Keep getting "Resource ID #4"

[Markus]

I'm trying to compare the random password i generated against the current password to see whether it needs changing or not.

But i hit this, and can't see what's wrong with it.

I'm using the mysql_fetch_array so i don't see why its saying Resuorce ID #4

:(
[php]
<?php
include("../library/config.php");
include("../library/opendb.php");
$pass = mysql_query("SELECT `ziprar_loginPass` FROM `ziprar_users` WHERE `ziprar_loginName` =

'markusn00b'");
$row = mysql_fetch_array($pass);
$change = mysql_query("SELECT `ziprar_passChange` FROM `ziprar_users` wHERE

`ziprar_loginName` = 'markusn00b'");
$change = mysql_fetch_array($change);
echo $row['ziprar_loginPass'];
/*
if($pass['ziprar_loginPass'] === $change['ziprar_passChange']){
echo "Change password";
} else{
echo "Password changed";
}
*/
?>
[/php]

[ak1dnar]

Your first query string will execute first and it will store the Result resource in $pass variable with the help of mysql_query function. And then by using mysql_fetch_array($pass) you are going to store that result resource on $row variable.Correct. BUT again with your IF block why you trying get a OUTPUT from $pass variable. You can’t do that.

What you have to do

$pass = mysql_query("SELECT `ziprar_loginPass` FROM `ziprar_users` WHERE `ziprar_loginName` = 'markusn00b'");
$row_pass = mysql_fetch_array($pass);
$var_pass = $row_pass ['ziprar_loginPass']
 
$change= mysql_query("SELECT `ziprar_passChange` FROM `ziprar_users` wHERE `ziprar_loginName` = 'markusn00b'");
$row_change = mysql_fetch_array($change);
$var_change = $row_change['ziprar_passChange']
 

Then Compare $var_pass and $var_change

[Markus]

Your first query string will execute first and it will store the Result resource in $pass variable with the help of mysql_query function. And then by using mysql_fetch_array($pass) you are going to store that result resource on $row variable.Correct. BUT again with your IF block why you trying get a OUTPUT from $pass variable. You can’t do that.

What you have to do

$pass = mysql_query("SELECT `ziprar_loginPass` FROM `ziprar_users` WHERE `ziprar_loginName` = 'markusn00b'");
$row_pass = mysql_fetch_array($pass);
$var_pass = $row_pass ['ziprar_loginPass']
 
$change= mysql_query("SELECT `ziprar_passChange` FROM `ziprar_users` wHERE `ziprar_loginName` = 'markusn00b'");
$row_change = mysql_fetch_array($change);
$var_change = $row_change['ziprar_passChange']
 

Then Compare $var_pass and $var_change

Legend! Thanks a bunch.

Do you know why it has to be like that though?

Cheers.

[ak1dnar]

Legend! Thanks a bunch.

Do you know why it has to be like that though?

Cheers.

After taking the result resource from mysql_query, you must use one of the mysql functions that take mysql results resources as input and processes them. exaples for these are mysql_fetch_assoc,mysql_fetch_array
You have not done this for second sql query