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Parse Error, but why?

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digatle

Technical User
Oct 31, 2003
85
US
<code>
<?php
echo $blockopen;
echo $siteID.&quot; Dept.&quot;;
echo $blockbody;
print &quot;<form name=\&quot;jump\&quot;>&quot;;
print &quot;<select name=\&quot;menu\&quot; onChange=\&quot;location=document.jump.menu.options[document.jump.menu.selectedIndex].value;\&quot; value=\&quot;GO\&quot;>&quot;;
print &quot;<option value=\&quot;Select a Department\&quot;>Select a Department</option>&quot;;
while ($a_row = mysql_fetch_array( $category ) ) {
$field=$a_row['category_label'];
$sid2=$a_row['category_blog_id'];
$blogid2=$a_row['category_id'];
print &quot;<option value=&quot;.FS_WEB_NAME.&quot;home.php?sid=$sid2&blogid=$blogid2>$field</option>&quot;;
}
print &quot;</form>&quot;;
echo $blockend;

while ($a_row = mysql_fetch_array( $divapp ) ) {
$app_url=$a_row['url'];
$app_image=$a_row['image_location'];
$app_alt=$a_row['description'];
if (isset($app_url)) {
echo $blockopen;
print &quot;Apps&quot;;
echo $blockbody;
print &quot;<a href=\&quot;$app_url\&quot;><img src=\&quot;$app_image\&quot; alt=\&quot;$app_alt\&quot; border=\&quot;0\&quot;></a><br> &quot;;&quot;;
echo $blockend;
}
}
?>

</code>

Why would I be getting the following error on this page?

Parse error: parse error in /var/ on line 29

If I remove the isset its fine but doing an isset gives me this. I thought this would be a good solution to the MySQL and PHP delima from earlier.

Digatle
 
this line :

print &quot;<a href=\&quot;$app_url\&quot;><img src=\&quot;$app_image\&quot; alt=\&quot;$app_alt\&quot; border=\&quot;0\&quot;></a><br> &quot;;&quot;;

has too many &quot;; at the end :)

______________________________________________________________________
There's no present like the time, they say. - Henry's Cat.
 
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