I have this code:

if (mysqli_connect_error()) { die("Can't connect: " . mysqli_connect_error()); }


       //$dbname = 'billing';
       $sql = "SHOW TABLES";

       $result = mysql_query($sql); // line 53

Now mysql, What are you doing?

Yes. 3 lashing. Thanks. I am not likely to neglect again remembering that mysql and mysqli are different and have different syntax.

unfortunately I am still in over my head enough to have to ask..

Here is what I have now:

if (!$db_billing) { die('Could not connect: ' . mysql_error()); }

        $sql = "SHOW TABLES";
        $result = mysql_query($sql);
        foreach(mysql_fetch_assoc($result) as $k => $v) { //line 62
        $ssql = "DESCRIBE ".mysql_real_escape_string($v);
        $rresult = mysql_query($ssql);
        echo "<b>".$k."</b>:<br />\n";
        echo "<pre>\n";
        echo "</pre>\n";
        echo "<br />\n";

giving this error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/meee/public_html/somedir/test.php on line 62

I read about:

but do not see why this should be failing. Why isn't $result a ' valid MySQL result resource'?


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