Hi Ralph,
Here is how I do it, which is how you suggested.
Regards,
Chris...
#--
CREATE TABLE `test` (
`id` int(11) NOT NULL auto_increment,
`browser` varchar(255) NOT NULL default '',
PRIMARY KEY (`id`)
);
INSERT INTO test VALUES (1,
looks like your include path and/or file don't exist. Double check that
colon before /usr - surely that's incorrect? The undefined function is
probably because the include file can't be found.
Dave
-Original Message-
From: Al Moote [mailto:[EMAIL PROTECTED]]
Sent: 02 August 2001 20:06
I have a simple script in php :
putenv("ORACLE_HOME=D:\\ORANT");
putenv("TNS_ADMIN=D:\\ORANT\\NETWORK\\ADMIN");
putenv("ORACLE_SID=x");
echo "Oracle : ". getenv("ORACLE_HOME")." \n";
echo "TNS_ADMIN : ". getenv("TNS_ADMIN")." \n";
echo "ORACLE_SID : ". getenv("ORACLE_SID")." \n";
$conn
JavaScript would allow them to immediately be sent to the page.
--
phill
"Webmaster" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I have a form and based off of this form if my user selects option a then
I
> want them to be redirected to another page. If the
Hi everyone !
I've installed Apache 1.3.20, MySQL 3.23.38 and PHP 4.0.6 with MySQL support but I
still have : "MySQL a répondu: Can't connect to local MySQL server through socket
'/var/lib/mysql/mysql.sock' (111) " when trying to connect to the database with
phpMyAdmin 2.2
Is there a way to ch
I've changed what I originally answered with, assuming option 1 has the
value 12 (as you describe) and option 2 has value 14...
then you can get that value with this...
(top of some_page_or_other.php)
$user_selected_value = getenv("QUERY_STRING");
(top of some_other_page_you_choose.p
I have this table :
service| netid |
+---+---+
| Email | netid1 |
| Print | netid2 |
| network | netid3 netid4 |
| Database | netid2 |
I am displayin
Hi there,
In the past 24 hours, my php pages for one site have gone from 2 seconds
loading time to nearly 2 minutes. What may have caused this degradation? I
have not modified any of my php code, so think it may be a problem on the
server, or with the database. However, not all my php database
Have you tried with NULL instead of '' the ContactID?
On Thu, 2 Aug 2001, Steve Fitzgerald wrote:
> I keep getting the following SQL error:
>
> You have an error in your SQL syntax near ')' at line 14
>
> Line 14 is: '$FirstName', so the error must correspond to a different part
> of the code.
I keep getting the following SQL error:
You have an error in your SQL syntax near ')' at line 14
Line 14 is: '$FirstName', so the error must correspond to a different part
of the code.
Any ideas?
$add_contact_sql = "INSERT INTO $table_name
(ContactID, FirstName, LastName, Title, WorkPhone,
Hom
"Steve Fitzgerald" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I keep getting the following SQL error:
>
The last fieldname is CompanyID not ContactID.
You have an error in your SQL syntax near ')' at line 14
>
> Line 14 is: '$FirstName', so the error must
Hi,
[...]
> Any ideas?
>
> $add_contact_sql = "INSERT INTO $table_name
> (ContactID, FirstName, LastName, Title, WorkPhone,
> HomePhone,Mobile,EmailName,Birthday,CompanyID)
> VALUES
> ('',
> '$FirstName',
> '$LastName',
> '$Title',
> '$WorkPhone',
> '$HomePhone',
> '$Mobile',
> '$EmailName',
> '$
Hi,
> I have a form and based off of this form if my user selects option a then
I
> want them to be redirected to another page. If they select option B then I
> want to continue with my php scrips as it does now.
>
> Does any one have any ideas on how I can accomplish this?
use the header() func
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