Develop Reference

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.}

How to fix Ajax fetch nothing from database?

I have a hard time finding my mistake in this code. I have a search bar and I'd use ajax so that the data will fetch automatically.
This is my html file.
<!DOCTYPE html>
<html>
<head>
<title></title>
<script src="bootstrap4/jquery/jquery.js"></script>
<script>
function loadproducts()
{
var name = document.getElementById("search");
if(name)
{
$.ajax({
type: 'post',
contentType:false,
cache: false,
processData:false,
data: {
products:name,
},
url: 'loadproducts.php',
success: function (response){
$('#product_area').html(response);
}
});
}
else
{
$('#product_area').html("No Product found!");
}
}
</script>
</head>
<body>
<input type="text" name="search" id="search" onkeyup="loadproducts();">
<div id="product_area">
</div>
</body>
</html>
----------
This is my loadproducts.php file
<?php
include('server/connection.php');
if (isset($_POST['products'])){
$name = mysqli_real_escape_string($db,$_POST['products']);
$show = "SELECT product_name,sell_price FROM products WHERE product_name LIKE '$name%' ";
$query = mysqli_query($db,$show);
if(mysqli_num_rows($query)>0){
while($row = mysqli_fetch_array($query)){
echo "<p>".$row['product_name']."</p>";
echo "<p>".$row['sell_price']."</p>";
}
}
}
Ill tried putting alert("called"); function on the ajax success and the alert is activated but still no output show. I also edit the var name = document.getElementById("search"); to var name = document.getElementById("#search"); but it pass straight to the else statement.Can someone site the problem of this code?

Currently, you're accessing the actual HTML element. You want the value, so use .value:
var name = document.getElementById("search").value;
Or if you prefer, you can simplify this down with jQuery:
var name = $("#search").val();

How To Send and Receive JSON With XAMPP PHP

I have a XAMPP 7.1.10-0 server running with this index.php
<?php
if(isset($_POST["username"])) {
echo $_POST;
header("Location:getbooks.php");
exit;
} else {
echo file_get_contents("login.html");
}
?>
This initially works and displays the login.html file. Then I want to make a ajax call to this file again to switch to another html file. So login.js does this.
window.onload = function() {
var button = document.getElementById("submit");
button.onclick = function() {
var user = document.getElementById("username").value;
var pass = document.getElementById("password").value;
var data = {"username": user, "password": pass};
data = $(this).serialize() + "&" + $.param(data);
$.post('index.php', data, function(response) {
console.log(response);
});
}
}
As you can imagine there is a button that is clicked and when it clicks the Ajax call is made. When I click the button it does not change the header and therefore does not change the page. Would anybody know what I am doing wrong?

If you want to redirect, you can do it in AJAX success method.
login.html:
<!DOCTYPE html>
<html>
<head>
<title>Login</title>
</head>
<body>
<div class="form">
Username: <input id="username" type="text"></input> <br>
Password: <input id="password" type="password"></input> <br>
<button id="submit">Submit</button>
</div>
<!-- jQuery -->
<script
src="https://code.jquery.com/jquery-1.12.4.min.js"
integrity="sha256-ZosEbRLbNQzLpnKIkEdrPv7lOy9C27hHQ+Xp8a4MxAQ="
crossorigin="anonymous"></script>
<script type="text/javascript">
$(document).ready(function(){
var button = document.getElementById("submit");
button.onclick = function() {
var user = document.getElementById("username").value;
var pass = document.getElementById("password").value;
$.ajax({
url: 'index.php',
data: {
"username" : user,
"password" : pass
},
type: 'POST',
success: function(response) {
window.location = response;
},
error: function(error) {
}
});
}
})
</script>
</body>
</html>
index.php:
<?php
if(isset($_POST["username"])) {
//process the array
echo "getbooks.php";
} else {
echo file_get_contents("login.html");
}
?>
Aside this, you can also handle the response from PHP in AJAX to update the view of the page without redirection.

ajax submit form why it cannot echo $_POST

I'm test using ajax submit form (submit to myself page "new1.php")
The thing that I want is, after click submit button, it will echo firstname and lastname. But I don't know why I do not see the firstname and lastname after submit.
here is new1.php page
<?php
echo $_POST['firstname']."<br>";
echo $_POST['lastname']."<br>";
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
</head>
<body>
<form id="myform" action="new1.php" method="post">
Firstname : <input type="text" name="firstname"> <br>
Lastname : <input type="text" name="lastname"> <br>
<input type="submit" value="Submit">
</form>
<script>
// this is the id of the form
$("#myform").submit(function(e) {
$.ajax({
type: "POST",
url: 'new1.php',
data: $("#myform").serialize(), // serializes the form's elements.
success: function(data)
{
alert('yeah'); // show response from the php script.
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});
</script>
</body>
</html>

In your case the best option to retrieve values as JSON format using json_encode in your PHP code and then accessing these values through data object.
Example:
PHP code:
if($_POST)
{
$result['firstname'] = $_POST['firstname'];
$result['lastname'] = $_POST['lastname'];
echo json_encode($result);
die(); // Use die here to stop processing the code further
}
JS code:
$("#myform").submit(function (e) {
$.ajax({
type : "POST",
url : 'new1.php',
dataType : 'json', // Notice json here
data : $("#myform").serialize(), // serializes the form's elements.
success : function (data) {
alert('yeah'); // show response from the php script.
// make changed here
$('input[name="firstname"]').text(data.firstname);
$('input[name="lastname"]').text(data.lastname);
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});

