Passing PHP variable in a javascript function

Question

<button class="addToPlaylist" onclick="javascript:myPopup(<?php echo $videos[$counter]?>);
return false;">+</button>

I have a button on an image as a html hyperlink. I want to perform different actions on hyperlink and button. The above code works whenever I do not pass the PHP variable using echo. When i pass PHP variable, the button also performs the same action as of the hyperlink, that means return false does not work.

Any idea why the return false; does not work when i pass PHP variable?

Is `$videos[$counter]` a string? If so, you need quotes around it... — Brad, Jan 16 '12 at 17:12

score 2 · Accepted Answer · answered Jan 16 '12 at 17:14

This should be:

<button class="addToPlaylist" onclick="javascript:myPopup('<?php echo $videos[$counter];?>');return false;">+</button>

Note the single quotes in myPopup. As you pass a string to myPopup, you will need to enclose it with single quotes. (Double won't work as there is already double quotes for the onclick)

score 1 · Answer 2 · answered Jan 16 '12 at 17:14

I am quite sure $videos[$counter] is not numeric, but a string. In this case you have to write the quotes:

onclick="javascript:myPopup('<?php echo $videos[$counter]?>');

And make sure, $videos[$counter] doesn't contain any, something like

onclick="javascript:myPopup('<?php echo addslashes($videos[$counter])?>');

comes to mind.

anjalis · Answer 3 · 2012-01-16T17:23:27.513

0

onclick="javascript:myPopup("";return false;" . This should work and i think it's more clear where you have javascript code and php code.

edited Jan 16 '12 at 17:23

answered Jan 16 '12 at 17:16

anjalis

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Passing PHP variable in a javascript function

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