json - 将来自多行的 JSON 对象组合成一个对象

Question

CREATE TABLE t(Id int, typee nvarchar(50), jsonStr nvarchar(max));
INSERT INTO t(Id, typee, jsonStr) VALUES
(3786, 'APV', '{"1":1,"3":3,"4":24,"5":95}'),
(3786, 'VN', '{"1":3,"5":25}');

-- Expected result
-- {"APV": {"1":1,"3":3,"4":24,"5":95}, "VN":{"1":3,"5":25} }

SELECT Id,(
    SELECT CASE WHEN typee = 'VN'  THEN jsonStr END AS [VN]
         , CASE WHEN typee = 'VO'  THEN jsonStr END AS [VO]
         , CASE WHEN typee = 'APV' THEN jsonStr END AS [APV]
    FROM t AS x
    WHERE x.Id = t.Id
    FOR JSON AUTO, WITHOUT_ARRAY_WRAPPER
) AS TEST1

FROM t
GROUP BY Id

DB<>小提琴

我想得到如下输出：

{
  "APV": {
    "1": 1,
    "3": 3,
    "4": 24,
    "5": 95
  },
  "VN": {
    "1": 3,
    "5": 25
  }
}

Answer 1

FOR JSON将字符串视为字符串，即使它表示有效的 JSON。您需要使用JSON_QUERY将 JSON 字符串转换为实际的 JSON 对象。MIN需要将多行合并为一：

SELECT Id, (
  SELECT JSON_QUERY(MIN(CASE WHEN typee = 'APV' THEN jsonStr END)) AS [APV]
       , JSON_QUERY(MIN(CASE WHEN typee = 'VN'  THEN jsonStr END)) AS [VN]
       , JSON_QUERY(MIN(CASE WHEN typee = 'VO'  THEN jsonStr END)) AS [VO]
  FROM t AS x
  WHERE x.id = t.id
  FOR JSON AUTO, WITHOUT_ARRAY_WRAPPER
)
FROM t
GROUP BY Id

db<>fiddle 上的演示

Answer 2

问题是您正在存储 JSON，然后将其格式化为 JSON 进一步返回。您需要存储非 JSON 数据才能按照您想要的方式进行操作。因此，将其视为字符串似乎更容易：

SELECT t.id,
       '{' + STRING_AGG( '"' +t.typee + '": ' + t.jsonStr,',') WITHIN GROUP (ORDER BY typee) + '}'
FROM t
GROUP BY t.id;

使用FOR XML PATH：

SELECT t1.id,
       '{' + STUFF((SELECT ',"' + t2.typee + '": ' + t2.jsonStr
                    FROM t t2
                    WHERE t2.Id = t1.id
                    FOR XML PATH(N''),TYPE).value('.','nvarchar(MAX)'),1,1,'') + '}'
FROM t t1
GROUP BY t1.id;