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KazeFlame

12-27-2012 02:46 PM

[HELP]mysql_fetch_array() expects parameter 1 to be resource, boolean given in....

Error:
mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\panel.php on line 12

Code:

PHP Code:


		
			
7. //Get Account Information

8. $getuaccount = mysql_query("SELECT * from members where username = '$uname'");

9. $useraccount = mysql_fetch_array($getuaccount);

10. //Get Profile Information

11. $getuprofile = mysql_query("SELECT * from members-profile-info where username = '$uname'");

12. $userprofile = mysql_fetch_array($getuprofile);

13.

14. echo 'Welcome ' . $useraccount['username'] . '!';

15. echo '<br/>Fullname:' . $userprofile['fullname'];

Thanks in advance for those who will help :)

Fou-Lu

12-27-2012 02:57 PM

You have no error handling in here.
Your query has simply failed. The easiest way to determine why is to tack on an or die(mysql_error()); to the end of the mysql_query call. It will be a structural fault on it, not a data one (data/logic errors would produce empty resultsets, not false).

Old Pedant

12-27-2012 07:43 PM

Actually, the reason is obvious:

Code:

SELECT * from members-profile-info where username = '$uname'"

MySQL will read that supposed table name as

Quote:

members MINUS profile MINUS info

Just as PHP would. Just as JavaScript would. Etc.

YOU CAN NOT USE A MINUS SIGN IN THE MIDDLE OF A NAME!

Well, fortunately, with MySQL, you can.

But if you do so, you *MUST* then tell MySQL that you are using a non-standard name by enclosing the entire name in back ticks (the ` character...shares a key with ~ normally).

So:

Code:

$getuprofile = 

    mysql_query("SELECT * from `members-profile-info` where username = '$uname'")

    or die( mysql_error() );

Incidentally, you don't show the rest of your PHP page, but almost surely you have goofed by making two separate queries in the code you showed us, insted of using a single query that JOINs the two tables.

KazeFlame

12-28-2012 05:10 AM

Thanks :D
I replaced

PHP Code:


		
			
 $getuprofile = mysql_query("SELECT * from members-profile-info where username = '$uname'");

PHP Code:


		
			
 $getuprofile = mysql_query("SELECT * from members_profile_info where username = '$uname'");

Old Pedant

12-28-2012 05:38 AM

But read what I said: There is a very very good chance that you should be making only *ONE* SQL query there, *NOT* two!!!

Without seeing more of the page, I can't be sure, but I would give 3 to 1 odds, sight unseen.

Old Pedant

12-28-2012 05:43 AM

Just for example, try something like:

Code:

$sql = "SELECT m.userame, p.fullname 

        FROM members AS m, members_profile_info AS p

        WHERE m.username = p.username

        AND m.username = '$uname'";

$getinfo = mysql_query( $sql ) or die( mysql_error );

$userinfo = mysql_fetch_array($getinfo); 

echo 'Welcome ' . $userinfo['username'] . '!'; 

echo '<br/>Fullname:' . $userinfo['fullname'];

...

If you need other fields from the two tables, LIST THEM EXPLICITLY in your SELECT statement. DO *NOT* USE select * if at all possible.

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