Question

我需要从一个查询创建一个JSON输出，该查询使用具有一对多关系的两个表之间的内连接我希望将辅助表的值嵌套为主表的数组属性。

考虑以下示例：

DECLARE @Persons AS TABLE
(
    person_id int primary key,
    person_name varchar(20)
)

DECLARE @Pets AS TABLE
(
    pet_owner int, -- in real tables, this would be a foreign key
    pet_id int  primary key,
    pet_name varchar(10)
)

INSERT INTO @Persons (person_id, person_name) VALUES
(2, 'Jack'),
(3, 'Jill')

INSERT INTO @Pets (pet_owner, pet_id, pet_name) VALUES
(2, 4, 'Bug'),
(2, 5, 'Feature'),
(3, 6, 'Fiend')

并查询：

DECLARE @Result as varchar(max)
SET @Result =
(
SELECT  person_id as [person.id],
        person_name as [person.name],
        pet_id as [person.pet.id],
        pet_name as [person.pet.name]
FROM @Persons 
JOIN @Pets ON person_id = pet_owner
FOR JSON PATH, ROOT('pet owners')
)

PRINT @Result

这将打印以下JSON：

{
    "pet owners":
    [
    {"person":{"id":2,"name":"Jack","pet":{"id":4,"name":"Bug"}}},
    {"person":{"id":2,"name":"Jack","pet":{"id":5,"name":"Feature"}}},
    {"person":{"id":3,"name":"Jill","pet":{"id":6,"name":"Fiend"}}}
    ]
}

但是，我想将宠物数据作为所有者数据中的数组：

{
    "pet owners":
    [
        {
            "person":
            {
                "id":2,"name":"Jack","pet":
                [
                    {"id":4,"name":"Bug"},
                    {"id":5,"name":"Feature"}
                ]
            }
        },
        {
            "person":
            {
                "id":3,"name":"Jill","pet":
                {"id":6,"name":"Fiend"}
            }
        }
    ]
}

我该怎么做？

Answer 1

您可以使用以下查询：

SELECT pr.person_id AS [person.id], pr.person_name AS [person.name],
    (
        SELECT pt.pet_id AS id, pt.pet_name AS name 
        FROM @Pets pt WHERE pt.pet_owner=pr.person_id 
        FOR JSON PATH
    ) AS [person.pet]
FROM @Persons pr 
FOR JSON PATH, ROOT('pet owners')

有关详细信息，请参阅https://blogs.msdn.microsoft.com/sqlserverstorageengine/2015/10/09/returning-child-rows-formatted-as-json-in-sql-server-queries/

Answer 2

使用深度嵌套的数组，子查询很快就无法处理：

d = {1: 2,
2:3,
3: 4,
5:6,
6:7}

我创建了一个关系（非json）视图，该视图连接了我的所有表，并在列别名中嵌入了json结构，就像 for json path 一样。但是我也有[]表示json节点是一个数组。像这样：

d= { 1: 4 , 2 :4 , 3:4 , 5:7 , 6:7}

我编写了一个存储过程，该过程在非json视图中创建一个json视图，该视图解析关系视图的列名并使json变得漂亮。见下文。用您的关系视图的名称调用它，它会创建一个视图。尚未经过全面测试，但对我有用。唯一需要说明的是，表需要具有称为 id的 id 列。它使用 string_agg（）和 json_array（） sql版本需要非常新。它还设置为在根目录中返回一个数组。需要调整以返回对象。

select id,foo, (select id, bar, (select ... for json path) things, (select...) more_things) yet_more, select(...) blarg

Answer 3

我按照@Razvan Socol 制作了以下 json 格式。

JSON

  [
      "domain_nm": "transactions",
      "tables": [
        {
          "tableName": "transactions_details",
          cols: [
            {
              "col_nm": "audit_transactions_details_guid",
              "col_data_typ": "string"
            }
          ]
        }
      ]
    ]

SQL

select outer1.DOMAIN_NM as domain_nm,
    (select inner2.TBL_NM as tableName,
            (select inner1.COL_NM as col_nm, inner1.COL_DATA_TYP as col_data_typ
            from ONBD_MTDT.CDM_TBL inner1
            where inner1.TBL_NM=inner2.TBL_NM
            FOR JSON PATH ) as cols
    from ONBD_MTDT.CDM_TBL inner2 
    where inner2.DOMAIN_NM=outer1.DOMAIN_NM
    group by inner2.DOMAIN_NM,inner2.TBL_NM
    FOR JSON PATH ) as tables
from ONBD_MTDT.CDM_TBL outer1 
group by outer1.DOMAIN_NM
FOR JSON PATH

使用FOR JSON PATH创建嵌套的JSON数组

3 个答案: