Javascript Syntax Error with embedded PHP code

Question

I have made some silly error, but can not work out what I have done.

I am attempting to test passing variables from PHP to Javascript and if it is an array, json_encode it

My file is a PHP file ie .php

The php line of code that seems to be causing the error I have added to the original PHP and it works OK

<?php

$php_var = 'lol';
$php_array = array ();
$php_array["lady"] = "mary";
$php_array["gent"] = "joseph";
echo is_array($php_array) ? json_encode($php_array) : $php_array;  // same as faulty line in javascript
?>

<html>
<body>

<script type="text/javascript" charset="utf-8">

var php_var = "<?php if (is_array($php_var)) {echo json_encode($php_var); } else { echo $php_var;}; ?>";
document.write(php_var + ' ifElse<br>');

// THE FOLLOWING LINE GIVES  Uncaught SyntaxError: Unexpected identifier 
var php_var2 = "<?php echo is_array($php_array) ? json_encode($php_array) : $php_array; ?>";

document.write (php_var2 + ' EitherOR<br>');

alert(php_var + php_array);

</script>
<h1> Testing Jscript variables</h1>
</body>
</html>

JSON contains double quotes to begin with, which is breaking your JS. — Blender
– Blender, Commented Jan 25, 2013 at 11:35
I guess it is because of quotes in the JSON string. Remove the surrounding quotes. — VisioN
– VisioN, Commented Jan 25, 2013 at 11:35
Guys, he specified it in the source // THE FOLLOWING LINE GIVES Uncaught SyntaxError: Unexpected identifier . — h2ooooooo
– h2ooooooo, Commented Jan 25, 2013 at 11:36
@h2ooooooo true, I've overlooked it, though it's still ... to post an error in the source and not mentioning it in the text — dualed
– dualed, Commented Jan 25, 2013 at 11:39

vedarthk · Accepted Answer · 2013-01-25 11:47:50Z

2

As you have specified the error is at :

var php_var2 = "<?php echo is_array($php_array) ? json_encode($php_array) : $php_array; ?>";

The error may be due to your using double quotes ("") use single quotes ('') in Javascript.

This may solve your error : var php_var2 = '<?php echo is_array($php_array) ? json_encode($php_array) : $php_array; ?>';

Or you can directly create an Javascript Object from the JSON string using eval().

http://jsfiddle.net/jduGp/

answered Jan 25, 2013 at 11:47

vedarthk

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1 Comment

mcl Over a year ago

Thanks for all the comments/answers - the single quote solved it. I have learnt a bit more about interfacing PHP and Javascript.

Techie · Accepted Answer · 2013-01-25 11:36:54Z

0

Try this code. Replace the two lines as shown below.

var php_var = <?php if (is_array($php_var)) {echo json_encode($php_var); } else { echo $php_var;}; ?>;

var php_var2 = <?php echo is_array($php_array) ? json_encode($php_array) : $php_array; ?>;

answered Jan 25, 2013 at 11:36

Techie

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Comments

And Finally · Accepted Answer · 2013-01-25 11:43:11Z

0

I did it like this

<?php

$php_var = 'lol';
$php_array = array ();
$php_array["lady"] = "mary";
$php_array["gent"] = "joseph";
?>

<html>
<body>

<script type="text/javascript" charset="utf-8">

var php_var = <?php if (is_array($php_var)) {echo json_encode($php_var); } else { echo '"' . $php_var . '"';} ?>;
document.write(php_var + ' ifElse<br>');

// THE FOLLOWING LINE GIVES  Uncaught SyntaxError: Unexpected identifier
var php_var2 = <?php echo is_array($php_array) ? json_encode($php_array) : $php_array; ?>;

document.write (php_var2 + ' EitherOR<br>');

alert(php_var);

</script>
<h1> Testing Jscript variables</h1>
</body>
</html>

Not sure why you're trying to alert out php_array when Javascript isn't aware of that variable. You also don't need the quotes unless you're outputting a string. If you put quotes around an object Javascript will think it's a string.

answered Jan 25, 2013 at 11:43

And Finally

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2 Comments

mcl Over a year ago

When I tried it without quotes it gave the following Uncaught ReferenceError: lol is not defined

And Finally Over a year ago

When I remove the quotes from around the outer braces: var php_var2 = <?php echo is_array($php_array) ? json_encode($php_array) : $php_array; ?>; JS seems to be happy - it just complains that php_array is undefined, which is fair enough because as far as JS is concerned it isn't.

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