When you use form as a serialize you have to retrieve like this.
Edit your ajax like this :
data: { formData: $("#myform").serialize()},
Then you can retrieve like this in your controller:
parse_str($_POST['formData'], $var);
echo "<pre>";
print_r($var);
exit;

Make some changes in javascript here:
success: function(data)
{
$('#response').html(data); // show response from the php script.
}
And in html code make a div with id response
<div id="response"></div>

Change from
alert('yeah'); // show response from the php script.
to
alert(data); // show response from the php script.

the value firstname, lastname will not display because you called the new1.php via ajax and the data (firstname, lastname and the page code) is returned to java script variable (data) you need to inject the data to your document
Try this
$.ajax({
type: "POST",
url: 'new1.php',
data: $("#myform").serialize(), // serializes the form's elements.
success: function(data) {
document.documentElement.innerHTML = data;
}
});

php ajax form submit ..nothing happens

I have a PHP Ajax form that I'm trying to submit a Zendesk API call. Whenever I use the ajax part, in order to keep the user on the same page, it doesn't work. When I remove the <script> part, it works fine, but obviously redirects to contact.php from contact.html so I'm thinking the problem is in the Ajax part, not in the PHP part.
Here is my HTML form:
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
</head>
<body>
<div class="box_form">
<form id="zFormer" method="POST" action="contact.php" name="former">
<p>
Your Name:<input type="text" value="James Duh" name="z_name">
</p>
<p>
Your Email Address: <input type="text" value="duh#domain.com" name="z_requester">
</p>
<p>
Subject: <input type="text" value="My Subject Here" name="z_subject">
</p>
<p>
Description: <textarea name="z_description">My Description Here</textarea>
</p>
<p>
<input type="submit" value="submit" id="submitter" name="submit">
</p>
</form>
</div>
<div class="success-message-subscribe"></div>
<div class="error-message-subscribe"></div>
<script>
jQuery(document).ready(function() {
$('.success-message-subscribe').hide();
$('.error-message-subscribe').hide();
$('.box_form form').submit(function() {
var postdata = $('.box_form form').serialize();
$.ajax({
type: 'POST',
url: 'contact.php',
data: postdata,
dataType: 'json',
success: function(json) {
if(json.valid == 1) {
$('.box_form').hide();
$('.error-message-subscribe').hide();
$('.success-message-subscribe').hide();
$('.subscribe form').hide();
$('.success-message-subscribe').html(json.message);
$('.success-message-subscribe').fadeIn();
}
}
});
return false;
});
});
</script>
</body>
</html>
And the PHP Part:
You can probably ignore most of this since it works when I don't use the Ajax. Only the last few lines gives the response $array['valid'] = 1; which should then be catched by if(json.valid == 1) above.
<?php
( REMOVED API CALL CODE FROM ABOVE HERE )
if (isset($_POST['submit'])) {
foreach($_POST as $key => $value){
if(preg_match('/^z_/i',$key)){
$arr[strip_tags($key)] = strip_tags($value);
}
}
$create = json_encode(array('ticket' => array(
'subject' => $arr['z_subject'],
'comment' => array( "body"=> $arr['z_description']),
'requester' => array('name' => $arr['z_name'],
'email' => $arr['z_requester'])
)));
$return = curlWrap("/tickets.json", $create, "POST");
$array = array();
$array['valid'] = 1;
$array['message'] = 'Thank you!';
echo json_encode($array);
?>
Any ideas why this isn't working?

I expect your use of contact.php as a relative URL isn't resolving properly. Check your JavaScript console and you should see an error that shows the post failing. Change contact.php to www.your_domain.com/contact.php and it should work fine

Replace jQuery(document).ready(function() { by
$(document).ready(function() {
Secondly from Jquery documentation:
Note: Only "successful controls" are serialized to the string. No
submit button value is serialized since the form was not submitted
using a button. For a form element's value to be included in the
serialized string, the element must have a name attribute. Values from
checkboxes and radio buttons (inputs of type "radio" or "checkbox")
are included only if they are checked. Data from file select elements
is not serialized.
Therefore submit button won't serialize through jQuery.serialize() function.
A solution below:
<script>
$(document).ready(function() {
$('.success-message-subscribe').hide();
$('.error-message-subscribe').hide();
$('#submitter').click(function(e) {
e.preventDefault();
$myform = $(this).parent('form');
$btnid = $(this).attr('name');
$btnval = $(this).attr('value');
var postdata = $myform.serialize();
$.ajax({
type: 'POST',
url: 'contact.php',
data: { "btnid" : $btnid, "btnval": $btnval, "form-data": $form.serialize() },
dataType: 'json',
success: function(json) {
if(json.valid == 1) {
$('.box_form').hide();
$('.error-message-subscribe').hide();
$('.success-message-subscribe').hide();
$('.subscribe form').hide();
$('.success-message-subscribe').html(json.message);
$('.success-message-subscribe').fadeIn();
}
}
});
return false;
});
});
</script>