Laravel list（）具有不推荐使用的函数的each（）函数错误

Question

由于不推荐使用each（）函数，我对此有些困惑，因为我已经意识到这一点，并在项目中修复了某些while( list() = each() )错误的情况。但是，在这种情况下，我还应该使用其他什么选项：

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();
}

Answer 1

您不使用键，所以只需获取当前值，然后移至下一个：

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，如果键在所有数组中都相同，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();

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