PHP:如何将一个数组中的键与另一个数组中的值进行比较,并返回匹配项?

我有以下两个数组:

$array_one = array('colorZero'=>'black', 'colorOne'=>'red', 'colorTwo'=>'green', 'colorThree'=>'blue', 'colorFour'=>'purple', 'colorFive'=>'golden');

$array_two = array('colorOne', 'colorTwo', 'colorThree');

我想要一个来自 $array_one 的数组,它只包含键是 $array_two 成员的键值对(通过创建一个新数组或删除其余的$array_one)

I want an array from $array_one which only contains the key-value pairs whose keys are members of $array_two (either by making a new array or removing the rest of the elements from $array_one)

我该怎么做?

我查看了 array_diffarray_intersect,但它们将值与值进行比较,而不是将一个数组的值与另一个数组的键进行比较.

I looked into array_diff and array_intersect, but they compare values with values, and not the values of one array with the keys of the other.

推荐答案

更新

查看 Michel 的答案:https://stackoverflow.com/a/30841097/2879722.这是一个更好、更简单的解决方案.

Check out the answer from Michel: https://stackoverflow.com/a/30841097/2879722. It's a better and easier solution.

原答案

如果我理解正确:

返回一个新数组:

$array_new = [];
foreach($array_two as $key)
{
    if(array_key_exists($key, $array_one))
    {
        $array_new[$key] = $array_one[$key];
    }
}

从 $array_one 中剥离:

Stripping from $array_one:

foreach($array_one as $key => $val)
{
    if(array_search($key, $array_two) === false)
    {
        unset($array_one[$key]);
    }
}

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