Question

在laravel中，是否有某种方法可以在表单中嵌套相关资源？

说我有这个：

class Person extends Eloquent {
  public function addresses() {
    return $this->hasMany("Address");
  }
}

class Address extends Eloquent {
  public function person() {
    return $this->belongsTo("Person");
  }
}

我想要一个Person表单来收集有关Person Address个accepts_nested_attributes_for :address的信息。 laravel是否以与Rails'fields_for :address和Address等效的方式为此提供了便利？

我只想简单地将Person字段与Address表单的结果包含在一起，因为Person除了{{{ Form::model(new Person, array("action" => "admin\PersonController@store", "method" => "POST")) }} {{ Form::text("name", array(...)) // <input name='person[name]' ... /> }} {{ Form::email("email", array(...)) // <input name='person[email]' ... /> }} {{ Form::fields_for("addresses"/* Would be name of relation */) }} {{ Form::text("street_address") // <input name='person[addresses][][street_address]' ... /> }} {{ Form::close_fields() }} {{ Form::close() }}之外并不存在1}}。这有意义吗？

==编辑==

这是假设代码

我正在寻找的是类似于此的东西：

{{1}}

Answer 1

您在输入名称的正确轨道上。

<强>表格

// Form open, Person fields, etc...

<h2>Addresses</h2>
@foreach ($addresses as $address)

    <fieldset>

        {{ Input::text('addresses['.$address->id.'][address_1]', $address->address_1) }}
        {{ Input::text('addresses['.$address->id.'][address_1]', $address->address_2) }}
        {{ Input::text('addresses['.$address->id.'][city]', $address->city) }}
        {{ Input::text('addresses['.$address->id.'][state]', $address->state) }}
        {{ Input::text('addresses['.$address->id.'][zip]', $address->zip) }}

    </fieldset>

@endforeach

// Form Close

如果要添加地址，则需要生成一些随机密钥而不是地址ID。这将使字段保持分组。

控制器逻辑

这是我处理输入的方式，使用'fillable'来过滤进入模型的数据。

// Get the Person model, fill, save, etc...

$addressIds = array();
foreach (Input::get('addresses', array()) as $id => $addressData)
{
    $address = Address::find($id) ?: new Address;
    $address->fill($addressData);
    $address->save();
    $addressIds[] = $address->id;
}

$changes = $person->addresses()->sync($addressIds);

// Delete the unused addresses
foreach ($changes['detached'] as $detachedAddressId)
{
    $address = Address::find($detachedAddressId);
    if (!empty($address)) $address->delete();
}

Answer 2

现在，您可以使用“modelform”包为各种模型甚至一组关系形式创建表单。

https://github.com/andersondanilo/modelform

Answer 3

您应该从控制器管理所有这些。从表单中收集数据并随意使用：

// Controller Code
$person = new Person(array(
    'field_1' => Input::get('field_1'),
    'field_2' => Input::get('field_2'),
    .............
));
$person->save();

$address = new Address(array(
    'field_x' => Input::get('field_x'),
    'field_y' => Input::get('field_y'),
    .............
));
$person->addresses()->save($address);

See it in the docs

嵌套模型关系的形式 - Laravel

3 个答案: