This codes third line is give an error message on all pages of my web site.

<?
$sql = "SELECT * FROM tb_config WHERE item='click' and howmany='1'";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
?>


The error message says the following

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/moneybux/public_html/index.php on line (number of line on page)

Can someone help me out to resolve this issue please.

Just checking.. is howmany an actual column in your table or is that how many results you want? LIMIT is the mysql syntax for specifying the maximum number of results.

PHP Code:

$sql = 'SELECT * FROM `tb_config` WHERE item = \'click\' LIMIT 1;'; 


I don't see 'howmany' in the table anywhere.

Well as the error is telling you. There is something wrong with you sql query. If there is no howmany column in your table then that's probably it. Try the query I gave you in place of the one you are using.

Try to
$sql = "SELECT * FROM `tb_config` WHERE `item`='click' and howmany='1'";

If howmany is an int type it probably doesn't need to be in inverted commas

i.e.
SELECT * FROM tb_config WHERE item='click' and howmany=1

Originally Posted by Monkey Do

If howmany is an int type it probably doesn't need to be in inverted commas

i.e.
SELECT * FROM tb_config WHERE item='click' and howmany=1

Exactly!

PHP Code:

$sql= "SELECT * FROM tb_config WHERE item='click' and howmany=1"
...