
<html STYLE="height:100%;">
<head></head>
<body>
<SCRIPT>
function zPostToTopic_ajax(){
var url='http://the_server/infospace/php/infospace2.php';
var formData2 = new FormData();
formData2.append('section', 'general');
formData2.append('action2', 'preview');
http_request=new XMLHttpRequest();//work for IE11 too, // code for IE7+, Firefox, Chrome, Opera, Safari
http_request.open("POST", url);
//------------------------------------
http_request.onreadystatechange = function() {
if(http_request.readyState == 4 && http_request.status == 200)
alert(http_request.responseText)
}
http_request.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
http_request.send(formData2);
}
</SCRIPT>
<FORM NAME=form_post_to_topic ID=form_post_to_topic METHOD=POST action="http://the_server/infospace/php/infospace2.php">
&nbsp <INPUT TYPE=BUTTON VALUE=Send onClick="zPostToTopic_ajax();return false;">
</FORM>
</body>
</html>
-------------------------- PHP script
<?php
print_r($_REQUEST);
?>
--------------------------- Received data:
Array
(
[-----------------------------276402058428
Content-Disposition:_form-data;_name] => "section"
general
-----------------------------276402058428
Content-Disposition: form-data; name="action2"
preview
-----------------------------276402058428--
)
Use jQuery's .ajax() function. Here's an example where I post a file upload too.
var jform = new FormData();
jform.append('supply_id',supply_id);
jform.append('fuel_usage',$('#fuel_usage').val());
jform.append('cost',$('#cost').val());
jform.append('currency',$('#currency').val());
jform.append('evidence',$('#evidence').get(0).files[0]);
$.ajax({
url: '/your-form-processing-page-url-here',
type: 'POST',
data: jform,
dataType: 'json',
contentType: false,
cache: false,
processData: false,
success: function(data, status, jqXHR){
alert('Hooray! All is well.');
console.log(data);
console.log(status);
console.log(jqXHR);
},
error: function(jqXHR,status,error){
// Hopefully we should never reach here
console.log(jqXHR);
console.log(status);
console.log(error);
}
});
Your problem is that you are setting the wrong content type for your request. When you use a formdata object the content type will be multi-part/formdata.
So when you are using a formdata object you do not set the content type and it is set for you.
function zPostToTopic_ajax(){
var url='http://the_server/infospace/php/infospace2.php';
var formData2 = new FormData();
formData2.append('section', 'general');
formData2.append('action2', 'preview');
http_request=new XMLHttpRequest();//work for IE11 too, // code for IE7+, Firefox, Chrome, Opera, Safari
http_request.open("POST", url);
//------------------------------------
http_request.onreadystatechange = function() {
if(http_request.readyState == 4 && http_request.status == 200)
alert(http_request.responseText)
}
http_request.send(formData2);
}

i'm trying to make a "link" in my php page so it make changes in database [duplicate]

I am trying to send data from a form to a database. Here is the form I am using:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
The typical approach would be to submit the form, but this causes the browser to redirect. Using jQuery and Ajax, is it possible to capture all of the form's data and submit it to a PHP script (an example, form.php)?
Basic usage of .ajax would look something like this:
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
jQuery:
// Variable to hold request
var request;
// Bind to the submit event of our form
$("#foo").submit(function(event){
// Prevent default posting of form - put here to work in case of errors
event.preventDefault();
// Abort any pending request
if (request) {
request.abort();
}
// setup some local variables
var $form = $(this);
// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// Serialize the data in the form
var serializedData = $form.serialize();
// Let's disable the inputs for the duration of the Ajax request.
// Note: we disable elements AFTER the form data has been serialized.
// Disabled form elements will not be serialized.
$inputs.prop("disabled", true);
// Fire off the request to /form.php
request = $.ajax({
url: "/form.php",
type: "post",
data: serializedData
});
// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
// Log a message to the console
console.log("Hooray, it worked!");
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
// Log the error to the console
console.error(
"The following error occurred: "+
textStatus, errorThrown
);
});
// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// Reenable the inputs
$inputs.prop("disabled", false);
});
});
Note: Since jQuery 1.8, .success(), .error() and .complete() are deprecated in favor of .done(), .fail() and .always().
Note: Remember that the above snippet has to be done after DOM ready, so you should put it inside a $(document).ready() handler (or use the $() shorthand).
Tip: You can chain the callback handlers like this: $.ajax().done().fail().always();
PHP (that is, form.php):
// You can access the values posted by jQuery.ajax
// through the global variable $_POST, like this:
$bar = isset($_POST['bar']) ? $_POST['bar'] : null;
Note: Always sanitize posted data, to prevent injections and other malicious code.
You could also use the shorthand .post in place of .ajax in the above JavaScript code:
$.post('/form.php', serializedData, function(response) {
// Log the response to the console
console.log("Response: "+response);
});
Note: The above JavaScript code is made to work with jQuery 1.8 and later, but it should work with previous versions down to jQuery 1.5.
To make an Ajax request using jQuery you can do this by the following code.
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
<!-- The result of the search will be rendered inside this div -->
<div id="result"></div>
JavaScript:
Method 1
/* Get from elements values */
var values = $(this).serialize();
$.ajax({
url: "test.php",
type: "post",
data: values ,
success: function (response) {
// You will get response from your PHP page (what you echo or print)
},
error: function(jqXHR, textStatus, errorThrown) {
console.log(textStatus, errorThrown);
}
});
Method 2
/* Attach a submit handler to the form */
$("#foo").submit(function(event) {
var ajaxRequest;
/* Stop form from submitting normally */
event.preventDefault();
/* Clear result div*/
$("#result").html('');
/* Get from elements values */
var values = $(this).serialize();
/* Send the data using post and put the results in a div. */
/* I am not aborting the previous request, because it's an
asynchronous request, meaning once it's sent it's out
there. But in case you want to abort it you can do it
by abort(). jQuery Ajax methods return an XMLHttpRequest
object, so you can just use abort(). */
ajaxRequest= $.ajax({
url: "test.php",
type: "post",
data: values
});
/* Request can be aborted by ajaxRequest.abort() */
ajaxRequest.done(function (response, textStatus, jqXHR){
// Show successfully for submit message
$("#result").html('Submitted successfully');
});
/* On failure of request this function will be called */
ajaxRequest.fail(function (){
// Show error
$("#result").html('There is error while submit');
});
The .success(), .error(), and .complete() callbacks are deprecated as of jQuery 1.8. To prepare your code for their eventual removal, use .done(), .fail(), and .always() instead.
MDN: abort() . If the request has been sent already, this method will abort the request.
So we have successfully send an Ajax request, and now it's time to grab data to server.
PHP
As we make a POST request in an Ajax call (type: "post"), we can now grab data using either $_REQUEST or $_POST:
$bar = $_POST['bar']
You can also see what you get in the POST request by simply either. BTW, make sure that $_POST is set. Otherwise you will get an error.
var_dump($_POST);
// Or
print_r($_POST);
And you are inserting a value into the database. Make sure you are sensitizing or escaping All requests (whether you made a GET or POST) properly before making the query. The best would be using prepared statements.
And if you want to return any data back to the page, you can do it by just echoing that data like below.
// 1. Without JSON
echo "Hello, this is one"
// 2. By JSON. Then here is where I want to send a value back to the success of the Ajax below
echo json_encode(array('returned_val' => 'yoho'));
And then you can get it like:
ajaxRequest.done(function (response){
alert(response);
});
There are a couple of shorthand methods. You can use the below code. It does the same work.
var ajaxRequest= $.post("test.php", values, function(data) {
alert(data);
})
.fail(function() {
alert("error");
})
.always(function() {
alert("finished");
});
I would like to share a detailed way of how to post with PHP + Ajax along with errors thrown back on failure.
First of all, create two files, for example form.php and process.php.
We will first create a form which will be then submitted using the jQuery .ajax() method. The rest will be explained in the comments.
form.php
<form method="post" name="postForm">
<ul>
<li>
<label>Name</label>
<input type="text" name="name" id="name" placeholder="Bruce Wayne">
<span class="throw_error"></span>
<span id="success"></span>
</li>
</ul>
<input type="submit" value="Send" />
</form>
Validate the form using jQuery client-side validation and pass the data to process.php.
$(document).ready(function() {
$('form').submit(function(event) { //Trigger on form submit
$('#name + .throw_error').empty(); //Clear the messages first
$('#success').empty();
//Validate fields if required using jQuery
var postForm = { //Fetch form data
'name' : $('input[name=name]').val() //Store name fields value
};
$.ajax({ //Process the form using $.ajax()
type : 'POST', //Method type
url : 'process.php', //Your form processing file URL
data : postForm, //Forms name
dataType : 'json',
success : function(data) {
if (!data.success) { //If fails
if (data.errors.name) { //Returned if any error from process.php
$('.throw_error').fadeIn(1000).html(data.errors.name); //Throw relevant error
}
}
else {
$('#success').fadeIn(1000).append('<p>' + data.posted + '</p>'); //If successful, than throw a success message
}
}
});
event.preventDefault(); //Prevent the default submit
});
});
Now we will take a look at process.php
$errors = array(); //To store errors
$form_data = array(); //Pass back the data to `form.php`
/* Validate the form on the server side */
if (empty($_POST['name'])) { //Name cannot be empty
$errors['name'] = 'Name cannot be blank';
}
if (!empty($errors)) { //If errors in validation
$form_data['success'] = false;
$form_data['errors'] = $errors;
}
else { //If not, process the form, and return true on success
$form_data['success'] = true;
$form_data['posted'] = 'Data Was Posted Successfully';
}
//Return the data back to form.php
echo json_encode($form_data);
The project files can be downloaded from http://projects.decodingweb.com/simple_ajax_form.zip.
You can use serialize. Below is an example.
$("#submit_btn").click(function(){
$('.error_status').html();
if($("form#frm_message_board").valid())
{
$.ajax({
type: "POST",
url: "<?php echo site_url('message_board/add');?>",
data: $('#frm_message_board').serialize(),
success: function(msg) {
var msg = $.parseJSON(msg);
if(msg.success=='yes')
{
return true;
}
else
{
alert('Server error');
return false;
}
}
});
}
return false;
});
HTML:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" class="inputs" name="bar" type="text" value="" />
<input type="submit" value="Send" onclick="submitform(); return false;" />
</form>
JavaScript:
function submitform()
{
var inputs = document.getElementsByClassName("inputs");
var formdata = new FormData();
for(var i=0; i<inputs.length; i++)
{
formdata.append(inputs[i].name, inputs[i].value);
}
var xmlhttp;
if(window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest;
}
else
{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
}
}
xmlhttp.open("POST", "insert.php");
xmlhttp.send(formdata);
}
I use the way shown below. It submits everything like files.
$(document).on("submit", "form", function(event)
{
event.preventDefault();
var url = $(this).attr("action");
$.ajax({
url: url,
type: 'POST',
dataType: "JSON",
data: new FormData(this),
processData: false,
contentType: false,
success: function (data, status)
{
},
error: function (xhr, desc, err)
{
console.log("error");
}
});
});
If you want to send data using jQuery Ajax then there is no need of form tag and submit button
Example:
<script>
$(document).ready(function () {
$("#btnSend").click(function () {
$.ajax({
url: 'process.php',
type: 'POST',
data: {bar: $("#bar").val()},
success: function (result) {
alert('success');
}
});
});
});
</script>
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input id="btnSend" type="button" value="Send" />
<script src="http://code.jquery.com/jquery-1.7.2.js"></script>
<form method="post" id="form_content" action="Javascript:void(0);">
<button id="desc" name="desc" value="desc" style="display:none;">desc</button>
<button id="asc" name="asc" value="asc">asc</button>
<input type='hidden' id='check' value=''/>
</form>
<div id="demoajax"></div>
<script>
numbers = '';
$('#form_content button').click(function(){
$('#form_content button').toggle();
numbers = this.id;
function_two(numbers);
});
function function_two(numbers){
if (numbers === '')
{
$('#check').val("asc");
}
else
{
$('#check').val(numbers);
}
//alert(sort_var);
$.ajax({
url: 'test.php',
type: 'POST',
data: $('#form_content').serialize(),
success: function(data){
$('#demoajax').show();
$('#demoajax').html(data);
}
});
return false;
}
$(document).ready(function_two());
</script>
In your php file enter:
$content_raw = file_get_contents("php://input"); // THIS IS WHAT YOU NEED
$decoded_data = json_decode($content_raw, true); // THIS IS WHAT YOU NEED
$bar = $decoded_data['bar']; // THIS IS WHAT YOU NEED
$time = $decoded_data['time'];
$hash = $decoded_data['hash'];
echo "You have sent a POST request containing the bar variable with the value $bar";
and in your js file send an ajax with the data object
var data = {
bar : 'bar value',
time: calculatedTimeStamp,
hash: calculatedHash,
uid: userID,
sid: sessionID,
iid: itemID
};
$.ajax({
method: 'POST',
crossDomain: true,
dataType: 'json',
crossOrigin: true,
async: true,
contentType: 'application/json',
data: data,
headers: {
'Access-Control-Allow-Methods': '*',
"Access-Control-Allow-Credentials": true,
"Access-Control-Allow-Headers" : "Access-Control-Allow-Headers, Origin, X-Requested-With, Content-Type, Accept, Authorization",
"Access-Control-Allow-Origin": "*",
"Control-Allow-Origin": "*",
"cache-control": "no-cache",
'Content-Type': 'application/json'
},
url: 'https://yoururl.com/somephpfile.php',
success: function(response){
console.log("Respond was: ", response);
},
error: function (request, status, error) {
console.log("There was an error: ", request.responseText);
}
})
or keep it as is with the form-submit. You need this only, if you want to send a modified request with calculated additional content and not only some form-data, which is entered by the client. For example a hash, a timestamp, a userid, a sessionid and the like.
Handling Ajax errors and loader before submit and after submitting success shows an alert boot box with an example:
var formData = formData;
$.ajax({
type: "POST",
url: url,
async: false,
data: formData, // Only input
processData: false,
contentType: false,
xhr: function ()
{
$("#load_consulting").show();
var xhr = new window.XMLHttpRequest();
// Upload progress
xhr.upload.addEventListener("progress", function (evt) {
if (evt.lengthComputable) {
var percentComplete = (evt.loaded / evt.total) * 100;
$('#addLoad .progress-bar').css('width', percentComplete + '%');
}
}, false);
// Download progress
xhr.addEventListener("progress", function (evt) {
if (evt.lengthComputable) {
var percentComplete = evt.loaded / evt.total;
}
}, false);
return xhr;
},
beforeSend: function (xhr) {
qyuraLoader.startLoader();
},
success: function (response, textStatus, jqXHR) {
qyuraLoader.stopLoader();
try {
$("#load_consulting").hide();
var data = $.parseJSON(response);
if (data.status == 0)
{
if (data.isAlive)
{
$('#addLoad .progress-bar').css('width', '00%');
console.log(data.errors);
$.each(data.errors, function (index, value) {
if (typeof data.custom == 'undefined') {
$('#err_' + index).html(value);
}
else
{
$('#err_' + index).addClass('error');
if (index == 'TopError')
{
$('#er_' + index).html(value);
}
else {
$('#er_TopError').append('<p>' + value + '</p>');
}
}
});
if (data.errors.TopError) {
$('#er_TopError').show();
$('#er_TopError').html(data.errors.TopError);
setTimeout(function () {
$('#er_TopError').hide(5000);
$('#er_TopError').html('');
}, 5000);
}
}
else
{
$('#headLogin').html(data.loginMod);
}
} else {
//document.getElementById("setData").reset();
$('#myModal').modal('hide');
$('#successTop').show();
$('#successTop').html(data.msg);
if (data.msg != '' && data.msg != "undefined") {
bootbox.alert({closeButton: false, message: data.msg, callback: function () {
if (data.url) {
window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
} else {
location.reload(true);
}
}});
} else {
bootbox.alert({closeButton: false, message: "Success", callback: function () {
if (data.url) {
window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
} else {
location.reload(true);
}
}});
}
}
}
catch (e) {
if (e) {
$('#er_TopError').show();
$('#er_TopError').html(e);
setTimeout(function () {
$('#er_TopError').hide(5000);
$('#er_TopError').html('');
}, 5000);
}
}
}
});
I am using this simple one line code for years without a problem (it requires jQuery):
<script src="http://malsup.github.com/jquery.form.js"></script>
<script type="text/javascript">
function ap(x,y) {$("#" + y).load(x);};
function af(x,y) {$("#" + x ).ajaxSubmit({target: '#' + y});return false;};
</script>
Here ap() means an Ajax page and af() means an Ajax form. In a form, simply calling af() function will post the form to the URL and load the response on the desired HTML element.
