Send a hidden ID input field with AJAX - javascript

With this AJAX script, I'm trying to send the content of contentText and contentID.
Just sending contentTEXT works, but I want to send the ID as well, so I can comment on the original post.
But it doesn't work!
myData looks like this when it semi works:
> var myData = '?content_txt='+$("#contentText").val(),
> '&content_id='+$("#contentId").val(); //build a post data structure
But i want it to be something like this, i think
<script type="text/javascript"> $(document).ready(function() {
//##### send add record Ajax request to response.php ######### $("#FormSubmit").click(function (e) { e.preventDefault(); if($("#contentText").val()==='') {
alert("Please enter some text!");
return false; }
$("#FormSubmit").hide(); //hide submit button $("#LoadingImage").show(); //show loading image
var myData = '?content_txt='+$("#contentText").val(), '&content_id='+$("#contentId").val(); //build a post data structure
jQuery.ajax({ type: "POST", // HTTP method POST or GET url: "response.php", //Where to make Ajax calls contentType: "application/x-www-form-urlencoded;charset=UTF-8", dataType:"text", // Data type, HTML, json etc. data:myData, //Form variables success:function(response){
$("#responds").append(response);
$("#contentText").val(''); //empty text field on successful
$("#FormSubmit").show(); //show submit button
$("#LoadingImage").hide(); //hide loading image
}, error:function (xhr, ajaxOptions, thrownError){
$("#FormSubmit").show(); //show submit button
$("#LoadingImage").hide(); //hide loading image
alert(thrownError); } }); });
//##### Send delete Ajax request to response.php ######### $("body").on("click", "#responds .del_button", function(e) { e.preventDefault(); var clickedID = this.id.split('-'); //Split ID string (Split works as PHP explode) var DbNumberID = clickedID[1]; //and get number from array var myData = 'recordToDelete='+ DbNumberID; //build a post data structure
$('#item_'+DbNumberID).addClass( "sel" ); //change background of this element by adding class $(this).hide(); //hide currently clicked delete button
jQuery.ajax({ type: "POST", // HTTP method POST or GET url: "response.php", //Where to make Ajax calls dataType:"text", // Data type, HTML, json etc. data:myData, //Form variables success:function(response){
//on success, hide element user wants to delete.
$('#item_'+DbNumberID).fadeOut(); }, error:function (xhr, ajaxOptions, thrownError){
//On error, we alert user
alert(thrownError); } }); });
}); </script>
My form I'm trying to use
<form class="form-horizontal" accept-charset="utf-8">
<fieldset>
<legend><?php echo WORDING_ADD_A_COMMENT; ?></legend>
<!-- Textarea -->
<div class="control-group">
<div class="controls">
<textarea name="content_txt" id="contentText" cols="45" rows="5" placeholder="<?php echo WORDING_COMMENT_PLACEHOLDER; ?>"></textarea>
<input type="hidden" name="content_id" id="contentId" value="<?php echo $_GET['topic_id']; ?>"/>
</div>
</div>
<!-- Button -->
<div class="control-group">
<label class="control-label" for="singlebutton"></label>
<div class="controls">
<button id="FormSubmit" class="btn btn-primary"><?php echo WORDING_BUTTON_COMMENT_BOX; ?></button>
<img src="images/loading.gif" id="LoadingImage" style="display:none" />
</div>
</div>
</fieldset>
</form>

var variable = {
'content_txt': $("#contentText").val(),
'content_id': $("#contentId").val()
};

try to use
var myData = '?content_txt='+$("#contentText").val()+'&content_id='+$("#contentId").val();

You only use ? and & when you manually build a query string and in that case you would need to encode your values as well.
When you send the data as a parameter, the easiest solution is to generate an object. That will be encoded automatically by jQuery:
//build a post data structure
var myData = { 'content_txt': $("#contentText").val(),
'content_id': $("#contentId").val() };
As you are using a form, you can also serialize the form:
var myData = $('.form-horizontal').serialize();

data:'content_id='+ $("#contentId").val()+ '&content_txt='+ $("#contentText").val() ,

Related

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.}

ajax [file 1] -> php [file 2] -> $_POST [file 1]

