PHP如何检查关联数组的键是否存在于另一个数组中？

错误原因 #

你当前代码的核心问题是误用了in_array()函数：

• in_array()的作用是判断某个值是否存在于目标数组的值集合中，但你的需求是判断第一个数组的键是否作为键存在于第二个数组里，这个函数完全不适用当前场景。

修正后的代码实现 #

方法1：修复原有while循环

将in_array替换为专门检查键存在性的array_key_exists()函数：

<?php
$team_member = ["Person_1" => 0, "Person_2" => 0, "Person_3" => 0, "Person_4" => 0];
$today_active_member = ["Person_1" => 0, "Person_3" => 0, "Person_4" => 0];

$i = count($team_member);

while($i--){
    $currentKey = key($team_member);
    if(array_key_exists($currentKey, $today_active_member)) {
        echo $currentKey . " is present today.<br/>";
    } else {
        echo $currentKey . " isn't present today.<br/>";
    }
    next($team_member);
}

方法2：使用foreach遍历（推荐）

foreach可以直接遍历关联数组的键值对，代码更简洁易读：

<?php
$team_member = ["Person_1" => 0, "Person_2" => 0, "Person_3" => 0, "Person_4" => 0];
$today_active_member = ["Person_1" => 0, "Person_3" => 0, "Person_4" => 0];

foreach($team_member as $name => $_) {
    if(array_key_exists($name, $today_active_member)) {
        echo "$name is present today.<br/>";
    } else {
        echo "$name isn't present today.<br/>";
    }
}

方法3：用isset提升性能

如果第二个数组中不存在键对应值为null的情况，isset()的性能比array_key_exists更优：

foreach($team_member as $name => $_) {
    if(isset($today_active_member[$name])) {
        echo "$name is present today.<br/>";
    } else {
        echo "$name isn't present today.<br/>";
    }
}

批量提取存在的键 #

如果只需要获取第一个数组中存在于第二个数组的所有键，不需要逐个输出，可以用array_intersect_key一步完成：

$presentMembers = array_intersect_key($team_member, $today_active_member);
// 结果为 ["Person_1" => 0, "Person_3" => 0, "Person_4" => 0]
print_r($presentMembers);

内容的提问来源于stack exchange，提问作者Hasib Hosen