Weird problem...

kdoggy

Hi,

I am getting an error:

Warning: Supplied argument is not a valid MySQL result resource in /home/sites/site108/web/shopping/view.php on line 28

When the following code is accessed:

###############################
	# Connect to the MySQL server #

$connection = @mysql_connect("217.158.42.18", "####", "####");

if (!$connection) {
	echo "Fatal Error!<br>Connection Failed!<br><br>";
	echo mysql_error();
	exit();
}

#####################
# Test the database #

if (!@mysql_select_db("amazing")) {
	echo "Fatal Error!<br>Database Selection Failed!<br><br>";
	echo mysql_error();
	exit();
}

###################
# The MySQL query # 

$myquery = @mysql_query("SELECT * FROM 'fb-vendors' WHERE id = '$id'");

$data = mysql_fetch_array($myquery);

(the line it is complaining about is the last line above)

Now this is strange, because I have the exact same code on two different web sites that are hosted on the same machine and they are not causing errors.

Any suggestions?

Thanks!

nnichols

$sql = "SELECT * FROM fb-vendors WHERE id = '$id'";
$myquery = @mysql_query($sql)
   or die('MySQL error: ' . mysql_error() . '<br>Query: ' . $sql);

Run it like this and it will tell you what's wrong.

kevinsequeira

is the column id in your table fb-vendors is an integer you may want to try this

$myquery = @("SELECT * FROM fb-vendors WHERE id = $id");

reg
kevin

kdoggy

Thanks guys!

I figured out the problem, for some reason my script does not like the table name with a "-" in it. I renamed the table to fb_vendors and it works fine with no modifications to the code.

Weird :rolleyes:

Thanks again!