php inside javascript

Question

why does the following code not work?

<?php
$j= "<option value='kj' name='dd'>" ;
?>

<script type="text/javascript">
    function getinfo(a){     
    if(a==2){
    var k ="<?php echo $j;?>";
    alert(k);
    }


}
</script>

It works when I say $j="option";

Any errors? Have you checked your page's generated source? What does it say in there? — Alex Turpin
– Alex Turpin, Commented Sep 7, 2012 at 18:57
Well it does not show the alert box so I assume it's not working — zista
– zista, Commented Sep 7, 2012 at 18:57
You need to debug it. Is your function even being called? Do you have any errors in your console? What does your generated source show? — Alex Turpin
– Alex Turpin, Commented Sep 7, 2012 at 18:58
@sam no you won't, in the context of putting a value in a JS variable from PHP, you don't need to do any deserialization in the JS. It will be the value as-is. — Alex Turpin
– Alex Turpin, Commented Sep 7, 2012 at 19:08
@sam I'm sorry, I don't understand where deserialization would ever be necessary in this. Can you show me an example? — Alex Turpin
– Alex Turpin, Commented Sep 7, 2012 at 19:34

thatidiotguy · Accepted Answer · 2012-09-07 18:59:33Z

4

There are a couple possibilities here:

1) getInfo is never called.

2) a is never equal to 2

3) There are errors in other javascript code on your page.

EDIT 4): Other people have pointed out that the content of $j may not be allowed in an alert box.

There is no way for anyone here to tell which it is.

answered Sep 7, 2012 at 18:59

thatidiotguy

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1 Comment

BNL Over a year ago

+1 W/o seeing how/if getInfo is called, this is unanswerable.

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