Re[2]: [PHP] session_start() || shell access problem......

From: Date: Sun, 21 Sep 2003 13:41:30 +0000
Subject: Re[2]: [PHP] session_start() || shell access problem......
References: 1 2 3 4 5  Groups: php.general 
Request: Send a blank email to php-general+get-163347@lists.php.net to get a copy of this message
Hi, Monday, September 22, 2003, 2:13:57 AM, you wrote: CH> Hmmmm, CH> Well, is there a way to pass params to file_to_be_executed in command line? CH> For example: CH> <? CH> $my_param = 'my_include_path'; CH> $text = usr/local/bin/php /path/to/my/php/page.php; ?>> CH> Somehow I need $my_param to be passed to page.php (the file to be processed CH> in command line). CH> Any ideas? CH> --Noah CH> "Marek Kilimajer" <kilimajer@webglobe.sk> wrote in message CH> news:3F6D7638.6050400@webglobe.sk... >> Sessions don't work on command line and that is how you run your script >> - from command line. >> >> CF High wrote: >> > Hey Robert. >> > >> > Indeed, hard to find the problem. >> > >> > I don't believe it's a whitespace issue, or even a "Headers sent >> > issue", >> > despite the fact that I'm receiving that error. >> > >> > Check it out: >> > >> > test.php contains just one line: <?$text = `usr/local/bin/php >> > /path/to/my/php/test1.php`;?> >> > >> > test1.php, the file to be executed, contains just one line: >> > <?session_start();?> >> > >> > There are no line breaks, spaces, etc. >> > >> > Still get "Headers already sent". >> > >> > Pretty strange, right? >> > >> > Correct me if I'm wrong, but I think the problem may be related to the CH> fact >> > that when files are executed from the command line, php now looks for >> > include_paths, session_paths, etc. relative to the server root; not the CH> site >> > root. >> > >> > I can think of no other reason why include paths, starting sessions, and CH> so >> > on, return errors from the command line but ork perfectly fine when run CH> in a >> > browser..... >> > >> > Feel free to clue me in -- I know didly about shell access issues. >> > >> > --Noah >> > >> > >> > "Robert Cummings" <robert@interjinn.com> wrote in message >> > news:1064091763.19204.1.camel@blobule.suds... >> > >> >>On Sat, 2003-09-20 at 15:46, CF High wrote: >> >> >> >>>Hey all. >> >>> >> >>>I'm running a script from the command-line php interpreter as follows: >> >>>(thanks to D. Souza for lead) >> >>> >> >>>$text = usr/local/bin/php >> >>>/path/to/my/php/page.php; >> >>> >> >>>within the read file I want to enable sessions, so I session_start() at >> > >> > the >> > >> >>>top of the page: >> >>> >> >>><? >> >>>session_start(); >> >>>?> >> >>><? >> >>> code to execute here....... >> >>>?> >> >>> >> >>>Regardless of how I mess around with placement of session_start(), I CH> get >> > >> > a >> > >> >>>"Headers already sent". >> >>> >> >>>Why? Nothing has been output to the browser within the read file! >> >>>Furthermore, if I create a test page with just: >> >>> >> >>><?$text = usr/local/bin/php >> >>>/path/to/my/php/page.php;?> >> >>> >> >>>Still receive "Headers already sent". >> >>> >> >>>My eyes are completely fried -- anyone feel like saving my vision? >> >> >> >>This often is difficult to detect when there's is implicit output >> >>outside of the <? tag. Check the top of the included file or start php >> >>script and see if there is any whitespace or newlines preceding the tag. >> >> >> >>HTH, >> >>Rob. >> >>-- >> >>.------------------------------------------------------------. >> >>| InterJinn Application Framework - >> >>http://www.interjinn.com | >> >>:------------------------------------------------------------: >> >>| An application and templating framework for PHP. Boasting | >> >>| a powerful, scalable system for accessing system services | >> >>| such as forms, properties, sessions, and caches. InterJinn | >> >>| also provides an extremely flexible architecture for | >> >>| creating re-usable components quickly and easily. | >> >>`------------------------------------------------------------' >> > >> > Try this <?php $args = '"hello world"'; $text = /usr/bin/php /usr/local/apache/htdocs/in.php $args; echo $text; ?> in.php contains: <?php echo '<pre>'; phpinfo(32); echo '</pre>'; echo $_SERVER['argv'][1]; ?> That should do what you want if I understand the problem :) -- regards, Tom

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