trying to dump out table info
| From: | Govinda | Date: | Mon, 13 Jul 2009 20:53:16 +0000 |
| Subject: | trying to dump out table info | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-46290@lists.php.net to get a copy of this message | ||
Hi all
I am not sure why this is failing/giving error. I am quite new at PHP/Mysql, and apologize for what will undoubtedly end up being a trivial error on my part.
Here is my code:
$db_billing=mysql_connect(localhost,metheuser,mypass,billing);
if (!$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>".$k."</b>:<br />\n";
echo "<pre>\n";
print_r(mysql_fetch_assoc($rresult));
echo "</pre>\n";
echo "<br />\n";
}
which is 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:
-mysql_fetch_assoc
-mysql_query
-SHOW TABLES
but do not see why this should be failing. Why isn't $result a ' valid MySQL result resource'?
-G
------------
Govinda
govinda.webdnatalk@gmail.com