On Tue, 13 Mar 2012 17:33:43 +0530, Gu®u <nagendra802000@xxxxxxxxx> sent:
Hi,
Please help me with this code. I have 2 different fields in mysql table.
What I want is if the field is empty don't show the image. Please look at
the code below.
<?php
if($search->plugin->ListViewValue()=="")
{
echo '<a href="'.$search->facebook->ListViewValue().'"><img
src="images/facebook.gif" width="22" height="23"/></a></a>';
}
if($search->facebook->ListViewValue()=="")
{
echo '<a href="'.$search->plugin->ListViewValue().'"><img
src="images/twitter.gif" width="22" height="23"/></a></a>';
}
else if($search->plugin->ListViewValue()=="" &&
$search->facebook->ListViewValue()=="")
{
echo "";
}
else
{
echo '<a href="'.$search->plugin->ListViewValue().'"><img
src="images/twitter.gif" width="22" height="23"/></a></a>'.'<a
href="'.$search->facebook->ListViewValue().'"><img
src="images/facebook.gif" width="22" height="23"/></a></a>';
}
?>
--
*Best,
*
*Gu®u*
I think we really need to see a lot more than this before we can help.
What is the output that is generated? What is $search and how is it
set prior to entering this bit of code? As it stands, I can see no
reference to MySQL tables in this, nor any idea what values you're
expecting and not seeing. If you don't already, please set
error_reporting to the most detailed, and turn on display_errors in
your output. Before each of the if's echo a var_dump of the values
you're testing so we can see their exact values.
--
Tamara Temple
aka tamouse__
May you never see a stranger's face in the mirror
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