Re: PHP/mysql equivalent of PEAR's tableInfo()??

[Govinda <govinda.webdnatalk@xxxxxxxxx>]

> I have this code:
>
>        $db_billing=mysqli_connect(localhost,metheuser,mypass,billing);
>        if (mysqli_connect_error()) { die("Can't connect: " .  
> mysqli_connect_error()); }
>
> mysqli
>
>
>        //$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:

$db_billing=mysql_connect(localhost,metheuser,mypass,billing);
if (!$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";
	}

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

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