having problem getting php to connect

24/06/2007, 21h33

Warning: mysql_query(): supplied argument is not a valid MySQL-Link
resource in /home/smith/public_html/personal/ab_functions.php on line
30

line 30 is [mysql_query ($sql, $connect) or die(mysql_error());]

function addPerson ()
{
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
if( $_POST{'dobMonth'} and $_POST{'dobDay'})
$dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'.
$_POST{'dobYear'});
else
$dob = ' ';
$category = $_POST['category'];
$address1 = $_POST['address1'];
$address2 = $_POST['address2'];
$city = $_POST['city'];
$state = $_POST['state'];
$zipcode = $_POST['zipcode'];
$telephone = $_POST['telephone'];
$cellphone = $_POST['cellphone'];
$email1 = $_POST['email1'];
$email2 = $_POST['email2'];
$aimid = $_POST['aimid'];
$url = $_POST['url'];
$notes = $_POST['notes'];

$sql = "INSERT INTO $db_table4
(firstName, lastName, dob, category, address1, address2, city,
state, zipcode, telephone, cellphone, email1, email2, aimid, url,
notes)
VALUES
('$firstName', '$lastName', '$dob', '$category', '$address1',
'$address2', '$city', '$state', '$zipcode', '$telephone',
'$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')";

mysql_query ($sql, $connect) or die(mysql_error());

echo "added";
}

function is called on this page

HTML>
<HEAD>
<?PHP include("includes/headsectionp.txt")?>
</HEAD>

<BODY>
<?PHP include("includes/connect.txt");?>
<?PHP include("ab_functions.php");?>
<?PHP
if(isset($_POST['do_addperson']))
{
addPerson ();
}
?>

24/06/2007, 21h34

my connect page is

<?PHP
$db_server = "localhost";
$db_user = " ";
$db_pass = " ";
$db_name = "smith_website";
$db_table = "blog_category";
$db_table1 = "blog_entry";
$db_table2 = "quotes";
$db_table3 = "ab_category";
$db_table4 = "ab_person";

$connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I
can not connect to server because: ' . mysql_error());

mysql_select_db($db_name, $connect) OR DIE ('I can not connect to
database because: ' . mysql_error());
?>

24/06/2007, 22h09

> function addPerson ()
> {
> $firstName = $_POST['firstName'];
> $lastName = $_POST['lastName'];
> if( $_POST{'dobMonth'} and $_POST{'dobDay'})
> $dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'.
> $_POST{'dobYear'});
> else
> $dob = ' ';
> $category = $_POST['category'];
> $address1 = $_POST['address1'];
> $address2 = $_POST['address2'];
> $city = $_POST['city'];
> $state = $_POST['state'];
> $zipcode = $_POST['zipcode'];
> $telephone = $_POST['telephone'];
> $cellphone = $_POST['cellphone'];
> $email1 = $_POST['email1'];
> $email2 = $_POST['email2'];
> $aimid = $_POST['aimid'];
> $url = $_POST['url'];
> $notes = $_POST['notes'];
>
> $sql = "INSERT INTO $db_table4
> (firstName, lastName, dob, category, address1, address2, city,
> state, zipcode, telephone, cellphone, email1, email2, aimid, url,
> notes)
> VALUES
> ('$firstName', '$lastName', '$dob', '$category', '$address1',
> '$address2', '$city', '$state', '$zipcode', '$telephone',
> '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')";
>
> mysql_query ($sql, $connect) or die(mysql_error());
>
> echo "added";
> }
>
> function is called on this page
>
> HTML>
> <HEAD>
> <?PHP include("includes/headsectionp.txt")?>
> </HEAD>
>
> <BODY>
> <?PHP include("includes/connect.txt");?>
> <?PHP include("ab_functions.php");?>
> <?PHP
> if(isset($_POST['do_addperson']))
> {
> addPerson ();
> }
> ?>

In your function you are using the variable $connect. This variable is not
in the scope of the function.

