Question

我有客户端表和client_address信息表。我需要在更新下面给出的client.my模型类时更新两个表，

        class Client extends Model {
             public function addressInfo() {
              return $this->hasOne('App\Model\ClientAddressInfo');
             }
        }

      class ClientAddressInfo extends Model {    
            protected $table = 'client_address_info';    
              public function client() {
               return $this->belongsTo('App\Model\Client');
            }
     }

我的控制器更新如下。

$client = Client::findOrFail($id);
$client->name = rand(0, 1222222);
$address = ClientAddressInfo::where('client_id', '=', $id)->get();
$address->street = "new street";
$address->save();

但它不起作用，请你解释一下更新模型及其相关模型的最佳实践。

Answer 1

你可以做得更简单：

$client = Client::findOrFail($id);
$client->name = rand(0, 1222222);
$client->addressInfo->street = 'new street';
$client->addressInfo->save();
$client->save();

您可以使用save()来保存模型及其相关模型，而不是在两个模型上调用push()：

$client = Client::findOrFail($id);
$client->name = rand(0, 1222222);
$client->addressInfo->street = 'new street';
$client->push(); // save client and addressInfo

Answer 2

我们也可以在@lukasgeiter的答案中使用如下所示的质量分配：

$client = Client::findOrFail($id);
$client->fill($request->all());
$client->addressInfo->fill($request->all());
$client->push();

在laravel中更新表格及其相关模型？

2 个答案: