Question

我正在使用laravel 5.6，我需要过滤输入。 laravel有相当漂亮的验证系统。我想知道有什么方法可以消除那些不正确的结果，而不是在输入不符合规则时显示错误。像下面的例子这是我的规则

$this->validate($request , [
        'room_attributes.*.title'      => 'required' ,
        'room_attributes.*.value'      => 'required' ,

    ]);

这些是我从表单中收到的数据

  "room_attributes" => array:6 [▼
0 => array:2 [▼
  "title" => "title 1"
  "value" => "val1"
]
1 => array:2 [▼
  "title" => "title2"
  "value" => "val2"
]
2 => array:2 [▼
  "title" => "title3"
  "value" => null
]
3 => array:2 [▼
  "title" => null
  "value" => "val4"
]
4 => array:2 [▼
  "title" => null
  "value" => null
]
5 => array:2 [▼
  "title" => null
  "value" => null
]
]

根据当前规则，我会收到这些错误

The room_attributes.3.title field is required.
The room_attributes.4.title field is required.
The room_attributes.5.title field is required.
The room_attributes.2.value field is required.
The room_attributes.4.value field is required.
The room_attributes.5.value field is required.

但是我想过滤我的数据，所以我过滤我的输入并接收此数组作为结果（仅有两个同时填充了标题和值的数组）

  0 => array:2 [▼
  "title" => "title 1"
  "value" => "val1"
]
1 => array:2 [▼
  "title" => "title2"
  "value" => "val2"
]

剂量laravel提供类似功能吗？ laravel本身还是一些第三方库？

预先感谢

Answer 1

您可以尝试这样的事情：

$data = $request->all();

$data['room_attributes'] = collect($data['room_attributes'])->filter(function($item){
    return !is_null($item['title']) && !is_null($item['value']);
})->toArray();

Validator::make($data, [
    'room_attributes.*.title' => 'required',
    'room_attributes.*.value' => 'required',
])->validate();

有关过滤器收集方法的更多信息：https://laravel.com/docs/5.7/collections#method-filter

laravel 5.6过滤表单中的输入数据

1 个答案: