Resource ID #14

[DeathStar]

HI i have a script and its not giveing an mysq_error()

but this

Resource ID #14

 

HEre is my sql part:

$gunfor=mysql_query("SELECT inv_itemid FROM inventory WHERE inv_userid=$userid", $c) or die("Error: gunfor");
$gunfor2=mysql_fetch_array($gunfor, $c);
$sm=mysql_query("SELECT * FROM items WHERE itmid={$gunfor2['inv_itemid']} && itmtype=4 ORDER BY itmname ASC", $c) or die("Error: sm");

if (!$_GET['action']) {
while($r=mysql_fetch_array($sm) or die("$sm"))
{
echo "<tr><td>{$r['itmname']}</td><td><a href='?action=buy&gun={$r['itmid']}</td></tr>";
}
echo "</table>";

 

The $c call a global that is the connection to database

[corbin]

Try

while($r = mysql_fetch_array($sm))

 

I think for some reason, when ever it's assigning the variable it's dying, but it's trying to echo out a query result which will be a resource id.

[tauchai83]

this means there are something wrong with query...try check the query that you wrote. Incorrect query result in data cannot be fetched correctly.

[corbin]

"SELECT * FROM items WHERE itmid={$gunfor2['inv_itemid']} && itmtype=4 ORDER BY itmname ASC"

 

MySQL doesn't allow &&.  Use AND instead .

[DeathStar]

thats worked...

but, not displaying anything now :s

[DeathStar]

Nevermid.

I used a $_GET to bypass that

 

THanks everybody!