<form id="form_id">
...
<input type="button" onclick="af('form_id','load_response_id')"/>
</form>
<div id="load_response_id">this is where response will be loaded</div>
Since the introduction of the Fetch API there really is no reason any more to do this with jQuery Ajax or XMLHttpRequests. To POST form data to a PHP-script in vanilla JavaScript you can do the following:
async function postData() {
try {
const res = await fetch('../php/contact.php', {
method: 'POST',
body: new FormData(document.getElementById('form'))
})
if (!res.ok) throw new Error('Network response was not ok.');
} catch (err) {
console.log(err)
}
}
<form id="form" action="javascript:postData()">
<input id="name" name="name" placeholder="Name" type="text" required>
<input type="submit" value="Submit">
</form>
Here is a very basic example of a PHP-script that takes the data and sends an email:
<?php
header('Content-type: text/html; charset=utf-8');
if (isset($_POST['name'])) {
$name = $_POST['name'];
}
$to = "test#example.com";
$subject = "New name submitted";
$body = "You received the following name: $name";
mail($to, $subject, $body);
Please check this. It is the complete Ajax request code.
$('#foo').submit(function(event) {
// Get the form data
// There are many ways to get this data using jQuery (you
// can use the class or id also)
var formData = $('#foo').serialize();
var url = 'URL of the request';
// Process the form.
$.ajax({
type : 'POST', // Define the type of HTTP verb we want to use
url : 'url/', // The URL where we want to POST
data : formData, // Our data object
dataType : 'json', // What type of data do we expect back.
beforeSend : function() {
// This will run before sending an Ajax request.
// Do whatever activity you want, like show loaded.
},
success:function(response){
var obj = eval(response);
if(obj)
{
if(obj.error==0){
alert('success');
}
else{
alert('error');
}
}
},
complete : function() {
// This will run after sending an Ajax complete
},
error:function (xhr, ajaxOptions, thrownError){
alert('error occured');
// If any error occurs in request
}
});
// Stop the form from submitting the normal way
// and refreshing the page
event.preventDefault();
});
Pure JS
In pure JS it will be much simpler
foo.onsubmit = e=> {
e.preventDefault();
fetch(foo.action,{method:'post', body: new FormData(foo)});
}
foo.onsubmit = e=> {
e.preventDefault();
fetch(foo.action,{method:'post', body: new FormData(foo)});
}
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
This is a very good article that contains everything that you need to know about jQuery form submission.
Article summary:
Simple HTML Form Submit
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get the form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = $(this).serialize(); // Encode form elements for submission
$.ajax({
url : post_url,
type: request_method,
data : form_data
}).done(function(response){ //
$("#server-results").html(response);
});
});
HTML Multipart/form-data Form Submit
To upload files to the server, we can use FormData interface available to XMLHttpRequest2, which constructs a FormData object and can be sent to server easily using the jQuery Ajax.
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="file" name="my_file[]" /> <!-- File Field Added -->
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = new FormData(this); // Creates new FormData object
$.ajax({
url : post_url,
type: request_method,
data : form_data,
contentType: false,
cache: false,
processData: false
}).done(function(response){ //
$("#server-results").html(response);
});
});
I hope this helps.
