1

因此，假设我收集了用户和公司。然后我有一个名为works_in 的边缘集合，它将用户与公司联系起来。我正在使用以下 aql 查询：

FOR user IN Users
   LET companies = (FOR company IN (NEIGHBORS(Users, works_in, user._id, 'outbound', [], {includeData:true}))
       RETURN {company_name: company.name, company_id: company._id})
RETURN {user, companies}

我得到的是：

[
  {
    "user": {
      "_id": "Users/45645",
      "_key": "45645",
      "_rev": "45645",
      "name": "user1",
      "city": "london",
      "age": 23
    },
    "companies": [
        {
          company_name: "company1",
          company_id: "Companies/7897"
        },
        {
          company_name: "company2",
          company_id: "Companies/7878"
        }
    ]
  },
  {
    "user": {
      "_id": "Users/465454",
      "_key": "465454",
      "_rev": "465454",
      "name": "user2",
      "city": "Paris",
      "age": 42
    },
    "companies": [
        {
          company_name: "company1",
          company_id: "Companies/7897"
        },
        {
          company_name: "company3",
          company_id: "Companies/788233"
        }
    ]
  }
]

但是，我想获取不嵌套在“用户”中的“用户”信息，但如下所示：

[
  {
    "_id": "Users/45645",
    "_key": "45645",
    "_rev": "45645",
    "name": "user1",
    "city": "london",
    "age": 23,
    "companies": [
        {
          company_name: "company1",
          company_id: "Companies/7897"
        },
        {
          company_name: "company2",
          company_id: "Companies/7878"
        }
    ]
  },
  {
    "_id": "Users/465454",
    "_key": "465454",
    "_rev": "465454",
    "name": "user2",
    "city": "Paris",
    "age": 42,
    "companies": [
        {
          company_name: "company1",
          company_id: "Companies/7897"
        },
        {
          company_name: "company3",
          company_id: "Companies/788233"
        }
    ]
  }
]

我知道我可以通过硬编码属性来做到这一点，例如

RETURN {_id: user.id, _key:user._key, companies}

但问题是我有很多属性要显示，而且用户可能没有某些特定属性（因此显示为“null”）

那么，有谁知道我怎样才能正确地“取消嵌套”用户？谢谢

4

1 回答 1

1

您可以为此使用MERGE。让我使用 return 语句来演示这一点：

db._query(`RETURN MERGE({
      "_id": "Users/465454", "_key": "465454",
      "_rev": "465454",      "name": "user2",
      "city": "Paris",       "age": 42
    }, {"companies": [
        {
          company_name: "company1",
          company_id: "Companies/7897"
        },
        {
          company_name: "company3",
          company_id: "Companies/788233"
        }
    ]})`
).toArray() =>
[ 
  { 
    "_id" : "Users/465454", 
    "_key" : "465454", 
    "_rev" : "465454", 
    "age" : 42, 
    "city" : "Paris", 
    "companies" : [ 
      { 
        "company_id" : "Companies/7897", 
        "company_name" : "company1" 
      }, 
      { 
        "company_id" : "Companies/788233", 
        "company_name" : "company3" 
      } 
    ], 
    "name" : "user2" 
  } 
]

由于公司给了你一个列表，你需要将它包装到一个对象中以便MERGE可以完成它的工作：

FOR user IN Users
  LET companiesList = (FOR company IN
                           (NEIGHBORS(Users, works_in, user._id, 'outbound', [],
                                     {includeData:true}))
  RETURN MERGE(user, {companies: companiesList})
于 2016-05-17T08:41:37.823 回答