You have 2 options you can do this:-

function addPerson ()
{
global $connect

I you change the start of your function to this then you will be able to use
the $connect variable within your function.

The better option however is to do this:-

function addPerson ($connect)
{

and when calling the function do addPerson($connect), this makes the
$connect variable a parameter of the function. The reason the $_POST array
works without doing this is because they are already global.
http://www.php.net/global

24/06/2007, 22h22

New problem now
I made change and got:

Warning: Missing argument 1 for addperson() in /home/smith/public_html/
personal/ab_functions.php on line 20
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near '(firstName, lastName, dob, category, address1, address2, city,
state, zipcode, t' at line 2

I'm sure insert code works because if I just run it from page with
form it is ok.

This is the new add function

function addPerson ($connect)
{
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
if( $_POST{'dobMonth'} and $_POST{'dobDay'})
$dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'.
$_POST{'dobYear'});
else
$dob = ' ';
$category = $_POST['category'];
$address1 = $_POST['address1'];
$address2 = $_POST['address2'];
$city = $_POST['city'];
$state = $_POST['state'];
$zipcode = $_POST['zipcode'];
$telephone = $_POST['telephone'];
$cellphone = $_POST['cellphone'];
$email1 = $_POST['email1'];
$email2 = $_POST['email2'];
$aimid = $_POST['aimid'];
$url = $_POST['url'];
$notes = $_POST['notes'];

$sql = "INSERT INTO $db_table4
(firstName, lastName, dob, category, address1, address2, city,
state, zipcode, telephone, cellphone, email1, email2, aimid, url,
notes)
VALUES
('$firstName', '$lastName', '$dob', '$category', '$address1',
'$address2', '$city', '$state', '$zipcode', '$telephone',
'$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')";

mysql_query ($sql) or die(mysql_error());

echo "$db_name";
echo "added";
}

24/06/2007, 22h41

> New problem now
> I made change and got:
>
> Warning: Missing argument 1 for addperson() in /home/smith/public_html/
> personal/ab_functions.php on line 20
> You have an error in your SQL syntax; check the manual that
> corresponds to your MySQL server version for the right syntax to use
> near '(firstName, lastName, dob, category, address1, address2, city,
> state, zipcode, t' at line 2
>
> I'm sure insert code works because if I just run it from page with
> form it is ok.

Did you change the following code as instructed:-

if(isset($_POST['do_addperson']))
{
addPerson ();
}

to the following:-

if(isset($_POST['do_addperson']))
{
addPerson ($connect);
}

The error message you posted suggests you did not.

24/06/2007, 22h42

> $sql = "INSERT INTO $db_table4
> (firstName, lastName, dob, category, address1, address2, city,
> state, zipcode, telephone, cellphone, email1, email2, aimid, url,
> notes)
> VALUES
> ('$firstName', '$lastName', '$dob', '$category', '$address1',
> '$address2', '$city', '$state', '$zipcode', '$telephone',
> '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')";

Also just noticed I presume that line is meant to actually read:-

$sql = "INSERT INTO $db_table

you have an eroneous 4 at the end of the table name

24/06/2007, 22h46

Sorry, I didn't add it in original file. After adding it, i got:

You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near '(firstName, lastName, dob, category, address1, address2, city,
state, zipcode, t' at line 2

function addPerson ($connect)
{
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
if( $_POST{'dobMonth'} and $_POST{'dobDay'})
$dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'.
$_POST{'dobYear'});
else
$dob = ' ';
$category = $_POST['category'];
$address1 = $_POST['address1'];
$address2 = $_POST['address2'];
$city = $_POST['city'];
$state = $_POST['state'];
$zipcode = $_POST['zipcode'];
$telephone = $_POST['telephone'];
$cellphone = $_POST['cellphone'];
$email1 = $_POST['email1'];
$email2 = $_POST['email2'];
$aimid = $_POST['aimid'];
$url = $_POST['url'];
$notes = $_POST['notes'];

