involve other php-file by condition

[fluidsharp]

Hello.

How I can involve immediately for execute other php-file from main php-file by condition.

I've tried to use require() and include() but they just share variable.

And looked at forum but didn't find solution, or didn't understand ).

Thanks in advance!

[Deoctor]

the php files which u are including will do any action on the variables that u are having from the main.php file

[ChemicalBliss]

Also, you need to state what "conditions" you require beofre including the file, give us some info on what you are trying to do .

 

-CB-

[fluidsharp]

I've found way, i've changed rules and involve those files from form and use POST, and conditions (if-else) is in involved file.

sorry for distraction but yours reply set thinking me.

Thanks.

 

ps. go to next step... 

[ChemicalBliss]

I'm glad you found your solution .

 

Just a note on security;

 

Never do this:

<?php

include($_POST['filename']);

?>

 

Instead do this:

<?php

Switch($_POST['filename']){
   default:
      include('index.php');
   break;
   
   case 'file1':
      Include('path/to/file1.php');
   break;
}
?>

 

I'm sure you will understand why

 

-CB-