I have the following line in a shell script:
/usr/local/bin/php /home/script_to_run.php;
This works fine with our setup, until I add an argument like so:
/usr/local/bin/php /home/script_to_run.php?needed_variable=1;
At which point, I get the "Could not open input file" error.
Ideas on how to get this to work?
Thanks!
There is no query string in the command line; you need to use arguments instead:
/usr/local/bin/php /home/script_to_run.php 1;
You then access the value with the $argv
variable :
$value = $argv[1];
For more advanced command line argument parsing, take a look at getopt
.
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