Re: Help with If else if

On Tue, 13 Mar 2012 17:33:43 +0530, Gu®u <[email protected]> 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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