admin_user_update.php / Adding info / Joining tables

 Hi Olaf,

I am adding info to the admin_user_update.php page.

The goal is that when the user searches for a username to update the result will display the profile info in addition to what's already there.

I have no problem when adding fields that are in the user table. When I change the query to left join the user_profile table I get  the following error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

I suspect the problem is $this->table_name but I'm not able to change it to a value that will work.



                                                $sql = sprintf("SELECT id, login, real_name, email, access_level, active FROM %s Left Join `users_profile` ON `users_profile`.`users_id` = `users`.`id` WHERE login = '%s'", $this->table_name, trim($for_user));
      }
      else {
         $sql = sprintf("SELECT id, login, real_name, email, access_level, active FROM %s Left Join `users_profile` ON `users_profile`.`users_id` = `users`.`id`  WHERE id = %d", $this->table_name, intval($for_user));
      }
      $result = mysql_query($sql);
      if (mysql_num_rows($result) == 1) {
         $obj = mysql_fetch_object($result);
         $this->user_id = $obj->id;
         $this->user_name = $obj->login;
         $this->real_name = $obj->real_name;
         $this->old_user_email = $obj->email;
         $this->user_access_level = $obj->access_level;
         $this->activation = $obj->active;
         if ($this->user_name != $_SESSION['user']) {
            $this->user_found = true;
         } else {
            $this->user_found = false;
            $this->the_msg = "You cannot change your own data!";
         }
         mysql_free_result($result);
      } else {
         $this->the_msg = "Sorry, no login data for this name!";
      }   
   }

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replace this code to get an error:

$result = mysql_query($sql) or die(mysql_error());

Thanks Olaf.

That gave me the error message to answer the problem. When I joined the user_profile table it created an ambiguous column name on the 'id' column.

Regards, Jeff

[Guest]

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