mysql_fetch_assoc(): supplied argument is not a valid MySQL

[techm3]

Hi there, Im using this code:

<?php
	function fetch_user_info($name){	
		$sql = "SELECT 
				'bname' AS 'name',
				'email' AS 'email',
				'website' AS 'website',
			FROM 'cInfo'
			WHERE 'name' = {$name}";
		$result = mysql_query($sql);
		return mysql_fetch_assoc($result);
	}
?>

but I'm getting "Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/content/core/user.inc.php on line 15" which is the return line.

Any ideas?

Thanks!

[Temor]

you need to use backticks ( ` ) instead of semiqoutes ( ' ).

$sql = "SELECT
                                `bname` AS `name`,
                                `email` AS `email`,
                                `website` AS `website`,
                        FROM `cInfo`
                        WHERE `name` = {$name}";

You also don't need the AS part on email and website seeing as you change the name to the exact same thing.


Try this query and report back

$sql = "SELECT
                                `bname` AS `name`,
                                `email`,
                                `website`,
                        FROM `cInfo`
                        WHERE `name` = {$name}";

[jacek]

If you add an

echo mysql_error();

after the failing query it will tell you what's wrong.

[techm3]

I tried using the following code:

<?php
	//fetches profile information for the given user
	function fetch_user_info($cname){	
		$sql = "SELECT 
				`cname`,
				`address`,
				`tel`,
				`email`,
				`wsite`
			FROM `cInfo`
			WHERE `cname` = {$cname}";

		$result = mysql_query($sql);

		return mysql_fetch_assoc($result);
	}
?>

but still getting an error that says:
"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 8"

[EcazS]

You should add semiquotes around {$cname}

// stuff...
WHERE `cname` = '{$cname}'

Also make sure you're actually passing in something into the cname variable when using the function,

fetch_user_info("Name");

Also, did you try with

echo mysql_error();

after the query?

[techm3]

Thanks for all your replies.

@EcazS '{$name}' solve it
Thank You!