CODEWITHSUNDEEP
phpjqueryajaxundefined-index
Abhishek Kaushik
Abhishek Kaushik

Reputation: 1169

Undefined Index in Post request

I have no idea why I am getting this weird error!

PHP Notice: Undefined index: refId in /var/www/echo.php on line 5

I am getting Console output, but cant echo refId. Have I done anything wrong here? 

<?php
    $rollUrl = 34;
    $refId = $_POST['refId'];
    echo $refId;
?>

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
    $.ajax({
        url:'echo.php',
        type: 'POST',
        data: { 'refId': "<?php echo $rollUrl ?>" },
        success: function(response){
            console.log('Getting response');
        }
    }); 
</script>

Upvotes: 2

Views: 1119

Answers (2)

user1932079
user1932079

Reputation:

Please see the comments in the code below:

<?php
$rollUrl = 34;

//Only try to process POST if there is something posted *and* refId exists
if (count($_POST) > 0 && isset($_POST['refId'])) { 
  $refId = $_POST['refId'];
  echo $refId;
  //Exit after echoing out the refId so that the HTML below does not also get returned.
  exit();
}
?>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
    $.ajax({
        url:'echo.php',
        type: 'POST',
        data: { 'refId': "<?php echo $rollUrl ?>" },
        success: function(response) {
            //Updated log to show the actual response received.
            console.log('Getting response of "' + response + '"');
        }
    }); 
</script>

That works for me when I tested without any errors being thrown and the Ajax being executed.

Upvotes: 1

Moeed Farooqui
Moeed Farooqui

Reputation: 3622

This is happening because your variable is not set. Use isset

<?php

$rollUrl=34;
if(isset($_POST['refId'])) {
$refId=$_POST['refId'];
echo $refId;
}

?>

Update: You should assign the refId as a name attribute to any input field to revive the input from user.

<input type="text" name="refId" />

Upvotes: 0

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