javascript ajax uncaught referenceerror function

Question

Im getting the tittle error when im calling this function.

<script type="text/javascript">     
    function upvote(user, id){
        fazer = <?php echo $doornot ?>;
        if( fazer == 'true'){
            window.location = "http://www.dawnsource.com/forums/register.php";
        }else {
            user = <?php echo $user; ?>;
            id = <?php echo $_GET[id]; ?>;
             $.ajax({
             url: 'scripts/upvote.php',
             type: 'post',
             data: 'user='+user+'&id='+id,
            success: function() 
             {
              alert('success, upvote completed ');
          }, error: function()
          {
              alert('something went wrong, rating failed');
          }
        });
        }

    }
</script>

The error is going on:

user = <?php echo $user; ?>;

But its getting me the right value in the echo. Why is this happening?

try user = "<?php echo $user; ?>";. note that quotes around php tags — bystwn22
– bystwn22, Commented Jun 23, 2013 at 21:52
refer to @MarcB's answer, some of your javascript variables need quotes around, otherwise you will get a syntax error — bystwn22
– bystwn22, Commented Jun 23, 2013 at 21:58
Dunno if its simple, but can I change the sucess function to reload the div thats calling the javascript? — user2469440
– user2469440, Commented Jun 23, 2013 at 22:49

Musa · Accepted Answer · 2013-06-23 21:59:02Z

1

You're missing quotes around your php data, so you're generating invalid javascript. Remember... whatever PHP is outputting has to be correct in a javascript context.

<?php
$foo = 'bar';
?>

<script>
baz = <?php echo $foo ?>;
</script>

is going to generate

baz = bar;

and bar will be interpreted as an undefined variable name.

The proper solution is to simply pass everything through as json from PHP:

 bar = <?php echo json_encode($foo); ?>;

which will ALWAYS produce valid javascript, no matter what's in $foo. and produces

baz = "bar";

edited Jun 23, 2013 at 21:59

Musa

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answered Jun 23, 2013 at 21:54

Marc B

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Comments

StackSlave · Accepted Answer · 2013-06-23 22:14:08Z

true should be boolean if $doornot is. Additionally, in both PHP and JavaScript String quotes are peeled off when they return a value. Your JavaScript sees another variable, not a String. Try:

<script type="text/javascript">     
  function upvote(){
    var fazer = <?php echo $doornot ?>;
    if(fazer == true){
        window.location = 'http://www.dawnsource.com/forums/register.php';
    }
    else {
      var user = '<?php echo $user; ?>';
      var id = '<?php echo $_GET[id]; ?>';
      $.ajax({
        url: 'scripts/upvote.php',
        type: 'post',
        data: 'user='+user+'&id='+id,
        success: function(){
          alert('success, upvote completed');
        },
        error: function(){
          alert('something went wrong, rating failed');
        }
      });
    }
  }
</script>

Side note:

I do not see a use for your user or id arguments in your upvote function, since you are getting those values from PHP, and only under a condition. Also, you should use the keyword var in JavaScript, unless you want those variables to be global.

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