[SOLVED] resource id#

[chocopi]

with my code it gives resource id# 16 or whatever it is and nomally to change it to the value i would use $blah = $row['name']; but for some reason it gives blank, but i know that it is not blank so heres my code:

 

<?php
$enemy_id = $_SESSION['enemy_id'];
// Work out enemy drops
$enemy_query = mysql_query("SELECT `drop1` FROM `Enemies` WHERE `id`='$enemy_id'");
$row = @mysql_fetch_assoc($enemy_query) or die("Invalid query, please contact an administrator!");
list($table1, $id1) = split("_", $row['drop1']);
$array = array('Body','Hats','Items','Pants','Weapons');
sort($array);
foreach ($array as $key => $table1)
{
$drop1 = "SELECT `id` FROM `$table1` WHERE `id`='$id1'";
$drop1 = mysql_query($drop1) or die(mysql_error() . ' ON Query ' . $drop1);
}
?>

 

Thanks,

 

~ Chocopi

[chocopi]

Nevermind it wasnt working as:

 

<?php
$enemy_id = $_SESSION['enemy_id'];
// Work out enemy drops
$enemy_query = mysql_query("SELECT `drop1` FROM `Enemies` WHERE `id`='$enemy_id'");
$row = @mysql_fetch_assoc($enemy_query) or die("Invalid query, please contact an administrator!");
list($table1, $id1) = split("_", $row['drop1']);
$array = array('Body','Hats','Items','Pants','Weapons');
sort($array);
foreach ($array as $key => $table1)
{
$drop1 = "SELECT `id` FROM `$table1` WHERE `id`='$id1'";
$drop1 = mysql_query($drop1) or die(mysql_error() . ' ON Query ' . $drop1);
}
?>

 

should have been

<?php
$enemy_id = $_SESSION['enemy_id'];
// Work out enemy drops
$enemy_query = mysql_query("SELECT `drop1` FROM `Enemies` WHERE `id`='$enemy_id'");
$row = @mysql_fetch_assoc($enemy_query) or die("Invalid query, please contact an administrator!");
list($table1, $id1) = split("_", $row['drop1']);
$array = array('Body','Hats','Items','Pants','Weapons');
sort($array);
foreach ($array as $key => $table1)
{
$drop1 = mysql_query("SELECT `id` FROM `$table1` WHERE `id`='$id1'");
$drop1 = mysql_fetch_assoc($drop1) or die(mysql_error() . ' ON Query ' . $drop1);
$drop1 = $drop1['id'];
}
?>

[paul2463]

if the resource id #16 is returned and not the data then you have not made a fetch statement

<?php
$query = "SELECT * table ";
$answer = mysql_query($query) of die ("Cannot get the stuff" . mysql_error());

echo $answer; //would output Resource id# 16 or some such number

$row = mysql_fetch_row($answer);

echo $row[0]; //would out put the first key in the row

?>

[chocopi]

yea like a muppet, i had used a query instead