Question 36423 in Jquery Question posted in
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I use ajax to send data but cannot receive data back from php file – Jquery

This function will get the value from the input and sent data with ajax to another php file. But I cannot get the result back from php file when I echo and store in ajax success function result.

<script>

$(document).ready(function(){
    $('.edit-button').click(function(){
        let clickpostid = $(this).prev().val();
        console.log(clickpostid);// I can see the input value
        $.ajax({
            url: "post/edit.php",
            method: "POST",
            data: { clickpostid },
            success: function(result){
                alert(result);
                //When I alert the result is blank not null or undefied.
            }
        })
    });
})
    

</script>

Below code will be in another file such as edit.php.

<?php 
include('../db.php');
if(isset($POST_['clickpostid'])){
      $id = $POST_['clickpostid'];
      echo $id;
   
}

?>
Tags:

2

Answers


  1. 0

    Your edit.php should be this:
    You should use $_POST instead of $POST_

    <?php 
include('../db.php');
if(isset($_POST['clickpostid'])){
      $id = $_POST['clickpostid'];
      echo $id;
   
}

?>
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  2. 0

    Please try this:

    js:

    $(document).ready(function(){
    $('.edit-button').click(function(){
        let clickpostid = $(this).prev().val();
        console.log(clickpostid);// I can see the input value
        $.ajax({
            url : "post/edit.php",
            type: "POST",
            data: { clickpostid: clickpostid },
            success : function (result) {
               alert(result);
            }
        });
    });
});

    php:

    <?php 
include('../db.php');
if(isset($_POST['clickpostid'])){
      $id = $_POST['clickpostid'];
      echo $id;
   
}

?>
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