$sql = "INSERT INTO $db_table4
(firstName, lastName, dob, category, address1, address2, city,
state, zipcode, telephone, cellphone, email1, email2, aimid, url,
notes)
VALUES
('$firstName', '$lastName', '$dob', '$category', '$address1',
'$address2', '$city', '$state', '$zipcode', '$telephone',
'$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')";

mysql_query ($sql) or die(mysql_error());

echo "$db_name";
echo "added";
}

<?PHP include("includes/connect.txt");?>
<?PHP include("ab_functions.php");?>
<?PHP
if(isset($_POST['do_addperson']))
{
addPerson ($connect);
}
?>

24/06/2007, 22h48

On Jun 24, 5:42 pm, "peter" <sub...@flexiwebhost.com> wrote:
> > $sql = "INSERT INTO $db_table4
> > (firstName, lastName, dob, category, address1, address2, city,
> > state, zipcode, telephone, cellphone, email1, email2, aimid, url,
> > notes)
> > VALUES
> > ('$firstName', '$lastName', '$dob', '$category', '$address1',
> > '$address2', '$city', '$state', '$zipcode', '$telephone',
> > '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')";
>
> Also just noticed I presume that line is meant to actually read:-
>
> $sql = "INSERT INTO $db_table
>
> you have an eroneous 4 at the end of the table name

$db_server = "localhost";
$db_user = "smith_admin";
$db_pass = "password";
$db_name = "smith_website";
$db_table = "blog_category";
$db_table1 = "blog_entry";
$db_table2 = "quotes";
$db_table3 = "ab_category";
$db_table4 = "ab_person";

$connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I
can not connect to server because: ' .mysql_errno(). " -
" .mysql_error());
mysql_select_db($db_name, $connect) OR DIE ('I can not connect to
server because: ' .mysql_errno(). " - " .mysql_error());

24/06/2007, 23h46

25/06/2007, 00h02

> $db_server = "localhost";
> $db_user = "smith_admin";
> $db_pass = "password";
> $db_name = "smith_website";
> $db_table = "blog_category";
> $db_table1 = "blog_entry";
> $db_table2 = "quotes";
> $db_table3 = "ab_category";
> $db_table4 = "ab_person";
>
> $connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I
> can not connect to server because: ' .mysql_errno(). " -
> " .mysql_error());
> mysql_select_db($db_name, $connect) OR DIE ('I can not connect to
> server because: ' .mysql_errno(). " - " .mysql_error());

Ahh yes my fault never noticed that.

The problem you have is similar to the $connect problem, the tables being
defined outside of the function, you will need to also pass them as a
parameter

25/06/2007, 00h07

On Jun 24, 7:02 pm, "peter" <sub...@flexiwebhost.com> wrote:
> > $db_server = "localhost";
> > $db_user = "smith_admin";
> > $db_pass = "password";
> > $db_name = "smith_website";
> > $db_table = "blog_category";
> > $db_table1 = "blog_entry";
> > $db_table2 = "quotes";
> > $db_table3 = "ab_category";
> > $db_table4 = "ab_person";
>
> > $connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I
> > can not connect to server because: ' .mysql_errno(). " -
> > " .mysql_error());
> > mysql_select_db($db_name, $connect) OR DIE ('I can not connect to
> > server because: ' .mysql_errno(). " - " .mysql_error());
>
> Ahh yes my fault never noticed that.
>
> The problem you have is similar to the $connect problem, the tables being
> defined outside of the function, you will need to also pass them as a
> parameter

my gratitude forever that fixed it

24/06/2007, 21h33	#1
gagal Messages: n/a Hébergeur:	having problem getting php to connect Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/smith/public_html/personal/ab_functions.php on line 30 line 30 is [mysql_query ($sql, $connect) or die(mysql_error());] function addPerson () { $firstName = $_POST['firstName']; $lastName = $_POST['lastName']; if( $_POST{'dobMonth'} and $_POST{'dobDay'}) $dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'. $_POST{'dobYear'}); else $dob = ' '; $category = $_POST['category']; $address1 = $_POST['address1']; $address2 = $_POST['address2']; $city = $_POST['city']; $state = $_POST['state']; $zipcode = $_POST['zipcode']; $telephone = $_POST['telephone']; $cellphone = $_POST['cellphone']; $email1 = $_POST['email1']; $email2 = $_POST['email2']; $aimid = $_POST['aimid']; $url = $_POST['url']; $notes = $_POST['notes']; $sql = "INSERT INTO $db_table4 (firstName, lastName, dob, category, address1, address2, city, state, zipcode, telephone, cellphone, email1, email2, aimid, url, notes) VALUES ('$firstName', '$lastName', '$dob', '$category', '$address1', '$address2', '$city', '$state', '$zipcode', '$telephone', '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')"; mysql_query ($sql, $connect) or die(mysql_error()); echo "added"; } function is called on this page HTML> <HEAD> <?PHP include("includes/headsectionp.txt")?> </HEAD> <BODY> <?PHP include("includes/connect.txt");?> <?PHP include("ab_functions.php");?> <?PHP if(isset($_POST['do_addperson'])) { addPerson (); } ?>

24/06/2007, 21h34	#2
gagal Messages: n/a Hébergeur:	Re: having problem getting php to connect my connect page is <?PHP $db_server = "localhost"; $db_user = " "; $db_pass = " "; $db_name = "smith_website"; $db_table = "blog_category"; $db_table1 = "blog_entry"; $db_table2 = "quotes"; $db_table3 = "ab_category"; $db_table4 = "ab_person"; $connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I can not connect to server because: ' . mysql_error()); mysql_select_db($db_name, $connect) OR DIE ('I can not connect to database because: ' . mysql_error()); ?>

24/06/2007, 22h09	#3
peter Messages: n/a Hébergeur:	Re: having problem getting php to connect > function addPerson () > { > $firstName = $_POST['firstName']; > $lastName = $_POST['lastName']; > if( $_POST{'dobMonth'} and $_POST{'dobDay'}) > $dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'. > $_POST{'dobYear'}); > else > $dob = ' '; > $category = $_POST['category']; > $address1 = $_POST['address1']; > $address2 = $_POST['address2']; > $city = $_POST['city']; > $state = $_POST['state']; > $zipcode = $_POST['zipcode']; > $telephone = $_POST['telephone']; > $cellphone = $_POST['cellphone']; > $email1 = $_POST['email1']; > $email2 = $_POST['email2']; > $aimid = $_POST['aimid']; > $url = $_POST['url']; > $notes = $_POST['notes']; > > $sql = "INSERT INTO $db_table4 > (firstName, lastName, dob, category, address1, address2, city, > state, zipcode, telephone, cellphone, email1, email2, aimid, url, > notes) > VALUES > ('$firstName', '$lastName', '$dob', '$category', '$address1', > '$address2', '$city', '$state', '$zipcode', '$telephone', > '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')"; > > mysql_query ($sql, $connect) or die(mysql_error()); > > echo "added"; > } > > function is called on this page > > HTML> > <HEAD> > <?PHP include("includes/headsectionp.txt")?> > </HEAD> > > <BODY> > <?PHP include("includes/connect.txt");?> > <?PHP include("ab_functions.php");?> > <?PHP > if(isset($_POST['do_addperson'])) > { > addPerson (); > } > ?> In your function you are using the variable $connect. This variable is not in the scope of the function. You have 2 options you can do this:- function addPerson () { global $connect I you change the start of your function to this then you will be able to use the $connect variable within your function. The better option however is to do this:- function addPerson ($connect) { and when calling the function do addPerson($connect), this makes the $connect variable a parameter of the function. The reason the $_POST array works without doing this is because they are already global. http://www.php.net/global

