将一个bash数组作为argv传递给php

问题描述：

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

我将shell脚本数组传递给php文件，如下所示。 p>

  php file.php $ {variable [@]} 
  code>  pre> 
 
  file.php用于查找 给定shell脚本数组的所有排列。   p> 
 
 
 ＆lt;？php 
function pc_array_power_set（$ array）{
 //通过添加空集初始化
 $ results = array（array（））; 
  
 \ foreach（$ array as $ element）
 foreach（$ results as $ combination）
 array_push（$ results，array_merge（array（$ element），$ combination））; 
返回$ results; 
} \  n $ set = $ argv; 
 $ power_set = pc_array_power_set（$ set）; 
foreach（pc_array_power_set（$ set）as $ combination）{
 if（2 == count（$ combination））{
 print join（  “\ t”，$组合）。  “
”; 
} 
} 
？＆gt; 
  code>  pre> 
 
 但是，因为我使用argv作为php文件的命令行参数， 输出正在考虑文件名也作为数组的元素。   p> 
 
 
 输出： strong>  p> 
 
 
  echo $ {variable [@]} 
 nphp checking 
php file.php  $ {variable [@]} 
  code>  pre> 
 
 输出结果为， p> 
 
 
  php done.php 
checking  done.php 
checking php 
  code>  pre> 
 
 正如我们所看到的，我在输出中也得到了文件名，而我只希望输出为， p> \  n 
 
 检查php 
  code>  pre> 
  div>

答

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);

将一个bash数组作为argv传递给php

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