通过比较另一个数组的键从一个数组中删除数组值

Question

假设我有第一个数组$ aAllCities作为

Array
(
   [21] => London
   [9]  => Paris
   [17] => New York
   [3]  => Tokyo
   [25] => Shanghai
   [11] => Dubai
   [37] => Mumbai
)

另一个数组$ aNotSupportedCities为

Array
(
   [0] => 37
   [1] => 25
   [2] => 11
)

是否有可能得到这样的数组？

Array
(
   [21] => London
   [9]  => Paris
   [17] => New York
   [3]  => Tokyo
)

我想删除其他数组中存在的那些键的数组值

Answer 1

foreach($aAllCities as $key => $value) {
    if(in_array($key,$aNotSupportedCities)) {
        unset($aAllCities[$key]); 
    }

}

Answer 2

尝试这个：

$aAllCities = array_flip( $aAllCities );
$aAllCities = array_diff( $aAllCities, $aNotSupportedCities );
$aAllCities = array_flip( $aAllCities );

希望这可以帮助。

Answer 3

其他答案是正确的，但是更流畅，更快捷的方法是：
$supportedCities = array_diff_key($aAllCities, $aNotSupportedCities);

这将从$aAllCities返回所有在$aAllCities中没有键的$aNotSupportedCities

注意，这通过它们的键比较两个数组，因此您需要使$aNotSupportedCities看起来像这样：

Array
(
   [37] => something
   [25] => doesn't really matter
   [11] => It's not reading this
)

祝你好运。

Answer 4

$new = $aAllCities;
foreach($aNotSupportedCities as $id) {
  if (isset($new[$id]) {
    unset($new[$id]);
  }
}

Answer 5

$supportedCities = array_diff_key($aAllCities, array_values($aNotSupportedCities));