I am building a budget application with HTML, Javascript, and PHP. My goal is to have the user be able to add data into a database from a form provided. I already have a ton of php at the top of my 'dashboard.php' (which contains the form) so I didn't want to run dashboard.php on submit, so instead I created a button that preforms an AJAX call to a different php file 'addIncome.php'.
I have two different files...
dashboard.php &
addincome.php
dashboard.php contains my form, as well as my javascript to run an AJAX call.
addincome.php is using $_POST to grab the values from the form in dashboard.php and make a mysqli_query. However, at first nothing was happening so I decided to echo the value of one of the return values from my $_POST. And ended up getting this error...
undefined index iName in addIncome.php
undefined index iAmount in addIncome.php
So from there I though that maybe I didn't have access to the dashboard.php by default so I included...
include('dashboard.php');
Still no difference...
I'm really at a stand still here. Any thoughts?
Thanks
The form...
<form>
<input type="text" name="iName" placeholder="income name">
<input type="number" step="0.01" min="0" name="iAmount" placeholder="amount">
<input type="date" name="iDate">
</form>
The javascript...
<script>
$('.in-btn').click(function() {
$.ajax({
url: "addIncome.php",
type: "POST",
data: 'show=content',
success: function(data) {
$('.in-btn').html(data);
}
});
setTimeout(() => {
// location.reload();
}, 2000);
});
</script>
The php...
<?php
echo "adding...";
require_once('connection.php');
include('dashboard.php');
$iUser = $_SESSION["username"];
$iName = $_POST["iName"];
$iAmount = $_POST["iAmount"];
echo $iName;
$sql = "INSERT INTO income (user, name, amount, date) VALUE ('pmanke', '$iName', '$iAmount','1/16/19')";
mysqli_query($dbCon, $sql);
?>
You are not sending any post data with your AJAX call except for:
show=content. You want to send your form data. You can retrieve your form data with:
$("#id-of-form").serialize()
That way your PHP code is able to retrieve the correct values from your POST data.
An even more general way to do this is to just create a normal form with a submit button and an action and use javascript to catch the submit event and make an AJAX call instead:
HTML:
<form id="idForm" action="addIncome.php">
<input type="text" name="iName" placeholder="income name">
<input type="number" step="0.01" min="0" name="iAmount" placeholder="amount">
<input type="date" name="iDate">
<input type="submit" />
</form>
Javascript:
$("#idForm").submit(function(e) {
var form = $(this);
var url = form.attr('action');
$.ajax({
type: "POST",
url: url,
data: form.serialize(), // serializes the form's elements.
success: function(data) {
alert(data); // show response from the php script.
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});

get all form data by jquery/js input fields ajax

Meanwhile I'm getting stuck on this issue. Normally, it's pretty simple but somehow it doesn't work for what I'm trying to do. I want to get all data from my form input fields by either Jquery or JS and then send them through AJAX to the server sided script (PHP). Even by using append or do it by serialize, I only obtain the object from input field with ID #file. I'm not using a submit button to confirm the uploaded image - only select the file and send it off.
I already tried too add
formdata.append("_token", document.getElementById('_token').val());
but whenever I try to append another element to the formdata the entire script stops working
By using $('#picUploadForm').serialize(); I do not get the any result from the input element with ID #file.
HTML:
<form enctype="multipart/form-data" id="picUploadForm">
<input type="file" name="file" id="file" style="display:none;" >
<input type="hidden" name="_token" id="_token" value="<?php echo $_SESSION['_token']; ?>" />
</form>
<!-- Default Avatar Image -->
<div class="click-slide overlay">
<!-- Profile Image-->
<img src="<?php if(isset($avatar['filelink']) && $avatar['filelink'] !='') { echo $avatar['filelink']; } else { echo "assets/images/avatars/default_avatar_large.png"; }?>" alt="" class="img-full-width-tight" id="imagePreview" />
<!-- Image update link -->
<div id="editLink" >
<span>
<a href="javascript:void(0);" class="pop-inline ti ti-image" ></a>
</span>
</div>
</div><!--/ click-slide-->
JS:
//On select file to upload
$('#file').on('change', function(e){
e.preventDefault();
var formdata = new FormData();
// any other code here....
} else {
// Upload Image to backend
formdata.append("file", document.getElementById('file').files[0]);
// formdata.append("_token", document.getElementById('_token').val()); // does not work!!!
// $('#picUploadForm').serialize(); // only returns data from input #_token
$.ajax({
url: "./serversided.php",
type: "POST",
data: formdata,
dataType: 'json',
cache: false,
contentType: false,
processData: false,
beforeSend: function(){
$("#statusav").removeClass().html('');
$('.overlay').LoadingOverlay("show");
HideLoadingOverlay();
},
success: function(data){
if(data.status === true){
// alert(data.imgURL);
setTimeout(function(){$("#statusav").removeClass('alert alert-danger').addClass('alert alert-success').html(data.reply)}, 2000);
$("#imagePreview").attr('src', data.imgURL);
} else {
// alert(data.error);
setTimeout(function(){$("#statusav").removeClass('alert alert-success').addClass('alert alert-danger').html(data.error)}, 2000);
}
}
});
}
});
.val() is a jQuery method - it is not a vanilla JS method, so it doesn't work when called on a plain element. document.getElementById will return an element (or null); $('selectors here') will return a jQuery object, on which you can use jQuery functions.
Try this instead, with vanilla JS:
formdata.append("_token", document.querySelector('#_token').value);
Or select the element with jQuery and use the jQuery method:
formdata.append("_token", $('#_token').val());

How to display specific record after clicked on anchor tag using php?