That's the code that fills a select option tag in HTML using ajax and XMLHttpRequest with the API is written in PHP and PDO
conn.php
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$database = "db_event";
try {
$conn = new PDO("mysql:host=$servername;dbname=$database", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
echo "Connection failed: " . $e->getMessage();
}
?>
category.php
<?php
include 'conn.php';
try {
$data = json_decode(file_get_contents("php://input"));
$stmt = $conn->prepare("SELECT * FROM events ");
http_response_code(200);
$stmt->execute();
header('Content-Type: application/json');
$arr=[];
while($value=$stmt->fetch(PDO::FETCH_ASSOC)){
array_push($arr,$value);
}
echo json_encode($arr);
} catch(PDOException $e) {
echo "Error: " . $e->getMessage();
}
script.js
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function () {
if (this.readyState == 4 && this.status == 200) {
data = JSON.parse(this.responseText);
for (let i in data) {
$("#cars").append(
'<option value="' + data[i].category + '">' + data[i].category + '</option>'
)
}
}
};
xhttp.open("GET", "http://127.0.0.1:8000/category.php", true);
xhttp.send();
index.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<link rel="stylesheet" href="style.css">
<script src="https://code.jquery.com/jquery-3.6.0.min.js"
integrity="sha256-/xUj+3OJU5yExlq6GSYGSHk7tPXikynS7ogEvDej/m4=" crossorigin="anonymous"></script>
<title>Document</title>
</head>
<body>
<label for="cars">Choose a Category:</label>
<select name="option" id="option">
</select>
<script src="script.js"></script>
</body>
</html>
I have one other idea.
Which the URL that of PHP files which provided the download file.
Then you have to fire the same URL via ajax and I checked this second request only gives the response after your first request complete the download file. So you can get the event of it.
It is working via ajax with the same second request.}

Uploading File via FormData

I am trying to upload a file to a Node backend that uses Multer. Multer requires the form to be submitted a specific way. If it's not submitted that way, the request.file parameter will be undefined. I have created an approach that works by using brute force. That working approach looks like this:
index-1.html:
<form method="POST" enctype="multipart/form-data" id="fileUploadForm">
<input type="file" id="selectedFile" name="selectedFile" /><br/><br/>
<input type="submit" value="Submit" id="btnSubmit"/>
</form>
...
var btn = document.getElementById('btnSubmit');
btn.addEventListener('click', function(e) {
e.preventDefault();
var form = $('#fileUploadForm')[0];
var data = new FormData(form);
console.log(data);
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: "/upload",
data: data,
processData: false,
contentType: false,
cache: false,
timeout: 600000,
success: function (data) {
console.log("SUCCESS!");
},
error: function (e) {
console.log("ERROR : ", e);
}
});
});
The code above successfully posts the file to my server. That server has the following:
server.js
app.post('/upload', upload.single('selectedFile'), function(req, res) {
if (req.file) {
console.log('file uploaded');
} else {
console.log('no file');
}
res.send({});
});
Using the code above, "file uploaded" is displayed in the console window as expected. However, I need a more dynamic approach. For that reason, I need to programmatically build the form in JavaScript. In an attempt to do that, I've created the following:
index-2.html
[my UI]
var btn = document.getElementById('btnSubmit');
btn.addEventListener('click', function(e) {
var form = document.createElement('form');
form.action = '/upload';
form.method = 'POST';
form.enctype = 'multipart/form-data';
var node = document.createElement("input");
node.name = 'selectedFile';
node.value = GLOBAL_SELECTED_FILE;
form.appendChild(node);
var data = new FormData(form);
data.append('id', this.id);
console.log(data);
$.ajax({
type: 'POST',
url: '/profile-picture/upload',
enctype: 'multipart/form-data',
data: data,
contentType: false,
processData: false,
success: function(res) {
console.log('success!');
},
error: function(xhr, status, err) {
console.log('error');
}
});
});
This second approach does not work. To clarify, the GLOBAL_SELECTED_FILE variable is the data of a file that was selected from an input element. The data is loaded via the FileReader api. That looks like this:
var GLOBAL_SELECTED_FILE = null;
var fileReader = new FileReader();
fileReader.onload = function(e) {
GLOBAL_SELECTED_FILE = e.target.result;
}
fileReader.readAsDataURL(fileSelected); // file selected comes from the onchange event on a <input type="file">..</input> element
Basically, I'm loading a preview of the image. Anyways, when I hit the submit button in the working version (index-1.html), I notice in Fiddler that a different value is sent over the value sent in index-2.html.
