Javascript problem

[snakeplissken]

onmouseover="dropdownmenu(this, event, \' $x \')

How can I make the above work, I could never figure out how to use php variables in javascript.

[Jeff Lewis]

How are you generating the  onmouseover? is it in an echo or just plain HTML?

if it's an echo:

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echo '<img src="bah.png" onmouseover="dropdownmenu(this, event, \''. $x . ' \')">';


If it's in HTML:

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<img src="bah.png" onmouseover="dropdownmenu(this, event, '<?php echo $x; ?>')">


There's obviously various shortcuts and other ways but that should do you fine...need to know the code around it and how it's being printed (like if it's coming from an echo)

[snakeplissken]

Thanks Jeff,

It was actually from an echo.

But it was late, I wasn't thinking

[karlbenson]

Theres nothing I hate more than working with Javascript and having to test it in various browsers only to find it won't work in one because browser x doesnt support it as an object or whatever.