dtj4307
dtj4307
2013-09-03 21:20

使用参数在shell脚本中调用PHP

已采纳

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!

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1条回答

  • donkey199024 donkey199024 8年前

    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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