如何匹配两个数组的共有元素并合并，为city数组新增post_code字段

你可以先将邮编数组转换为城市名为键、邮编为值的映射表，再遍历城市数组匹配追加字段，该方案时间复杂度为O(n+m)，数据量大时效率远高于嵌套循环。

PHP 实现代码 #

// 示例数组定义（可替换为你自己的实际数组）
$city = [
    ['id' => 1, 'city' => 'Budapest', 'population' => 1700000],
    ['id' => 2, 'city' => 'Szeged', 'population' => 160000]
];

$city_postcode = [
    ['name' => 'Budapest', 'post_code' => 12345],
    ['name' => 'Szeged', 'post_code' => 33356]
];

// 1. 构造邮编映射表
$postcodeMap = [];
foreach ($city_postcode as $item) {
    $postcodeMap[$item['name']] = $item['post_code'];
}

// 2. 遍历城市数组追加字段
foreach ($city as &$item) {
    if (isset($postcodeMap[$item['city']])) {
        $item['post_code'] = $postcodeMap[$item['city']];
    }
    // 未匹配到需要默认值可改为：$item['post_code'] = $postcodeMap[$item['city']] ?? '默认邮编';
}
unset($item); // 解除引用避免后续逻辑异常

// 打印验证结果
print_r($city);

JavaScript 实现代码 #

如果是JS场景可以用以下实现：

// 示例数组定义
const city = [
    {id: 1, city: 'Budapest', population: 1700000},
    {id: 2, city: 'Szeged', population: 160000}
];
const cityPostcode = [
    {name: 'Budapest', post_code: 12345},
    {name: 'Szeged', post_code: 33356}
];

// 构造映射表
const postcodeMap = Object.fromEntries(cityPostcode.map(item => [item.name, item.post_code]));

// 追加字段
const result = city.map(item => ({
    ...item,
    post_code: postcodeMap[item.city]
    // 未匹配到需要默认值可改为：post_code: postcodeMap[item.city] || '默认邮编'
}));

// 打印验证
console.log(result);

内容的提问来源于stack exchange，提问作者Norbert Eper