bash script to loop through array and run php script with arg

Question

I have a bash script which runs in cron every day at 1:01am the bash script is:

array_of_clients=(2 187 317 927 1863 2993 3077 3440 3444 3457 3459 3469 3484 3487 3494 3497 3522 3544 3551 3553)

for i in "${array_of_clients[@]}"
do
    echo "\nRunning Client - $i"
    php -f "/mnt/www/bin/scheduled/import_client.php" $i
    echo "\nFinished Client - $i"
done

This issue is that I don't know if $i is being passed as a argument to the php script. Am I doing something wrong ? If I put the $i within "" it says it cannot find the file because the file name becomes /mnt/www/bin/scheduled/import_client.php 2 for example

Could anyone help ?

Joakim Nohlgård · Accepted Answer · 2012-11-14 09:08:08Z

1

You can access the command line arguments in your PHP script in the predefined global variable $argv. Your $i in this case would be found as $argv[1].

Try this script:

<?php
global $argv;
var_dump($argv);
?>

Running it with php -f test.php A B C defgh yields:

array(5) {
  [0]=>
  string(8) "test.php"
  [1]=>
  string(1) "A"
  [2]=>
  string(1) "B"
  [3]=>
  string(1) "C"
  [4]=>
  string(5) "defgh"
}

answered Nov 14, 2012 at 9:08

Joakim Nohlgård

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