cannot display data

Dillenger

Hello,

I'm trying to display some fields from my database, but I am getting this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/dhill/simplebit-www/testing2/users/dillenger/index.php on line 115

Can anyone help??? Here is my source code:

<?
require('../../Templates/config.inc');
// connect to database
$cid = mysql_connect($host,$usr,$passwd);
if (!$cid) {
	echo("ERROR: " . mysql_error() . "\n");
}

// setup SQL statement to retrieve subscriber that we want to edit  
$SQL = " SELECT * FROM users ";
$SQL = $SQL . " WHERE username = $username ";
// execute SQL statement
$ret = mysql_db_query($db,$SQL,$cid);
//retrieve values
$row = mysql_fetch_array($ret);
$id = $row["id"];
$username = $row["username"];
$type = $row["type"];

echo "<span class=\"submenu\"><font size=2>Welcome $username</font><br><br>";
echo "Account Type: <font color=#ff0000>$type</font></span>";

?>

jpmoriarty

you're getting that error because there are no results coming back from your query. if you do a echo mysql_num_rows($ret) you'll probably find it equals zero.

Try echoing out the SQL statement before you run it to check that it's okay.

and also i always use this for connecting and querying my db - you may find there's a problem with your use of mysql_connect within mysql_db_query, and i find this is more intuitive:

$db = mysql_pconnect("db", "user", "pword");
mysql_select_db("database");
$sql = "select * from table";
$result = mysql_query($sql);
$num_results = mysql_num_rows($result);
for($i=0;$i<$num_results;$i++)
{
$row = mysql_fetch_array($result)
}

xblue

username is likely to be a string I think?
then don't forget to put it in single quotes, or it is mistaken as being a column name. also you should add some basic error checking when coming across these kind of errors, like


// setup SQL statement to retrieve subscriber that we want to edit  
$SQL = " SELECT * FROM users ";
$SQL = $SQL . " WHERE username = '$username' ";
// execute SQL statement
$ret = mysql_db_query($db,$SQL,$cid) or die("Error: ".mysql_error()." in query: ".$SQL);