On Jul 13, 2009, at 5:42 PM, Daniel Brown wrote:
On Mon, Jul 13, 2009 at 19:28, Govinda<govinda.webdnatalk@xxxxxxxxx>
wrote:
$db_billing=mysql_connect(localhost,metheuser,mypass,billing);
if (!$db_billing) { die('Could not connect: ' .
mysql_error()); }
if(!mysql_select_db('dbname',$db_billing))
die(mysql_error());
//$sql = "SHOW TABLES FROM billing LIKE 'mytable'";
$sql = "SHOW TABLES"; //line 237
The fourth parameter you have in mysql_connect() only evaluates
within the engine as a boolean True statement. RTFM to see why. ;-P
Dan, I am really making the effort.. despite how it may look.
I read about mysql_connect again and see that 4th param is optional.
seemingly intended for when we want to force a new link and not just
use one that was already opened. Well I am just at this point trying
to use a first link, so I would assume I need to say what db I am
trying to talk to. Too bad the docs do not give a single example
where the 4th param is being used. Anyway I see in the function def.
that that 4th is a boolean. That does not make sense to me since I
think it should be a string (the name of the database). Anyway it was
working with PEAR DB with this:
$db_billing=DB::connect('mysql://metheuser:mypass@localhost/billing');
and still does seem to be connecting ok with this now (without pear db)-
$db_billing=mysql_connect(localhost,metheuser,mypass,billing);
...no error arises from the next line-
if (!$db_billing) { die('Could not connect: ' . mysql_error()); }
...so we must be talking to the db, right?
And to my real question, how can I get mysql_query what it needs in
this line:
$result = mysql_query($sql) or die(mysql_error());
I learn the best by *seeing* example code. I wish I could see a
working example.
-G
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
