CODEWITHSUNDEEP
jqueryajax
john snow
john snow

Reputation: 21

Jquery ajax post, can't get post data

so I am trying to make a ajax post so I made a simple post:

 <!DOCTYPE html>
 <html>
 <head>
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
 </head>
 <body>
 <script type="text/javascript">
 $(function(){
$.ajax({
  type: "POST",
  url: 'example1.php',
  data: ({Imgname:"13"}),
  success: function(data) {
    //alert(data);
  }
 });
 });

 </script>
 <?php 
 $temp = $_GET['Imgname'];
 echo $temp;
 ?>
 </body>
 </html>

But it I don't get anything, what am I doing wrong? please help.

Upvotes: 0

Views: 1098

Answers (2)

Mohamed-Yousef
Mohamed-Yousef

Reputation: 24001

  • data: ({Imgname:"13"}), should be data: {Imgname:"13"},

  • The default type of ajax is get ..But while you use type: "POST", use $_POST['Imgname']; instead of $_GET['Imgname'];

  • If console.log(data); after changes still returns an error so check the path of url: 'example1.php',

  • If the ajax and the php code in the same file so use url: '/', and put the php code on top of the html code and in php after echo return false;

Upvotes: 1

Abdul Rehman Sheikh
Abdul Rehman Sheikh

Reputation: 939

Try this, it should must work for your case: Here is your HTML Code:

<!DOCTYPE html>
<html>
    <head>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
    </head>
    <body>
        <script type="text/javascript">
            $(function () {
                $.ajax({
                    type: "POST",
                    url: 'example1.php',
                    data: {Imgname: "13"},
                    success: function (result) {
                        alert(result);
                    }
                });
            });

        </script>
    </body>
</html>

INSIDE your example1.php File you have to check with:

<?php 
        $temp = $_POST['Imgname'];
        echo $temp; exit; 
?>

Your mistake was, you was sending the POST request and you was trying to get data in $_GET variable. Another mistake with your data posted as well. Try my code now.

Upvotes: 0

Related Questions