IF Statement to display image

[mrquestions]

Hey, new to the forum, have looked everywhere for some information of the following query but can't find it anywhere...

Within a site I am building a division calls on an include function to display a webpage ($page), so only the one template page is used throughout the entire site. It has been all smooth sailing up until I wanted to have a MouseOver image (image.src), become "selected" upon requesting its specific page.

[PHP]<? if ($page = 'pagename') {
echo
'<img src="imageM.jpg name="image" border="0">';
}
else {
echo
'<a onMouseOver="im age.src='imageM .jpg';" onMouseOut="ima ge.src='image.j pg';"><img src="image.src name="image" border="0"></a>';
}
?>[/PHP]


It is within a table division if that means anything...
oh, and its all on one line, not like above...

Cheers

[TheMadMidget]

[PHP]<? if ($page = 'pagename') {
echo
'<img src="imageM.jpg " name="image" border="0">';
}
else {
echo
'<a onMouseOver="im age.src=&#39imageM.jpg'" onMouseOut="ima ge.src=&#39image.jpg'"> <img src="image.src" name="image" border="0"></a>';
}
?>[/PHP]
I changed the internal 's to &#39 . This allows the html to have the apostrophe but doesn't interfere with the PHP code.
Also in both <img> tags, the src attribute only had a quotation before it, non after it, so I fixed that too.