24/06/2007, 22h22	#4
gagal Messages: n/a Hébergeur:	Re: having problem getting php to connect New problem now I made change and got: Warning: Missing argument 1 for addperson() in /home/smith/public_html/ personal/ab_functions.php on line 20 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(firstName, lastName, dob, category, address1, address2, city, state, zipcode, t' at line 2 I'm sure insert code works because if I just run it from page with form it is ok. This is the new add function function addPerson ($connect) { $firstName = $_POST['firstName']; $lastName = $_POST['lastName']; if( $_POST{'dobMonth'} and $_POST{'dobDay'}) $dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'. $_POST{'dobYear'}); else $dob = ' '; $category = $_POST['category']; $address1 = $_POST['address1']; $address2 = $_POST['address2']; $city = $_POST['city']; $state = $_POST['state']; $zipcode = $_POST['zipcode']; $telephone = $_POST['telephone']; $cellphone = $_POST['cellphone']; $email1 = $_POST['email1']; $email2 = $_POST['email2']; $aimid = $_POST['aimid']; $url = $_POST['url']; $notes = $_POST['notes']; $sql = "INSERT INTO $db_table4 (firstName, lastName, dob, category, address1, address2, city, state, zipcode, telephone, cellphone, email1, email2, aimid, url, notes) VALUES ('$firstName', '$lastName', '$dob', '$category', '$address1', '$address2', '$city', '$state', '$zipcode', '$telephone', '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')"; mysql_query ($sql) or die(mysql_error()); echo "$db_name"; echo "added"; }

24/06/2007, 22h41	#5
peter Messages: n/a Hébergeur:	Re: having problem getting php to connect > New problem now > I made change and got: > > Warning: Missing argument 1 for addperson() in /home/smith/public_html/ > personal/ab_functions.php on line 20 > You have an error in your SQL syntax; check the manual that > corresponds to your MySQL server version for the right syntax to use > near '(firstName, lastName, dob, category, address1, address2, city, > state, zipcode, t' at line 2 > > I'm sure insert code works because if I just run it from page with > form it is ok. Did you change the following code as instructed:- if(isset($_POST['do_addperson'])) { addPerson (); } to the following:- if(isset($_POST['do_addperson'])) { addPerson ($connect); } The error message you posted suggests you did not.

24/06/2007, 22h42	#6
peter Messages: n/a Hébergeur:	Re: having problem getting php to connect > $sql = "INSERT INTO $db_table4 > (firstName, lastName, dob, category, address1, address2, city, > state, zipcode, telephone, cellphone, email1, email2, aimid, url, > notes) > VALUES > ('$firstName', '$lastName', '$dob', '$category', '$address1', > '$address2', '$city', '$state', '$zipcode', '$telephone', > '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')"; Also just noticed I presume that line is meant to actually read:- $sql = "INSERT INTO $db_table you have an eroneous 4 at the end of the table name

24/06/2007, 22h46	#7
gagal Messages: n/a Hébergeur:	Re: having problem getting php to connect Sorry, I didn't add it in original file. After adding it, i got: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(firstName, lastName, dob, category, address1, address2, city, state, zipcode, t' at line 2 function addPerson ($connect) { $firstName = $_POST['firstName']; $lastName = $_POST['lastName']; if( $_POST{'dobMonth'} and $_POST{'dobDay'}) $dob = ($_POST{'dobMonth'}.'/'.$_POST{'dobDay'}.'/'. $_POST{'dobYear'}); else $dob = ' '; $category = $_POST['category']; $address1 = $_POST['address1']; $address2 = $_POST['address2']; $city = $_POST['city']; $state = $_POST['state']; $zipcode = $_POST['zipcode']; $telephone = $_POST['telephone']; $cellphone = $_POST['cellphone']; $email1 = $_POST['email1']; $email2 = $_POST['email2']; $aimid = $_POST['aimid']; $url = $_POST['url']; $notes = $_POST['notes']; $sql = "INSERT INTO $db_table4 (firstName, lastName, dob, category, address1, address2, city, state, zipcode, telephone, cellphone, email1, email2, aimid, url, notes) VALUES ('$firstName', '$lastName', '$dob', '$category', '$address1', '$address2', '$city', '$state', '$zipcode', '$telephone', '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')"; mysql_query ($sql) or die(mysql_error()); echo "$db_name"; echo "added"; } <?PHP include("includes/connect.txt");?> <?PHP include("ab_functions.php");?> <?PHP if(isset($_POST['do_addperson'])) { addPerson ($connect); } ?>

