Dear Aleks:
What I usually do for a situation like this is...
1) User fills out form and data is INSERTed into the table. If you use
auto_increment for the row id, use PHP's mysql_insert_id to get the id and
use this for step 2.
2) Page confirms data by SELECTing the data from the MySQL table
I have an if statement which includes two mail() functions. Curiously, the second one
works,
in that it results in e-mails being received. But the first doesn't - and I cannot for
the life of me
figure out why.
The code is below. It is part of the final sequence for registering for an on-line
Forget my post of a few moments ago - sudenly, it is working. Must be
gremlins in the server.
Sorry for the wasted bandwidth.
Jeffrey Baumgartner
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Date sent: Thu, 04 Dec 2003 20:34:26 +0100
Priorit
> I am looking to create a "comp-time" calculator of sorts using
> PHP/MySQL. I am still in the learning process of how all of this
> works. I want to be able to create a web interface where my employee
> can login and enter her comp-time as she works extra. I also want to be
> able to have an a
I am puzzled.
I have built a php script to upload a document to a directory and save the info
in an MySQL table.
I created a sample document in OpenOffice and saved it as a .doc file.
When I access the page via an Opera browser, I can upload the file without
problem.
When I access the page
If you have already have a number of scripts and, especially, if you are doing
things with the variables inside the script (in other words, the variables
appear more than once), it can be more convenient to convert at the top of
each page, eg:
$id = $_REQUEST['id'];
$this = $_REQUEST['this'];
$