Question

所以，让我们说我收集了用户和公司。然后我有一个名为works_in的边集合，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"
        }
    ]
  }
]

然而，我想得到＆＃34;用户＆＃34;信息没有嵌套在＆＃34;用户＆＃34;但如下：

[
  {
    "_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"
        }
    ]
  }
]

我知道我可以通过诸如

之类的硬编码属性来实现

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

但问题是我要显示很多属性，而且用户可能没有某些特定属性（因此显示为＆＃34; null＆＃34;）

那么，有没有人知道我怎么能＆＃34;不需要＆＃34;用户好吗？谢谢

Answer 1

您可以使用MERGE。让我用return语句证明这一点：

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" 
  } 
]

由于公司为您提供了一个列表，您需要将其包装到一个对象中，以便MERGE能够完成它的工作：

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

＆＃34; UNNEST＆＃34; arangodb中的aql查询结果

1 个答案: