Re: calling php from perl

On Mon, 18 Dec 2000, peter beardsley wrote: > Hi, > What I'm trying to accomplish is including the output of a php script in a > perl CGI. The output of a perl/mod_perl CGI is not parsed by php, so I > tried calling the php executable and printing the output, ala: It probably won't help you much, but the situation you've described here works fine for me, using an admittedly dated 4.0.1pl2 cgi binary. > ## note: backticks in perl execute a command and returns stdout. Yes, this is true in PHP as well. > print /usr/local/bin/php -f /home/httpd/cgi-bin/test.php; > > The contents of the test.php file is a simple: > > <?php print "hello world.\n" ?> > > What happens is PHP is smart enough to know that it's being called in a CGI > context, and I get the following output in a browser: When I run php -f invalidfilename, I don't get the X-Powered-By line. > X-Powered-By: PHP/4.0.3pl1 Content-type: text/html No input file specified. > > If I run the perl script from the command line, it works fine. It's weird > because I'm definitely specifying an input file with the "-f". I've tried > many variations on this in perl including system(), various pipe and temp > file hacks to no avail. I've poked around in the sources a bit and I think > what's happening is that php is looking for a CGI parameter to specify an > input file, and it's not there. > > Does anyone know how to work around this or an entirely different approach > altogether? Well, one option would be to put #!/usr/local/bin/php -q at the top of your php script and chmod u+x it. Then from the perl side, $output = /path/to/php/script.php; Matt