Php verification display content

Php verification display content verification member identity

 GetUserType () = 10) {echo "";} else if ($ cuserLogin-> getUserType () = 5) {echo "" ;}?>

Query and display content

 SetQuery("SELECT * FROM `#@__arcatt` ORDER BY sortid ASC");         $dsql->Execute();         while($trow = $dsql->GetObject())         {         if($trow->att=='j')         {         $jumpclick = " onclick='ShowUrlTr()'";         }         else         {         $jumpclick = '';         }         if(preg_match("#".$trow->att."#", $arcRow['flag']))         {         echo "{$trow->attname}[{$trow->att}]";         }         else         {         echo "{$trow->attname}[{$trow->att}]";         }         }          ?>

Description: I want to verify the member identity before executing the query code. for example, if the member identity is "administrator", the query code is executed, if the member identity is "channel editor", nothing is displayed.

Reply to discussion (solution)

Directly if ($ cuserLogin-> getUserType () = 10) {}, it is not good to write the query code. if it is not, it will not be executed.

Directly if ($ cuserLogin-> getUserType () = 10) {}, it is not good to write the query code. if it is not, it will not be executed.
Thank you very much.

Directly if ($ cuserLogin-> getUserType () = 10) {}, it is not good to write the query code. if it is not, it will not be executed.
Points at a glance!

Directly if ($ cuserLogin-> getUserType () = 10) {}, it is not good to write the query code. if it is not, it will not be executed.
Points at a glance!

Grab a friend !!!

Go and reply to my post!

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