返回带有嵌套Json的SQL Server数据库查询

当我使用端点时，我试图得到这样的答案：

[  
   {  
      "McqID":"7EED5396-9151-4E3D-BCBF-FDB72CDD22B7",
      "Questions":[  
         {  
            "QuestionId":"C8440686-531D-4099-89E9-014CAF9ED054",
            "Question":"human text",
            "Difficulty":3,
            "Answers":[  
               {  
                  "AnswerId":"7530DCF4-B2D9-48B0-9978-0E4690EA0C34",
                  "Answer":"human text2",
                  "IsTrue":false
               },
               {  
                  "AnswerId":"5D16F17F-E205-42A5-873A-1A367924C182",
                  "Answer":"human text3",
                  "IsTrue":false
               },
               {  
                  "AnswerId":"64E78326-77C3-4628-B9E3-2E8614D63632",
                  "Answer":"human text4",
                  "IsTrue":false
               },
               {  
                  "AnswerId":"199241A9-0EF6-4F96-894A-9256B129CB1F",
                  "Answer":"human text5",
                  "IsTrue":true
               },
               {  
                  "AnswerId":"EDCCAC18-5209-4457-95F2-C91666F8A916",
                  "Answer":"human text6",
                  "IsTrue":false
               }
            ]
         }
      ]
   }
]

下面是我的问题（示例）：

我想要一个嵌套的JSON来表示我的数据，但它最终的格式如下（较小的示例）：

我已经尝试了一切，

用于JSON路径

，

用于JSON自动

，

JSON\u查询

，等等

什么都不管用<代码>对于JSON路径似乎不适用于多个嵌套集合

---

我如何得到这个结果

您需要像平常一样使用联接。 使用FOR JSON AUTO将选择连接别名，如果您想要更多控制，请使用FOR JSON路径

SELECT
    ent.Id AS 'Id',
    ent.Name AS 'Name',
    ent.Age AS 'Age',
    EMails = (
        SELECT
            Emails.Id AS 'Id',
            Emails.Email AS 'Email'
        FROM EntitiesEmails Emails WHERE Emails.EntityId = ent.Id
        FOR JSON PATH
    )
FROM Entities ent
FOR JSON PATH

我将给您一个通用示例，它将很容易映射到您的场景：

选项1-对于JSON自动： 联接别名将用作嵌套集合属性名称

SELECT
    ent.Id AS 'Id',
    ent.Name AS 'Name',
    ent.Age AS 'Age',
    Emails.Id AS 'Id',
    Emails.Email AS 'Email'
FROM Entities ent
LEFT JOIN EntitiesEmails Emails ON Emails.EntityId = ent.Id
FOR JSON AUTO

选项2-对于JSON路径： 处理所有内容并注意，内部select必须返回一个字符串，这里也使用FOR JSON PATH

SELECT
    ent.Id AS 'Id',
    ent.Name AS 'Name',
    ent.Age AS 'Age',
    EMails = (
        SELECT
            Emails.Id AS 'Id',
            Emails.Email AS 'Email'
        FROM EntitiesEmails Emails WHERE Emails.EntityId = ent.Id
        FOR JSON PATH
    )
FROM Entities ent
FOR JSON PATH

两者生成相同的结果：

[{
    "Id": 1,
    "Name": "Alex",
    "Age": 35,
    "Emails": [{
        "Id": 1,
        "Email": "abc@domain.com"
    }, {
        "Id": 2,
        "Email": "def@domain.com"
    }, {
        "Id": 3,
        "Email": "ghi@domain.net"
    }]
}, {
    "Id": 2,
    "Name": "Another Ale",
    "Age": 40,
    "Emails": [{
        "Id": 4,
        "Email": "user@skdfh.com"
    }, {
        "Id": 5,
        "Email": "asldkj@als09q834.net"
    }]
}, {
    "Id": 3,
    "Name": "John Doe",
    "Age": 33,
    "Emails": [{
        "Id": 6,
        "Email": "ooaoasdjj@ksjsk0913.org"
    }]
}, {
    "Id": 4,
    "Name": "Mario",
    "Age": 54,
    "Emails": [{}]
}]

---

干杯

如果有多个表在关系方面类似于

电子邮件

，则会导致

自动

方式的行为与预期不符，因此应为此类情况选择

路径

方式。更不用说我觉得它更优雅了。我们注意到，在这种特殊情况下，选项1的性能更好。Sql Server 2012怎么样？Sql Server 2012不支持“for JSON”，所以这些都不适用。此功能仅在SQL Server 2016中引入。不过，您可以使用“FOR XML”，并获得类似的结果，但可以用XML序列化。

[{
    "Id": 1,
    "Name": "Alex",
    "Age": 35,
    "Emails": [{
        "Id": 1,
        "Email": "abc@domain.com"
    }, {
        "Id": 2,
        "Email": "def@domain.com"
    }, {
        "Id": 3,
        "Email": "ghi@domain.net"
    }]
}, {
    "Id": 2,
    "Name": "Another Ale",
    "Age": 40,
    "Emails": [{
        "Id": 4,
        "Email": "user@skdfh.com"
    }, {
        "Id": 5,
        "Email": "asldkj@als09q834.net"
    }]
}, {
    "Id": 3,
    "Name": "John Doe",
    "Age": 33,
    "Emails": [{
        "Id": 6,
        "Email": "ooaoasdjj@ksjsk0913.org"
    }]
}, {
    "Id": 4,
    "Name": "Mario",
    "Age": 54,
    "Emails": [{}]
}]