将一个bash数组作为argv传递给php

Question

I am passing a shell script array to php file as below.

php file.php ${variable[@]}

The file.php is used to find all the permutations of the given shell script array.

<?php
function pc_array_power_set($array) {
    // initialize by adding the empty set
    $results = array(array( ));

    foreach ($array as $element)
        foreach ($results as $combination)
            array_push($results, array_merge(array($element), $combination));
    return $results;
}
$set = $argv;
$power_set = pc_array_power_set($set);
foreach (pc_array_power_set($set) as $combination) {
    if (2 == count($combination)) {
        print join("\t", $combination) . "
";
    }
}
?>

However, since I use argv as the command line argument for the php file, my output is considering the file name also as an element of the array.

Output:

echo ${variable[@]}
php checking
php file.php ${variable[@]}

The output is coming as,

php done.php
checking done.php
checking php

As we can see, I get the file name also in the output while I just expect the output as,

checking php

Answer 1

Simply perform an array_shift() on the array in your PHP script, before using it, to discard the first element:

$set = $argv;
array_shift($set);
$power_set = pc_array_power_set($set);