Date: Tue, 9 Mar 2010 10:18:12 -0800 From: Nerius Landys <nlandys@gmail.com> To: Darrell Betts <betts@norden1.com> Cc: freebsd-questions@freebsd.org Subject: Re: Trying run a php script from cron Message-ID: <560f92641003091018m1c1ffeeg229cc2fc29079d7@mail.gmail.com> In-Reply-To: <551AED7A-2336-40B6-B8B8-3EF2637D03B0@norden1.com> References: <551AED7A-2336-40B6-B8B8-3EF2637D03B0@norden1.com>
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> I am trying to run a php script from the cron tab and these are the errors I > receive: > > > /usr/local/bin/php php -q /home/xxxx/ripper.php result > Could not open input file: php > /usr/local/bin/php php -/home/xxxx/ripper.php result > > Could not open input file: php > > /usr/local/bin/php -/home/xxxx/ripper.php result > > This script must be called from the command line. > > Running Freebsd 8.0, Php 5.2.12 > I have chmod the script 644 still no luck tried it chmod 777 still no luck. > I have goggled this problem and followed the tutorials but still no luck. > Any ideas how I can get the script to run? > I can run run it from the command line without any problems. Instead of /usr/local/bin/php php -q /home/xxxx/ripper.php try /usr/local/bin/php -f /home/xxxx/ripper.php or just /usr/local/bin/php /home/xxxx/ripper.php You can also try a script like this one: #!/usr/local/bin/php -f <?php echo "foo\n"; ?> And running it like this: /home/xxxx/ripper.php after chmod'ing it to be executable.
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