Post Ajax Data to PHP and Return Data

How to return data from PHP to a jQuery ajax call

I figured it out. Need to use echo in PHP instead of return.

<?php 
  $output = some_function();
  echo $output;
?>

And the jQ:

success: function(data) {
  doSomething(data);
}

Returning data from Ajax to PHP

I would suggest a JSON-answer from your PHP-backend, something like:

{
    "filename": "my_image.jpg",
    "message": "Image uploaded!",
    "success": true
}

Which would be something like:

<?php
define('UPLOAD_DIR', 'uploads/');
$img = $_POST['base64image'];
$img = str_replace('data:image/jpeg;base64,', '', $img);
$img = str_replace(' ', '+', $img);
$data = base64_decode($img);
$file = UPLOAD_DIR .date("d-m-Y")."_".uniqid(). '.jpg';
$success = file_put_contents($file, $data);

$cliente_foto_webcam = $file;

if ($success) {
    echo json_encode([
        "filename" => $cliente_foto_webcam,
        "message" => "Image uploaded!",
        "success" => true
    ]);
} else {
    echo json_encode([
        "filename" => null,
        "message" => "Couldn't upload image",
        "success" => false
    ]);
}
?>

Using Jquery Ajax to retrieve data from Mysql

For retrieving data using Ajax + jQuery, you should write the following code:

 <html>
 <script type="text/javascript" src="jquery-1.3.2.js"> </script>

 <script type="text/javascript">

 $(document).ready(function() {

    $("#display").click(function() {                

      $.ajax({    //create an ajax request to display.php
        type: "GET",
        url: "display.php",             
        dataType: "html",   //expect html to be returned                
        success: function(response){                    
            $("#responsecontainer").html(response); 
            //alert(response);
        }

    });
});
});

</script>

<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
   <tr>
       <td> <input type="button" id="display" value="Display All Data" /> </td>
   </tr>
</table>
<div id="responsecontainer" align="center">

</div>
</body>
</html>

For mysqli connection, write this:

<?php 
$con=mysqli_connect("localhost","root","");

For displaying the data from database, you should write this :

<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);

echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";

while($data = mysqli_fetch_row($result))
{   
    echo "<tr>";
    echo "<td align=center>$data[0]</td>";
    echo "<td align=center>$data[1]</td>";
    echo "<td align=center>$data[2]</td>";
    echo "<td align=center>$data[3]</td>";
    echo "<td align=center>$data[4]</td>";
    echo "</tr>";
}
echo "</table>";
?>

post ajax data to PHP and return data

 $.ajax({
            type: "POST",
            data: {data:the_id},
            url: "http://localhost/test/index.php/data/count_votes",
            success: function(data){
               //data will contain the vote count echoed by the controller i.e.  
                 "yourVoteCount"
              //then append the result where ever you want like
              $("span#votes_number").html(data); //data will be containing the vote count which you have echoed from the controller

                }
            });

in the controller

$data = $_POST['data'];  //$data will contain the_id
//do some processing
echo "yourVoteCount";

Clarification

i think you are confusing

{data:the_id}

with

success:function(data){

both the data are different for your own clarity sake you can modify it as

success:function(vote_count){
$(span#someId).html(vote_count);

jQuery Ajax POST example with 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.