1

我正在使用 Symfony2、JMS Serializer Bundle、FOS Rest Bundle和Hateoas Bundle开发 RESTful 服务。有 2 个实体用户和公司，我想在序列化公司时获得更多细节。但是，当序列化与用户相关的公司时，仅显示公司 ID 和名称对象或仅将 ID 显示为整数。

我有如下的序列化策略。

用户

Acme\UserBundle\Entity\User:
exclusion_policy: ALL
xml_root_name: user
properties:
    id:
        expose: true
        type: integer
    company:
        expose: true
        type: Acme\CompanyBundle\Entity\Company
    name:
        expose: true
        type: string
    surname:
        expose: true
        type: string
    picture:
        expose: true
        type: string
relations:
    -
        rel: self
        href:
            route: acme_v1_get_user
            parameters:
                id: expr(object.getId())
            absolute: true

公司

Acme\CompanyBundle\Entity\Company:
exclusion_policy: ALL
xml_root_name: company
properties:
    id:
        expose: true
        type: integer
    name:
        expose: true
        type: string
    address:
        expose: true
        type: string
    phone:
        expose: true
        type: string
    web:
        expose: true
        type: string
    created_date:
        expose: true
        type: DateTime
    updated_date:
        expose: true
        type: DateTime
    status:
        expose: true
        type: integer
relations:
    -
        rel: self
        href:
            route: acme_v1_get_company
            parameters:
                id: expr(object.getId())
            absolute: true

预期产出

{
  "id": 1,
  "name": "Jenny",
  "surname": "Doe",
  "picture": "http://google.com/kittens.jpg",
  "info": [],
  "company": {
    "id": 1,
    "name": "Demo Company"
  }
}

或者

{
  "id": 1,
  "name": "Jenny",
  "surname": "Doe",
  "picture": "http://google.com/kittens.jpg",
  "info": [],
  "company": 1
}

我得到了什么

{
  "id": 1,
  "name": "Jenny",
  "surname": "Doe",
  "picture": "http://google.com/kittens.jpg",
  "info": [],
  "company": {
    "id": 1,
    "name": "Demo Company",
    "address": "Lorem ipsum dolor sit amet",
    "phone": "0902124440444",
    "web": "http://www.demo-company.com",
    "created_date": "2015-07-22T11:21:03+0300",
    "updated_date": "2015-07-24T01:50:39+0300",
    "status": 1
  }
}
4

2 回答 2

2

您可以使用群组

AppBundle\Entity\User\User:
    exclusion_policy: ALL
    properties:
        lastname:
            expose: true
            groups: [info]

并且通过注释，您可以定义哪个属性显示在哪个组上。最后，您可以为您使用的每条路线分配一个组。

或者您可以像这样使用虚拟属性：

AppBundle\Entity\User\User:
    exclusion_policy: ALL
    properties:
         […]
    virtual_properties:
        getCompanyId:
            serialized_name: company
            type: string
            groups: [info]

然后在您的 User 实体中创建一个getCompanyId()方法，该方法返回 companyId

于 2015-07-30T08:13:09.793 回答
0

更多的 Hateoas 方法是与关系。

Acme\UserBundle\Entity\User:
exclusion_policy: ALL
xml_root_name: user
properties:
    id:
        expose: true
        type: integer
    name:
        expose: true
        type: string
    surname:
        expose: true
        type: string
    picture:
        expose: true
        type: string
relations:
    -
        rel: self
        href:
            route: acme_v1_get_user
            parameters:
                id: expr(object.getId())
            absolute: true
    -
        rel: company
        href:
            route: acme_v1_get_company
            parameters:
                id: expr(object.getCompany().getId())
            absolute: true

会产生...

{
  "id": 1,
  "name": "Jenny",
  "surname": "Doe",
  "picture": "http://google.com/kittens.jpg",
  "info": []
  "_links": {
    "self": {
      "href": "http://server.com/api/user/1"
    },
    "company": {
      "href": "http://server.com/api/company/1"
    },
  }
}
于 2017-03-01T16:21:48.707 回答