i think i missed the basic sql lessons!

owaring

$passquery = "SELECT password FROM users WHERE login = 'owaring'";
		 $password = mysql_query($passquery)
			or die ("Couldn't execute password query.");


echo $password;

ok, this is slightly embarraisng, i seem to have missed the basics of sql!!

why does this echo "Resource id #4"??

i figured the obvious solution was that i sql returned an array, knowing there could only be one value i tried

$password[0]

this was blank.

if somone could point out the blazing obvious i would appreciate it, if any one could point me towards the sql basics i missed that would be ace too!!

laserlight

You didnt miss the basics of SQL, you missed the bases of using RDMS with PHP.
The reason for your problem is that mysql_query() returns a result resource.

$passquery = "SELECT password FROM users WHERE login = 'owaring'";
$password = mysql_query($passquery)
	or die ("Couldn't execute password query.");
$result = mysql_fetch_assoc($passquery);
echo $result['password'];

//or

$password = mysql_result($passquery, 0);
echo $password;

By the way, since password is (most probably) a reserved word, you should use:
SELECT password FROM users WHERE login = 'owaring'

owaring

thanks laserlight

RDMS ??

could you suggest some reading material??

owaring

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in H:\apachefriends\xampp\htdocs\willow\login.php on line 28

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in H:\apachefriends\xampp\htdocs\willow\login.php on line 33

i recieve these two errors when i try your two examples respectivly.

any idea why?

travelbuff

That means that the query returned FALSE. You did open a DB connection and select a DB, right? There IS a row in the DB that hasa value for 'password' and where login = 'owaring', right?

owaring

yes and yes!!!

i ran the exact same query in php myadmin and recieved the espected result

laserlight

RDMS ::= Relational Database Management System
e.g. mysql, postgresql, oracle, mssql, sqlite, firebird

i recieve these two errors when i try your two examples respectivly.

Like what travelbuff mentioned, check your database server connection, database selection, and that your query is indeed correct.

mtmosier

Looks like a typo in the example.

$passquery = "SELECT password FROM users WHERE login = 'owaring'";
$password = mysql_query($passquery)
    or die ("Couldn't execute password query.");
$result = mysql_fetch_assoc($password);  //  Change $passquery to $password
echo $result['password'];

owaring

the typo sorts out the first example.

anyone any ideas on getting laserlights second example working??

this would be a lot neater.

mtmosier

Sorry, I didn't bother including the second one since its such an uncommonly used form.

$passquery = "SELECT password FROM users WHERE login = 'owaring'";
$passresult = mysql_query($passquery)  // Changed from $password to $passresult to avoid confusion
    or die ("Couldn't execute password query.");
$password = mysql_result($passresult, 0);  // Changed from $passquery to $passresult
echo $password;

When retreiving more than one column of data in a single query, it's recommended that you use mysql_fetch_assoc, or a similar function which returns an entire row at a time, for performance reasons.

laserlight

Sorry about the typo, by convention I use $query for the SQL query string, $result for the result resource returned, and $row for the current row array retrieved, so I made a mistake when trying to accomodate your style.