NYCPHP Meetup

[PaulCheung]

Hi Micheal,

Using a search criteria, a MySQL call is made and the required MySQL record 
is returned.
The HTML FORM is made up of three forms. the first two forms are used to 
display (read-only) infomation to the enduser
The third and last FORM is used by the enduser to update the MySQL record.

All of the above works as required. Here it falls over

before passing the info to the next script to do the actual update I try to 
ECHO the user entered data and it is empty.

Just for this test and to show  what is happening I passed the entered info 
to the next script and this is what was received by the next script as you 
see NULL values were passed. whereas the ACCESS and TESTNO from the MySQL 
search are passed and picked up by the next script using $_SESSION.


$query = UPDATE test_record SET data = '' record = '' note = '' WHERE access 
= '73226318' AND testno = '002'

Paul



----- Original Message ----- 
From: "Michael Southwell" <michael.southwell at nyphp.com>
To: "NYPHP Talk" <talk at lists.nyphp.org>
Sent: Sunday, November 18, 2007 9:02 PM
Subject: Re: [nycphp-talk] Passing info entered into HTML FORMS into 
SESSIONvariables.


> PaulCheung wrote:
>>        $t = $row['data']; $u = $row['result']; $v = $row['note'];
>
> How are you populating the $row array?  I would have expected this to be:
>     $t = $_POST['data'];
> Doing it this way you should have no problem. And by the way, you don't 
> need the $t etc variables unless you are using them elsewhere. That would 
> make it this:
>     $_SESSION['data'] = $_POST['data'];
>
> -- 
> =================
> Michael Southwell
> Vice President, Education
> NYPHP TRAINING:  http://nyphp.com/Training/Indepth
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