24/06/2007, 22h48	#8
gagal Messages: n/a Hébergeur:	Re: having problem getting php to connect On Jun 24, 5:42 pm, "peter" <sub...@flexiwebhost.com> wrote: > > $sql = "INSERT INTO $db_table4 > > (firstName, lastName, dob, category, address1, address2, city, > > state, zipcode, telephone, cellphone, email1, email2, aimid, url, > > notes) > > VALUES > > ('$firstName', '$lastName', '$dob', '$category', '$address1', > > '$address2', '$city', '$state', '$zipcode', '$telephone', > > '$cellphone', '$email1', '$email2', '$aimid', '$url', '$notes')"; > > Also just noticed I presume that line is meant to actually read:- > > $sql = "INSERT INTO $db_table > > you have an eroneous 4 at the end of the table name $db_server = "localhost"; $db_user = "smith_admin"; $db_pass = "password"; $db_name = "smith_website"; $db_table = "blog_category"; $db_table1 = "blog_entry"; $db_table2 = "quotes"; $db_table3 = "ab_category"; $db_table4 = "ab_person"; $connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I can not connect to server because: ' .mysql_errno(). " - " .mysql_error()); mysql_select_db($db_name, $connect) OR DIE ('I can not connect to server because: ' .mysql_errno(). " - " .mysql_error());

24/06/2007, 23h46	#9
gagal Messages: n/a Hébergeur:	Re: having problem getting php to connect

25/06/2007, 00h02	#10
peter Messages: n/a Hébergeur:	Re: having problem getting php to connect > $db_server = "localhost"; > $db_user = "smith_admin"; > $db_pass = "password"; > $db_name = "smith_website"; > $db_table = "blog_category"; > $db_table1 = "blog_entry"; > $db_table2 = "quotes"; > $db_table3 = "ab_category"; > $db_table4 = "ab_person"; > > $connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I > can not connect to server because: ' .mysql_errno(). " - > " .mysql_error()); > mysql_select_db($db_name, $connect) OR DIE ('I can not connect to > server because: ' .mysql_errno(). " - " .mysql_error()); Ahh yes my fault never noticed that. The problem you have is similar to the $connect problem, the tables being defined outside of the function, you will need to also pass them as a parameter

25/06/2007, 00h07	#11
gagal Messages: n/a Hébergeur:	Re: having problem getting php to connect On Jun 24, 7:02 pm, "peter" <sub...@flexiwebhost.com> wrote: > > $db_server = "localhost"; > > $db_user = "smith_admin"; > > $db_pass = "password"; > > $db_name = "smith_website"; > > $db_table = "blog_category"; > > $db_table1 = "blog_entry"; > > $db_table2 = "quotes"; > > $db_table3 = "ab_category"; > > $db_table4 = "ab_person"; > > > $connect = mysql_connect($db_server, $db_user, $db_pass) OR DIE ('I > > can not connect to server because: ' .mysql_errno(). " - > > " .mysql_error()); > > mysql_select_db($db_name, $connect) OR DIE ('I can not connect to > > server because: ' .mysql_errno(). " - " .mysql_error()); > > Ahh yes my fault never noticed that. > > The problem you have is similar to the $connect problem, the tables being > defined outside of the function, you will need to also pass them as a > parameter my gratitude forever that fixed it

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