PHP为一组命令选项验证命令行argv数组

Question

I am doing command line script with php and trying to develop validate the set of commands to see if they were there, but i am not going anyware with the following code, i dont know why it alwasy goes to the help().. even if i have the argv options like

sh#php geninvoices.php --type=generate --w --bill_day=7


$commands=array('--type','--bill_day','--save');
if(!in_array($commands, $argv)){
    print_r($argv);
    help();
} 

any help ?

Answer 1

Just to show the idea

$argv = array('generate.php', '--type=generate', '--write', '--bill_day=7');
$args = array();
foreach($argv as $i => $str)
{
    if ($i == 0) continue;
    list($arg, $value) = explode('=', $str, 2);
    $args[ltrim($arg, '-')] = $value;
}

var_dump($args);

And now you have arguments with values and you can do whatever you want.

array(3) {
  ["type"]=>
  string(8) "generate"
  ["write"]=>
  NULL
  ["bill_day"]=>
  string(1) "7"
}