marksie1988 Posted July 12, 2007 Share Posted July 12, 2007 Hi, i have written some code that i want to display an image but only if a field in my sql has information in it i currently have: if($req_user_info['Skype'] > 0){ /* Skype */ echo "<a href='callto://".$req_user_info['Skype'] ."'><img src='../images/icons/skype.gif' border='0'/></a>"; } elseif($req_user_info['myspace'] > 0){ /* myspace */ echo "<a href='http://".$req_user_info['myspace'] ."'><img src='../images/icons/myspace.gif' border='0'/></a>";} else{echo "No Extras";} i think that the problem is if($req_user_info['myspace'] > 0){ ^^^ should i be saying is greater than 0 or not? im new to this so please help. Quote Link to comment Share on other sites More sharing options...
cmgmyr Posted July 12, 2007 Share Posted July 12, 2007 you could try isset() if(isset($req_user_info['Skype'])){ ... } Quote Link to comment Share on other sites More sharing options...
marksie1988 Posted July 12, 2007 Author Share Posted July 12, 2007 OMG i cant belive i didnt notice that sometimes i wonder what my mind is doing to me i totaly disreguarded the isset command Cheers Quote Link to comment Share on other sites More sharing options...
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