Re: Help with If else if

On Tue, 13 Mar 2012 17:33:43 +0530, Gu®u <nagendra802000@gmail.com> 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