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Resource id #5

Posted: Tue Feb 15, 2011 12:39 pm
by Linx
Hello. I have one code but its shows echo "Resource id #5". How to fix it?
Thanks

Code: Select all

$name = $_POST["player_name"]; 
$sql = mysql_query("SELECT * FROM player_extra WHERE name = '".$name."'");
while($row = mysql_fetch_assoc($sql)) {
$id = $row['player_id'];
$sql2 = mysql_query("SELECT * FROM player_race WHERE id = '".$id."' ORDER BY race_id");
$row2 = mysql_fetch_assoc($sql2);

echo "<table border='1'>
<tr>
<th>user</th>
<th>Points</th>
</tr>";

 echo "<tr>";
  echo "<td>". $name. "</td>";
  echo "<td>" .$sql2. "</td>";
  echo "</tr>";
echo "</table>";
}

Re: Resource id #5

Posted: Tue Feb 15, 2011 12:42 pm
by AbraCadaver
Don't echo $sql2.

Re: Resource id #5

Posted: Tue Feb 15, 2011 12:47 pm
by AbraCadaver
Also, one query might be better:

[text]SELECT * FROM player_extra, player_race
WHERE player_extra.name = '$name'
AND player_extra.player_id = player_race.id
ORDER BY race_id[/text]

Re: Resource id #5

Posted: Tue Feb 15, 2011 12:53 pm
by Linx
I chanded sql2 echo to row2 and now its says "Array"

Re: Resource id #5

Posted: Tue Feb 15, 2011 1:04 pm
by AbraCadaver
Linx wrote:I chanded sql2 echo to row2 and now its says "Array"
You did it properly before the loop:

Code: Select all

$id = $row['player_id'];
Maybe:

Code: Select all

echo $row2['something'];

Re: Resource id #5

Posted: Wed Feb 16, 2011 11:08 am
by Mordred
Also read about SQL injection and mysql_real_escape_string