ID: 14930 Updated by: edink Reported By: [EMAIL PROTECTED] Old Status: Analyzed Status: Closed Bug Type: Output Control Operating System: linux 2.4.9 PHP Version: 4.1.1 New Comment:
After some research on the subject I found out that there is no good and portable way of solving this. Some unix systems simply allow only one argument in the #! line. One of those systems is Linux. Some others limit the line length to 32 characters. For a good description of these limitations see http://groups.google.com/groups?q=shell+limit+one+argument&hl=en&selm=qij4tk2zgp9.fsf%40lambda.ai.mit.edu&rnum=7 Workarounds: * You can put all the options that do not require parameters as one, for example -qC * Use PHPRC env variable to set the php.ini path Previous Comments: ------------------------------------------------------------------------ [2002-01-10 16:32:04] [EMAIL PROTECTED] I don't think this is a glibc issue. Its (AFAIK) the kernel's responsibility to set this stuff up. Take a look at load_script() in fs/binfmt_script.c of the linux source. It seems that only one argument is allowed, and everything after the interpreter (/usr/local/bin/php) up until the EOL is considered one argument. I'm sure there is a reason for implementing it this way, I just don't know what it is. Sean ------------------------------------------------------------------------ [2002-01-08 19:29:05] [EMAIL PROTECTED] I was able to reproduce this. However it does not appear to be a php bug. Most likely a glibc bug. If I put this line on top of my script: #!/usr/bin/php -q -c /path/to/ini/file -C the following values get passed to php's main() function. argc=3 argv[0]: php argv[1]: -q -c /path/to/ini/file -C argv[2]: ./test (which is the name of the script) With parameters passed like that, php has no chance of passing them correctly. FreeBSD systems appear to be free of this problem. ------------------------------------------------------------------------ [2002-01-08 10:54:04] [EMAIL PROTECTED] I don't know if this should go under output control but anyways... The command line executable is having problems suppressing the headers using the -q option in combination with the -c option if you're running the script as, well, a script. For instance, take this small script: #!/usr/local/bin/php -c /path/to/ini/file -q <?php echo "hello world"; ?> The headers won't be suppressed if you try running the script with a command like $ ./smallscript.php X-Powered-By: PHP/4.1.1 Content-type: text/html hello world However, running the script like so: $ php -q -c /path/to/ini/file smallscript.php hello world produces the expected result, without the headers. Also, this may or may not be an associated bug, but if you put the -q option before the -c option in the command line for a script, i.e. #!/usr/local/bin/php -q -c /some/path <?php ... ?> An error is produced, which looks something like this: Error in argument 1, char 3: option not found Error in argument 1, char 4: option not found - Error in argument 1, char 3: option not found along with the output from "php -h". This doesn't happen when you execute directly from the command line, only when the command line is in the script file itself. J ------------------------------------------------------------------------ Edit this bug report at http://bugs.php.net/?id=14930&edit=1 -- PHP Development Mailing List <http://www.php.net/> To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
