[snip]
> Sorry my question... but why I must use JOIN if I can use SELECT * FROM
> table1 AND table2 AND table3 WHERE table1.id=table2.id=table3 not?
> thanks
>
> + _
[/snip]
select table1.field1, table1.field2, table2.field
I've had no success trying to execute any stored procedure from php.
I'm just trying to follow the example for multi-query in the php manual,
and have tried an example from Artfulsoftware.com. I have one input and
one output parameter. I've simplified the stored procedure to always
output one v
On Mon, Jul 13, 2009 at 11:52 AM, Patrick Moloney wrote:
> I've had no success trying to execute any stored procedure from php.
> I'm just trying to follow the example for multi-query in the php manual, and
> have tried an example from Artfulsoftware.com. I have one input and one
> output parameter
Bastien Koert wrote:
On Mon, Jul 13, 2009 at 11:52 AM, Patrick Moloney wrote:
I've had no success trying to execute any stored procedure from php.
I'm just trying to follow the example for multi-query in the php manual, and
have tried an example from Artfulsoftware.com. I have one input and one
On Mon, Jul 13, 2009 at 1:33 PM, Patrick Moloney wrote:
> Bastien Koert wrote:
>>
>> On Mon, Jul 13, 2009 at 11:52 AM, Patrick Moloney
>> wrote:
>>>
>>> I've had no success trying to execute any stored procedure from php.
>>> I'm just trying to follow the example for multi-query in the php manual,
Bastien Koert wrote:
On Mon, Jul 13, 2009 at 1:33 PM, Patrick Moloney wrote:
Bastien Koert wrote:
On Mon, Jul 13, 2009 at 11:52 AM, Patrick Moloney
wrote:
I've had no success trying to execute any stored procedure from php.
I'm just trying to follow the example for multi-query in the php manua
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: '
On Mon, Jul 13, 2009 at 16:53, Govinda wrote:
> 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.
Hey, I recognize that code ;-P
> Here is my code:
>
> $db_billing=m
On Jul 13, 2009, at 2:58 PM, Daniel Brown wrote:
On Mon, Jul 13, 2009 at 16:53, Govinda
wrote:
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.
Hey, I recognize that c
On Mon, Jul 13, 2009 at 17:24, Govinda wrote:
>
> "No database selected"
You're getting warmer
How would you select a MySQL database in PHP? (Hint: there's a
built-in function.)
--
[email protected] || [email protected]
http://www.parasane.net/ || http://www.pilotpig.net/
C
On Jul 13, 2009, at 4:40 PM, Daniel Brown wrote:
On Mon, Jul 13, 2009 at 17:24, Govinda
wrote:
"No database selected"
You're getting warmer
How would you select a MySQL database in PHP? (Hint: there's a
built-in function.)
To try and follow your lead and focus just on what yo
On Mon, Jul 13, 2009 at 19:28, Govinda 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 TAB
this:
$q=mysqli_query($db_billing,'INSERT INTO billing (clientFname,
clientLname) VALUES (?,?)', array($defaultsFormVars[clientFname],
$defaultsFormVars[clientLname]));
is giving
Warning: mysqli::query() expects parameter 2 to be long, array given
in /home/meee/public_html/somedir/test.ph
On Jul 13, 2009, at 5:42 PM, Daniel Brown wrote:
On Mon, Jul 13, 2009 at 19:28, Govinda
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
On Mon, Jul 13, 2009 at 20:17, Govinda wrote:
>
> Dan, I am really making the effort.. despite how it may look.
I believe you. I sent you the function and code that should work
for you, but it may have been lost in the chopped replies. So it'll
be something like this:
--
daniel.br...@p
> this:
>
> $q=mysqli_query($db_billing,'INSERT INTO billing (clientFname,
> clientLname) VALUES (?,?)', array($defaultsFormVars[clientFname],
> $defaultsFormVars[clientLname]));
>
> is giving
> Warning: mysqli::query() expects parameter 2 to be long, array given
> in /home/meee/public_html
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