Best way to send data to an include.

[RoninStretch]

Hey again all..

 

Different question here.

 

I'm using a switch with GET to determine which page to load in the main part of my site.

 

so for example.

main.php?action=WhateverPage

 

would include WhateverPage.php into my main.php design.

 

 

What i'm wondering is going from one included page to another. What's the best way to transfer data?

So on the first page, let's say main.php?action=one  we have a link to main.php?action=two

 

but we need to send $id along too for using in the second page.

 

I tried

main.php?action=two&id=2

and then trying to GET it on the second page but it doesn't work.

 

so how is this done?

[Jessica]

That will work.

Did you do

$id = $_GET['id'];

on your second page?

[RoninStretch]

yeah.. it will?

 

Ok i was getting quite confused.. If you say it will work i'll go back and try again.

 

...

 

Oh and thanks!!

[RoninStretch]

I can't seem to figure it out..

 

I keep getting this error

 

Warning: main(admin_News_del.php?id=1) [function.main]: failed to open stream: No such file or directory in /home/l/a/lanlounge_org/experimental/index.php on line 130

Warning: main() [function.include]: Failed opening 'admin_News_del.php?id=1' for inclusion (include_path='.:/usr/local/lib/php/') in /home/l/a/lanlounge_org/experimental/index.php on line 130"

 

Here's my code..

 

index.php

case "adminNewsDel":
include 'admin_News_del.php;
break;

 

main_news.php - First page

<a href='index.php?action=adminNewsDel&id=" . $storyID . "'>Delete</a>

 

 

admin_News_del.php - Second page.

$storyID = $_GET['id'];

$query = "DELETE FROM fed_news WHERE id='$storyID'";
$result = mysql_query($query);

echo ("News item " . $storyID . " deleted.");

 

 

 

the link on main_news.php works fine.. it is linking to

index.php?action=adminNewsDel&id=3

 

..

 

Lost!  ???