Calling another page ?

Anon

Hi,

I am developing a page with PHP + MySQL, which requires a lot of
information from FORMS.
As a newbie, I´ve the PHP manual, but couldn´t find what I need.
I would like to do the following : After the data received from a FORM is
processed and stored in my database by a PHP file, I want to call another page - a PHP file also, containing another form, but I don´t wanna have a "intermediary page", with a link or button to go to the next one, I want it to be automatic.

Can anyone help me ?

Priscilla.

thelad

you could have like this.
on pages one
<form action = "page2.php" method ="post">
when you submit the form on pages one,
it will go to page2.php.

This page can do what you wish if you do it like this.

at the top of the page have,

if($submit)

{
// process results into database

}

//then just put all your next form here.

The work is all being done, before the new page fully appears.

Hope it helps.

Anon

u can process ur form contents, then redirect the script:

Header("location: http://urdomain.com/urpage");

hope my recomendation is helpful...

Anon

Priscilla,

you can make it automatic without any problems...

for example:

You can use $PHP_SELF varialbe and store the form and the code to process the form in the same file.
This is simplified code:

Code to process the form

if($enter)
{
do this,
do that, and then
use <b>header("location: http://mysite.com/mysecondpage.php");</b>

you create the form

$form="
<form method=post action=$PHP_SELF>
form body..
<input type=submit value=Click <b>name=enter</b>>
</form>";

You display the form

echo "$form";

Of course you can sup it up, but conceptually it should work without any problem.

bansidhar

If you want to show the same form after the processing (Like some data entry work) you can easly do it in a single form also.

refer to the following. The file should be stored as mainpage.php, because i am using that name in the form action.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">

</form>

<?php

if($subm == "Add This News"){

$connection = mysql_connect($sqlserver,$sqluser,$sqlpassword) or die ("Unable to connect to MySQL server.");

/ your database operation here /
mysql_close($connection);
}
?>

</BODY>