What is wrong with this code?

[iimrii]

<?php
$db = mysql_connect("localhost", "xxxxx", "xxxxxx");
mysql_select_db("xxxxxx",$db);
$result = mysql_query("SELECT * FROM amx_regusers WHERE nick=$nick",$db);
$myrow = mysql_fetch_array($result);
echo "Nick: ".$myrow["nick"];
echo "<br>Pass: ".$myrow["pass"];
echo "<br>Flags: ".$myrow["flags"];
?>

it is giving me this error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/boyem/boyem-www/view.php on line 6

line 6 is $myrow = mysql_fetch_array($result);

thanks in advance

[scoppc]

enclose $nick within single-quote

$sql="SELECT * FROM amx_regusers WHERE nick='$nick'";

[twilson]

I find that it's usually good practice (when talking to a mySQL DB at least) to always inclose your comparative in single quotes, even if it is a field that doesn't require it (such as in INT field).

Example:

mysql_query("SELECT Date, IntField, TextField, CharOrVarcharField FROM InfoTable WHERE Date='$Date' and IntField='$Int' AND TextField='$Text'");

Hope this helps,
Tom

[iimrii]

thanks guys you are life savers 🙂)))