Re: RE : include interpreted php file

hi, thanks for your help again. i've noticed that file function have two different behaviors. i tested file function with the url and it seems that the file is interpreted. But i tried the code u give me and without the "http://myurl.com/" at the beginning, a warning is raised : Warning: file("test.php?foo=test") - Invalid argument in c:\program files\easyphp\www\un.php on line 10 otherwise, without the http string, it reads the file as text file and print exactly his content. You "Timo Stamm" <t.stamm@macnews.de> a écrit dans le message de news: 398A9046-BF8B-11D6-BB42-003065F6B29A@macnews.de... > @all: > > I just noticed that I was not replying to the list, but to the > posting email adresses. I just switched to another email client > and was not aware of this. My apologies :-( > > > --- > > > You, > > could it be that you want to do the following?: > > <?php > // primary php > > $foo = $_REQUEST["foo"] > // I guess you want to do something with $foo here. > $outputfromsecondaryphp = file("secondary.php?foo=".$foo) > echo implode('', $outputfromsecondaryphp); > ?> > > > Am Dienstag den, 3. September 2002, um 00:15, schrieb you: > > Thanks for ur help and i'm sorry to have a so bad english > > I guess my french is worse than your english :-) > > > Timo >