表:contact
,company
以及包含自定义透视属性company_contact (company_id, contact_id, is_main)
的关系表
公司和联系人有多对多关系(两个模型都belongsTo
)。
检索公司联系人时的预期输出:
{
"data": [
{
"id": 1,
"name": "JohnDoe",
"is_main": false
},
{
"id": 2,
"name": "JaneDoe",
"is_main": true
}
]
}
使用?include=companies
检索联系人列表时的预期输出:
{
"data": [
{
"id": 1,
"name": "John Doe",
"companies": {
"data": [
{
"id": 501,
"name": "My Company",
"is_main": true
},
{
"id": 745,
"name": "Another Company",
"is_main": false
}
]
}
},
{
"id": 2,
"name": "Jane Doe",
"companies": {
"data": [
{
"id": 999,
"name": "Some Company",
"is_main": true
}
]
}
}
]
}
添加数据透视表属性的最佳方法是什么?如果设置了属性,在公司变换器上添加is_main
似乎不是很干净。
对于第一个例子,我考虑使用参数?include=company_relationship:company_id(1)
,例如:
public function includeCompanyRelationship(Contact $contact, ParamBag $params) {
// .. retrieve the pivot table data here
$is_main = $company->is_main;
// but now I would need another transformer, when what I actually want is to push the value on the main array (same level)
return $this->item(??, ??);
}
我了解如何检索数据透视数据(相关:Laravel 5.1 - pivot table between three tables, better option?),但不了解在https://github.com/thephpleague/fractal变换器逻辑中添加数据的最佳方法。
我已经有了一个ContactTransformer和CompanyTransformer但是如果我将is_main
添加到CompanyTransformer,我所做的所有电话(与联系人相关或不相关)都会期望该属性。
答案 0 :(得分:3)
如果我正确地阅读了您,您可以使用单个CompanyTransformer
来处理您是否希望设置is_main
属性,但仅限于传递$contact
参数的情况它的构造函数,就是这些:
class CompanyTransformer extends TransformerAbstract
{
public function __construct(Contact $contact = null)
{
$this->contact = $contact;
}
public function transform(Company $company)
{
$output = [
'id' => $company->id,
'name' => $company->name,
];
if($this->contact) {
// This step may not be necessary, but I don't think the pivot data
// will be available on the $company object passed in
$company = $this->contacts->find($company->id);
// You may have to cast this to boolean if that is important
$output['is_main'] = $company->pivot->is_main;
}
return $output;
}
}
然后在includeCompanyRelationship
中使用参数:
CompanyTransformer
public function includeCompanyRelationship(Contact $contact)
{
$companies = $contact->companies;
return $this->collection($companies, new CompanyTransformer($contact));
}
无论您是直接呼叫companies
端点,还是在嵌入公司关系数据时呼叫联系人的端点,这都应该有效。
答案 1 :(得分:2)
我知道这已经过时了,但我刚遇到这个问题。以下是我如何解决它。
在我的案例Library_Kind
和dynamic
中为关系添加了Library_Standalone
。
在CategoryTransformer中,我定义了withPivot
方法:
category
然后在users
课程中,使用includeUsers
方法:
/**
* Include Users
* @param Category $category
* @return \League\Fractal\Resource\Collection
*/
public function includeUsers(Category $category)
{
# Just Add withPivot Here.
$users = $category->users()->withPivot('role')->get();
return $this->collection($users, new UserTransformer);
}
然后,当我为UserTransformer
打电话给我的api并包含transform()
时,我得到了:
public function transform($user)
{
return [
'username' => $user['username'],
'lastname' => $user['lastname'],
// ... truncated
'role' => isset($user['pivot']) ? $user['pivot']['role'] : null,
]
}
如您所见,我从枢轴中获得了我想要的categories
关系。否则,您只需获得users
该字段。在我看来,仍然不理想,但不那么混乱。