[SOLVED] variable and if

[onthespot]

Hey guys, cant get this working.

So I have a variable

 

     $ulevel = mysql_result($result,$i,"userlevel");

 

I want to have a second variable where it the first variable for the return of a number. So if ulevel = 9, echo Admin.

 

      $level = if $ulevel = 9; echo "Admin";

 

That didn't work.

Any ideas?

[KevinM1]

First, this is a simple syntax error that can be avoided by reading the manual, or any of the millions of beginner tutorials found through a google search.  Link to the manual page you want: http://us2.php.net/manual/en/control-structures.if.php

 

Second, you don't need to assign a variable here.  Simply:

 

if ($ulevel = 9) { echo "Admin"; }

[onthespot]

Thankyou, how would I fit it into the following?

 

   echo "<table cellspacing=\"3\" style='border: 1px dotted;' bgcolor=\"#eeeeee\">\n";
   echo "<tr><td><b>Username</b></td><td><b>Level</b></td><td><b>Email</b></td></tr>\n";
   for($i=0; $i<$num_rows; $i++){
      $uname  = mysql_result($result,$i,"username");
      $ulevel = mysql_result($result,$i,"userlevel");
      $email  = mysql_result($result,$i,"email");

      echo "<tr><td>$uname</td><td>[b]$ulevel[/b]</td><td>$email</td></tr>\n";
   }
   echo "</table><br>\n";

 

where i have put bold, I want the following to happen

      if ($ulevel = 9) { echo "Admin"; }

 

Also can I put several if statements in, for different numbers?

[KevinM1]

After you obtain the $ulevel, but before you echo the data, do something like:

 

$level = "";

if ($ulevel == 9) { $level = "admin"; }

 

Then, use $level as the echoed data.

 

And, you can have an infinite number of if statements, or if/else statements. 

[onthespot]

Perfect, thankyou.