[Resolved] what does this mean? and how to fix it.

[charles256]

I am in the beginning stages of understanding all this, this is what I get when I visit my page calling my data in mysql:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/xxxx/apache/xxxxxx/xxxx/httpdocs/xxxxng.php on line 15

[intenz]

Maybe the database doesn't exist?

[jayant]

There can be many reasons.

You didn't connect to MySQL server. You didnt select any database. The query you gave didnt return any results (0 rows selected), blah blah blah.

Though the chances of "The query you gave didnt return any results (0 rows selected)" are maximum.

[MrAlaska]

Couple of things you can try that can save you a ton of frustration in the long run:

Include something like this after each database hit:

or die ("<p>" . mysql_errno() . ": " . mysql_error() . "<p>");

(example)

$mysql_link = mysql_connect
	('localhost', 'root' , '')
	or die ("Connection failed!<p>".
	mysql_errno().": ".mysql_error());
mysql_select_db('test')
	or die ("Could not connect to db!<p>".
	mysql_errno().": ".mysql_error());
$result=mysql_query($sql)
	or die ("Sql prob:".$sql."<p>".
	mysql_errno().": ".mysql_error());

After the final one add a line like this, using your variable for the sql statement instead of $sql:

echo "<p>".$sql."<p>".
	mysql_errno() . ": " . mysql_error() 
	. "<p>";

[el_kab0ng]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/xxxx/apache/xxxxxx/xxxx/httpdocs/xxxxng.php on line 15

I have found this usually occurs AFTER the database has been queried, and the result set breaks. Not sure how to display more of an error than what you already have, except that the line number helps you to determine which result set puked. I'm dealing with one right now that has a similar confusing issue. I run the query manually against mysql and it runs fine, I use php to do so, and it puked.