基于另一个数组过滤 $_REQUEST 数组？

Question

数组 $ALLOWED_CALLS 包含函数名称和所需参数。我想过滤 $_REQUEST 数组以获得仅包含必需参数的 $params 数组。怎样做？

$call = 'translate';

$ALLOWED_CALLS = array(
    'getLanguages' => array(),
    'detect' => array('text'),
    'translateFrom' => array('text', 'from', 'to'),
    'translate' => array('text', 'to'),
);

$params = array(); // Should contain $_REQUEST['text'] and $_REQUEST['to']

Answer 1

看看 array_intersect()

Answer 2

我会像这样使用 array_intersect_key() ：

$params = array_intersect_key($_REQUEST, array_flip($ALLOWED_CALLS[$call]));

因此，整个事情：

$call = 'translate';

$ALLOWED_CALLS = array(
    'getLanguages' => array(),
    'detect' => array('text'),
    'translateFrom' => array('text', 'from', 'to'),
    'translate' => array('text', 'to'),
);

$params = array_intersect_key($_REQUEST, array_flip($ALLOWED_CALLS[$call]));

Answer 3

像这样的东西：

$contains_required = true;
foreach( $ALLOWED_CALLS[$call] as $key => $value )
{
    if(!in_array($value, $_REQUEST))
    {
        $contains_required = false;
    }
}

Answer 4

function getParams ($call, $allowedCalls) {
  // Return FALSE if the call was invalid
  if (!isset($allowedCalls[$call])) return FALSE;
  // Get allowed params from $_REQUEST into result array
  $result = array();
  foreach ($allowedCalls[$call] as $param) {
    if (isset($_REQUEST[$param])) {
      $result[$param] = $_REQUEST[$param];
    }
  }
  // Return the result
  return $result;
}

返回 $params 数组，失败时返回 FALSE。