subhomoy Posted January 14, 2013 Share Posted January 14, 2013 i am trying to catch the id from the url sent in this way... http://localhost/Form/Admin/members_details.php?uid=14<---------------- This Id I have tried using $_REQUEST but its not workinh... I have done this... $uid=$_REQUEST['uid']; $sql="SELECT * FROM students WHERE student_id='$uid'"; $result= mysql_query($result); while($row=mysql_fetch_array($result)){ .......... ........... ........ There are few codes.... } But i'm getting error in while loop... Error is Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Form\Admin\members_details.php on line 6 Any help would be grately appreciated... Quote Link to comment Share on other sites More sharing options...
Dharmender Posted January 14, 2013 Share Posted January 14, 2013 i am trying to catch the id from the url sent in this way... http://localhost/Form/Admin/members_details.php?uid=14<---------------- This Id I have tried using $_REQUEST but its not workinh... I have done this... $uid=$_REQUEST['uid']; $sql="SELECT * FROM students WHERE student_id='$uid'"; $result= mysql_query($result); while($row=mysql_fetch_array($result)){ .......... ........... ........ There are few codes.... } But i'm getting error in while loop... Error is Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Form\Admin\members_details.php on line 6 Any help would be grately appreciated... in Your select statement remove single quote from $uid. Query should be $sql="SELECT * FROM students WHERE student_id=$uid"; And in your While loop use mysql_fetch_assoc($result) and in your code write row['filedname'] And if you want to use mysql_fetch_array the the code would be while($row[]=mysql_fetch_array($result)) { ///use $row[0] $row[1] etc.... } Quote Link to comment Share on other sites More sharing options...
devWhiz Posted January 14, 2013 Share Posted January 14, 2013 (edited) $result= mysql_query($result); should be $result= mysql_query($sql); Edited January 14, 2013 by CLUEL3SS Quote Link to comment Share on other sites More sharing options...
subhomoy Posted January 14, 2013 Author Share Posted January 14, 2013 I've tried what u have said, but its not working... Any more suggestion.... Quote Link to comment Share on other sites More sharing options...
subhomoy Posted January 14, 2013 Author Share Posted January 14, 2013 $result= mysql_query($result); should be $result= mysql_query($sql); Shit man it was such a stupid mistake.... Anyways Thanks man for the help.... Quote Link to comment Share on other sites More sharing options...
devWhiz Posted January 14, 2013 Share Posted January 14, 2013 I've tried what u have said, but its not working... Any more suggestion.... Try using $uid = $_GET['uid']; instead of $_REQUEST[]; Quote Link to comment Share on other sites More sharing options...
devWhiz Posted January 14, 2013 Share Posted January 14, 2013 Shit man it was such a stupid mistake.... Anyways Thanks man for the help.... No problem. What ended up being the issue? Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.