Php Laravel list（）和each（）函数不推荐使用的函数出现错误

在这里，我对如何替换each（）函数有点困惑，因为它已被弃用，我知道这一点，并修复了项目中的一些

while（list（）=each（））

错误案例。但是，在这种情况下，我应该使用什么其他选项：

foreach($new_id as $new_ids) {
    list($key,$valueAddress) = each($address);
    list($key,$valueCity) = each($city);
    list($key,$valueState) = each($state);
    if(isset($_POST['publicOnly'])) {
        list($key,$valuePublicOnly) = each($publicOnly);
    } else {
        $valuePublicOnly = 0;
    }

    $propertyAddress = PropertyAddressManagement::find($new_ids);
        $propertyAddress->address = $valueAddress;
        $propertyAddress->city = $valueCity;
        $propertyAddress->state = $valueState;
        $propertyAddress->publicOnly = $valuePublicOnly;
    $propertyAddress->save();
}

---

您没有使用关键点，因此只需获取当前值，然后移动到下一个值：

foreach($new_id as $new_ids) {
    $propertyAddress = PropertyAddressManagement::find($new_ids);
        $propertyAddress->address = current($address);
        $propertyAddress->city = current($city);
        $propertyAddress->state = current($state);
        $propertyAddress->publicOnly = isset($_POST['publicOnly']) ? current($publicOnly) : 0;
    $propertyAddress->save();

    next($address); next($city); next($state); next($publicOnly);
}

但是，如果所有数组中的键都是相同的，那么我认为这确实应该起作用：

foreach($new_id as $key => $new_ids) {
    $propertyAddress = PropertyAddressManagement::find($new_ids);
        $propertyAddress->address = $address[$key];
        $propertyAddress->city = $city[$key];
        $propertyAddress->state = $state[$key];
        $propertyAddress->publicOnly = isset($_POST['publicOnly']) ? $publicOnly[$key] : 0;
    $propertyAddress->save();
}

---

我试过这个。。虽然它最终出现在这个错误消息中：

next（）期望参数1是数组，但如果给定null，我仍然遇到了这个
next（）
问题。虽然我发现页面正在保存。对不起，第二个示例不需要下一个。我明白了。你的答案终于奏效了。看到文本编辑器是空的，我很激动谢谢你的帮助，先生！