Problem with PHP and mouseover states

[dmacman1962]

Hi all,

I have a coding issue with the following:

<? print "<a href=\"mg_frame3.php?productsID=$productsID\"target=\"mainFrame\">$Name onMouseOut=\"MM_swapImgRestore()\" onMouseOver=\"MM_swapImage('Poster','','images/buttons/button1A.png',1)\"><img src=\"images/buttons/button1.png\" alt=\"Poster\" name=\"Poster\" width=\"153\" height=\"35\" border=\"0\"></a>"; ?>

The mouse over does not work and it displays the code "onMouseOut=......." with the image.

If I just do an HTML mouse over it works perfectly.

When I try to mix PHP and HTML using the \ to escape the "s, it fails.

I'm stumped.

Thanks in advance,

Don

[AdamBrill]

Your problem is in this part of that line: >$Name

Notice that this will end the <a tag prematurely... I don't know what's in the $Name variable, but it may work if you just delete what I said above so your code would look something like this:

echo "<a href=\"mg_frame3.php?productsID=$productsID\" target=\"mainFrame\" onMouseOut=\"MM_swapImgRestore()\" onMouseOver=\"MM_swapImage('Poster','','images/buttons/button1A.png',1)\"><img src=\"images/buttons/button1.png\" alt=\"Poster\" name=\"Poster\" width=\"153\" height=\"35\" border=\"0\"></a>";

You'll notice that I also changed your print to echo since there is no point in having it return a value(which is the difference between echo and print). If you care to know more about the differences between echo and print, look here: http://www.faqts.com/knowledge_base/view.phtml/aid/1/fid/40

[dmacman1962]

Awesome,

Thanks a lot!!!

I have been going back and forth with this and I knew it was some stupid little error, which makes it a big error when it wil not work.

I owe you one.

Thanks a bunch,

Don

[AdamBrill]

No problem, I'm glad it worked for you. 😉