PHP Javascript Uncaught SyntaxError: Unexpected token ILLEGAL

Question

I have this PHP script and it does not work properly. What is the mistake?

<?php
if ( isset($success) || isset($failure) ) {
?>
<script>
    $(document).ready(function(){
        $('div.aler').css('display','block');
        $("div.aler").html("<?php if($success){echo '<p class=\"success\">'.$success.'</p>';} elseif($failure) {echo '<p class=\"failure\">'.$failure.'</p>';}; ?>");
        setTimeout(function(){
            $("div.aler").fadeOut("slow", function (){
                $("div.aler").remove();
            });
        }, 5000);
    });
</script>
<?php }

I think there must be a problem with quotes:

" . $failure

has the message, but the JavaScript does not put it in HTML div:

div.aler

I get this error message in the Chrome console:

Uncaught SyntaxError: Unexpected token ILLEGAL

You have to double encode your quotes, \\", so that the result contains \" for the JavaScript string. Look at the generated HTML and you will see. — Felix Kling
– Felix Kling, Commented Jun 22, 2013 at 17:50
Try escaping any " inside PHP like this \\" see what that does. Or change them to single quotes '. Or maybe even \". — Funk Forty Niner
– Funk Forty Niner, Commented Jun 22, 2013 at 17:50
@FelixKling, It puts nothing to the div, so I can not check using html source. but I will give it a try using double encode my quotes. Thanks — Peyman Mohamadpour
– Peyman Mohamadpour, Commented Jun 22, 2013 at 18:26

inf3rno · Accepted Answer · 2013-06-22 17:59:47Z

6

The php output is not escaped for ", so instead of \" you have to use \\\" or \'.

Btw json_encode as a string would be much better...

$("div.aler").html(<?php
    if($success){
        echo json_encode('<p class="success">'.$success.'</p>');
    }
    elseif($failure){
        echo json_encode('<p class="failure">'.$failure.'</p>');
    };?>
);

edited Jun 22, 2013 at 17:59

answered Jun 22, 2013 at 17:53

inf3rno

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1 Comment

inf3rno Over a year ago

@Arash The problem is with your copy-paste skills, not with this code. (I tested before I published.)

bystwn22 · Accepted Answer · 2013-06-22 17:55:29Z

2

you forgot isset in below line, its required since your using "||" (OR) in your first if statement, php throwing an error and thats breaking your javascript

$("div.aler").html( "<?php if( $success ){ echo '<p class=\"success\">'.$success.'</p>';}elseif($failure){echo '<p class=\"failure\">'.$failure.'</p>';}; ?>");

change this to...

$("div.aler").html( "<?php echo ( isset( $success ) ) ? '<p class=\"success\">'.$success.'</p>' : '<p class=\"failure\">'.$failure.'</p>'; ?>");

answered Jun 22, 2013 at 17:55

bystwn22

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2 Comments

bystwn22 Over a year ago

yeah and also dont forget to escape the success or failure message quotes, use 'addslashes'

inf3rno Over a year ago

Ye this clearly depends on wrong server settings. (display_error etc...) Voted up because I got the undefined variable error message, but I declared the variables after that. Did not know that it is possible to not debug something like that...

Raythe · Accepted Answer · 2013-06-22 17:51:54Z

1

You're trying to put it into div.alert...yet in your code you wrote "div.aler" you're missing a T...

answered Jun 22, 2013 at 17:51

Raythe

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Comments

u_mulder · Accepted Answer · 2013-06-22 17:55:57Z

0

$("div.aler").html("<p class='<?=$success? 'success' : 'failure'?>'><?=$success? $success : $failure?></p>");

And of course escaping $success and $failure before output.

answered Jun 22, 2013 at 17:55

u_mulder

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