With the approach in index-1.html, Fiddler shows something like this in the "TextView" tab:
------WebKitFormBoundary183mBxXxf1HoE4Et
Content-Disposition: form-data; name="selectedFile"; filename="picture.PNG"
Content-Type: image/png
PNG
However, when I look in the "TextView" tab in Fiddler for the data sent via index-2.html, I see the following:
------WebKitFormBoundary9UHBP02of1OI5Zb6
Content-Disposition: form-data; name="selectedFile"
data:image/png;base64,iVBORw0KGg[A LOT MORE TO GO]
It's like the FormData is using two different encodings for the same value. Yet, I don't understand why. How do I get index-2.html to send the image in the same format as index-1.html so that Multer will populate the req.file property?
Thank you!
In index-1.html, you are using a file input:
<input type="file" id="selectedFile" name="selectedFile" />
In index-2.html, you are creating an ordinary form input (not a file input):
var node = document.createElement("input");
node.name = 'selectedFile';
node.value = GLOBAL_SELECTED_FILE;
form.appendChild(node);
To create a file input, you would need to add node.type = 'file'. However, you won't be able to set the value because browser security restrictions prevent setting the value of file inputs.
Instead, what you need to do is append the file that was selected by the user to the FormData object:
var data = new FormData(form);
data.append('id', this.id);
data.append('selectedFile', $('#fileInputElement')[0].files[0]);

How to post form to two different pages with a single button click?

I have a form with a single button like below:
<form name="sampleForm" id="sampleForm" method="post" action="" enctype="multipart/form-data">
<input type="text" id="biosample" name="biosample" class="sample"/>
<input type="text" id="library" name="library" class="sample"/>
<input type="submit" name="btnAdd" id="btnAdd" class="buttonsub" value="NEXT>>">
</form>
Ajax code is:
<script>
$(document).ready(function(){
var encoded_project_id = $('#encoded_project_id').val();
$('#sampleForm').on('submit', function(){
var target = 'windowFormTarget';
window.open('', target, 'width=500,height=300');
this.setAttribute('target', target);
$.post('postdata.php', $(this).serialize(), function(){
window.location.href = 'phases.php?edit='+encoded_project_id;
}).fail(function(){
window.location.href = 'sample.php?edit='+encoded_project_id;
});
});
});
</script>
Now when button is clicked, I want to post the data from the above form in 2 pages - handler.php and postdata.php
Handler.php should open in a new javascript window and postdata.php should open in same tab and same window.
How it can be achieved?
EDIT: It would seem you are using jQuery, so change this:
$(document).ready(function () {
document.getElementById('sampleForm').onsubmit = function (e) {
var req = new XMLHttpRequest();
req.open('POST', 'test.php', true);
req.send();
}
});
to this:
$(document).ready(function(){
$('#sampleForm').on('submit', function(){
$.post('postdata.php', $(this).serialize(), function(){
console.log('success');
}).fail(function(){
console.log('error');
});
});
});
You should do two things. First add
<form target="_blank" action="handler.php"></form>
This will ensure when the submit button is clicked that the form will open a new window.
Then you need to intercept the submit like so:
document.getElementById('sampleForm').onsubmit = function(e){
//xmlHTTPRequest function
//This is where you send your form to postdata.php
}
The code above will be called first and you can send your form with the asynchronous XMLHTTPRequest object to postdata.php . After that function ends, the default behavior of the form will start and your handler.php will receive the form.
you just to need two ajax call . Do something like this
$(document).ready(function(){
// if want to stop default submission
$("#sampleForm").submit(function(e){
e.preventDefault();
});
$(document).on('click','#btnAdd',function(){
send('page1.php',{'data1':'value'});
send('page2.php',{'data1':'value'});
});
});
function send(url,data)
{
$.ajax({
url: url,
type: 'POST',
datatype: 'json',
data: data,
success: function(data) {
// success
},
error: function(data) {
alert("There may an error on uploading. Try again later");
},
});
}

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