Dynamic $result based on IF statement, Help please

[Darkstars31]

list($game, $type)= explode('|', $_POST[game]);
if($type == "game"){$result = mysql_query( "SELECT TeamName,Owner,Wins,Loses FROM $dbname WHERE game_name = '$_POST[game]'")or die("SELECT Error: ".mysql_error());
} else {
$result = mysql_query ( "SELECT user_id FROM $dbnametwo WHERE Teamname = '$_POST[game]'")or die("SELECT Error: ".mysql_error());
}
$num_rows = mysql_num_rows($result);
print "There are $num_rows Teams/Players.<P>";
print "<table width=600 border=1>\n<tr><td>Team Names/Players</td><td>Owner</td><td>W</td><td>L</td></tr> </table>";
print "<table width=600 border=1>\n";
while ($get_info = mysql_fetch_row($result)){
      print "<tr>\n";
      foreach ($get_info as $field)
      print "\t<td>$field</font></td>\n";
      print "</tr>\n";
}
print "</table>\n";
?>
<form method="POST">
<Select name="game">
        <option></option>
        <?php
            while (list($game_selection) = mysql_fetch_row($find)) {
             $display = "<option value=\"$game_selection\">$game_selection</option>";
                      echo $display;
                            while (list($team_selection) = mysql_fetch_row($find2)) {
                                  $display2 = "<option value=\"$team_selection\">*$team_selection</option>";
                                            echo $display2; }
            }
            ?>

</select>

 

The IF statement at the top, always returns false, so I can never view the $result. I'm writing a simple Clan Ladder script. I want it to display the teams in the table based on user selection, thats the first $result. The Second Displays all the players on the team selected by the User from a Dropdown box. I can view the users on the teams, but i cant just view a master list of the teams. Help?

[Jessica]

First, you need to use $_POST['game'];

Secondly, what happens if you print $game before going into the loop. Why is it never == 'game'?

[Darkstars31]

so it should be:

 

if($type == $_POST[game]) ? i think I've tried that do i need '$_POST[game]' instead?

 

Not sure what you mean by printing $game before going to the loop?

[Jessica]

That's not what I wrote...you need $_POST['game'].

Before the if statement, add an echo or print to display the value of $game.

 

[AndyB]

Change:

list($game, $type)= explode('|', $_POST[game]);
if($type == "game"){$result = mysql_query( "SELECT TeamName,Owner,Wins,Loses FROM $dbname WHERE game_name = '$_POST[game]'")or die("SELECT Error: ".mysql_error());
} else {
$result = mysql_query ( "SELECT user_id FROM $dbnametwo WHERE Teamname = '$_POST[game]'")or die("SELECT Error: ".mysql_error());
}

 

to:

list($game, $type)= explode('|', $_POST['game']);
if($type == "game") {$result = mysql_query( "SELECT TeamName,Owner,Wins,Loses FROM $dbname WHERE game_name = '$game'")or die("SELECT Error: ".mysql_error());
} else {
$result = mysql_query ( "SELECT user_id FROM $dbnametwo WHERE Teamname = '$game'")or die("SELECT Error: ".mysql_error());
}