I have two links, Delete and View. Delete is working perfectly. I am getting the issue on view link. I have to display specific records after clicked on view link.If you notice that delete anchor tag I wrote href='delete.php?function=delete&del_id=$id'. It's calling the delete.php file and deleting a specific record. I want to know about how to write in view anchor tag to display records? Please check below image.I want to clicked on view and display the records in text fields.PHP file added. Please check my ajax code. It is also not working. Ajax code is not working here. I have to display the output in the text field.If i write directly $id=2 then it is display records in alert. Hope you guys can understand now.Would you help me in this?
//delete.php
function view($conn) {
$id=$_GET['view_id'];
$admin_view="SELECT * from view_table WHERE Id=$id";
$result = $conn->query($admin_view);
if (isset($result->num_rows) > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$parent_name=$row['Name'];
$parent_email=$row['Email'];
$admin_mobile=$row['Mobile_no'];
}
}
}
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<a href='delete.php?function=delete&del_id=$id' >Delete</a>
<a href='#Popup' onClick='a_onClick(<?php $id?>)' class='btn-view'>View</a>
<form action="#" method="post">
<input type="text" name="Name" placeholder="username" value="<?php echo $parent_name;?>" >
<input type="email" name="email" placeholder="email" value="<?php echo $parent_email;?>" >
<input type="text" name="Mobile_no" placeholder="Mobile no" value="<?php echo $admin_mobile;?>" >
Cancel <span class="close"></div>
</form>
<script>
function a_onClick(id) {
var id = id;
var Name=$('#name').val();
$.ajax({ //create an ajax request to load_page.php
type: "GET",
url: "delete.php?function=view&view_id=id",
data:'Name='+Name,
dataType: "html", //expect html to be returned
success: function(response) {
// $("#responsecontainer").html(response);
alert(response);
}
});
}
</script>
you need to change some changes
<a href='#Popup' onClick='a_onClick(<?= $id ?>)' class='btn-view'>View</a>
<script>
function a_onClick(id) {
var id = id;
$.ajax({ //create an ajax request to load_page.php
type: "GET",
url: "delete.php?function=view&view_id=id",
data:'Name='+Name,
dataType: "html", //expect html to be returned
success: function(response) {
// $("#responsecontainer").html(response);
alert(response);
}
});
}
</script>
hope its help you

How to submit data using AJAX and POST Method

How can I post the following form using Ajax. When submitting the form that page refreshes and I want to avoid that.
<form method="post" action="" >
<div class="rp-donation-block checkout_donation" >
<p class="message"><strong><?php echo $this->get_setting('message'); ?></strong></p>
<div class="input text">
<input type="text" value="<?php echo $amount; ?>" class="input-text text-donation" name="donation-amount">
<input type="submit" value="<?php echo $this->get_setting('btn_lable'); ?>" class="button" name="donate-btn">
</div>
</div>
</form>
I tried both of these with no luck.
<form method="post" action="" onSubmit="return false">
<form method="post" action="" action= "<?php echo $_SERVER['PHP_SELF']; ?>">
I suppose your form element has id of form as <form id="form"...
<script>
// Attach a submit handler to the form
$( "#form" ).submit(function( event ) {
// Stop form from submitting normally
event.preventDefault();
// Get some values from elements on the page:
var $form = $( this ),
donate_amount = $form.find( "input[name='donation-amount']" ).val(),
url = $form.attr( "action" );
// Send the data using post
var posting = $.post( url, { amount: donate_amount } );
// Put the results in message div
posting.done(function( data ) {
var message_returned = $( data ).find( "#message" ); //hope your are returning message, also you can return other things like message type and create message on frontend as accordingly
$( ".message" ).empty().append( message_returned ); //adding returned message to your message div
});
});
</script>
A form is not mandatory to post using Ajax. You can write an Ajax request using just a button click as well.
$(function () {
$('#buttonId').on('click', function (e) {
$.ajax({
type: "POST",
cache: false,
url: <url_name>, //some php file where you may hit the database and want to modify content
data:{'post1': <post1value>,'post2': <post2value>},
datatype: "JSON", // can be html or text as well
success: function (data) {
var actualData = $.parseJSON(data); // if in case of JSON
}
});
});
});
If you want to invoke only on form submission you can write your Ajax like so.
$('#formId').on('submit', function (e) {
});
On the other side use :
$post_1 = $_POST['post1'];
To Get post1 value.
The button you are using to trigger the event that causes the submit of the form via ajax post should not be of type submit! Else this will always fail.

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