JSON 每个选定日期的根

Question

我每天都在为嵌套根的问题而苦恼（这是我 table 的一个元素）。我想从 table Day.

获取嵌套的 Key: value 对

这是我的结果：

[
   {
      "date":"2022-01-10T00:00:00",
      "title":"Coloring",
      "start_time":"2022-01-10T12:00:00",
      "end_time":"2022-01-10T13:00:00"
   },
   {
      "date":"2021-12-28T00:00:00",
      "title":"Coloring",
      "start_time":"2021-12-27T15:20:00",
      "end_time":"2021-12-27T16:00:00"
   },
   {
      "date":"2021-12-28T00:00:00",
      "title":"Coloring",
      "start_time":"2021-12-27T12:20:00",
      "end_time":"2021-12-27T14:00:00"
   }
]

预期结果如下：

{
   "2022-01-10":[
      {
         "date":"2022-01-10T00:00:00",
         "title":"Coloring",
         "start_time":"2022-01-10T12:00:00",
         "end_time":"2022-01-10T13:00:00"
      }
   ],
   "2021-12-28":[
      {
         "date":"2021-12-28T00:00:00",
         "title":"Coloring",
         "start_time":"2021-12-27T15:20:00",
         "end_time":"2021-12-27T16:00:00"
      },
      {
         "date":"2021-12-28T00:00:00",
         "title":"Coloring",
         "start_time":"2021-12-27T12:20:00",
         "end_time":"2021-12-27T14:00:00"
      }
   ]
}

天table:

id  date

0   2021-12-01 00:00:00.0000000
1   2021-12-02 00:00:00.0000000
2   2021-12-03 00:00:00.0000000
... ...

这是我的活动 Table:

id  title       start_time                  end_time                 day_of_timetable  service_id
0   Coloring    2022-01-10 12:00:00.0000000 2022-01-10 13:00:00.0000000 0   0
1   Coloring    2021-12-27 15:20:00.0000000 2021-12-27 16:00:00.0000000 1   0
2   Coloring    2021-12-27 12:20:00.0000000 2021-12-27 14:00:00.0000000 1   0

这是我的 day_of_timetable table:

id  day_id  end_user_id
0   40      1
1   27      1

这是我的代码

    select date, e.title, e.start_time, e.end_time,  e.day_of_timetable_id
    from day 
    join day_of_timetable dot on day.id = dot.day_id
    join end_user eu on dot.end_user_id = eu.id
    join event e on dot.id= e.day_of_timetable_id
    where eu.id = 1 for json path

Answer 1

请尝试以下解决方案。

SQL

-- DDL and sample data population, start
DECLARE @tbl TABLE (id INT IDENTITY PRIMARY key, [date] DATETIME, title VARCHAR(20), start_time DATETIME, end_time DATETIME);
INSERT INTO @tbl (date, title, start_time, end_time) VALUES
('2022-01-10T00:00:00', 'Coloring','2022-01-10T12:00:00','2022-01-10T13:00:00'),
('2021-12-28T00:00:00', 'Coloring','2021-12-27T15:20:00','2021-12-27T16:00:00'),
('2021-12-28T00:00:00', 'Coloring','2021-12-27T12:20:00','2021-12-27T14:00:00');
-- DDL and sample data population, end

SELECT * FROM @tbl;

SELECT CONCAT(
    N'{',
    STUFF(
      (
        SELECT CONCAT(N',"', CAST(k.[date]AS DATE), '":', c.[Json])
        FROM @tbl AS k
        CROSS APPLY (
            SELECT [date], title, start_time, end_time
            FROM @tbl
            WHERE [date] = k.[date]
            FOR JSON PATH
        ) c([Json])
        GROUP BY [date], c.[Json]
        ORDER BY [date] DESC
        FOR XML PATH('')
        ), 1, 1, N''
   ),  
   N'}'
) 
AS JsonOutput;

输出

{
    "2022-01-10": [
        {
            "date": "2022-01-10T00:00:00",
            "title": "Coloring",
            "start_time": "2022-01-10T12:00:00",
            "end_time": "2022-01-10T13:00:00"
        }
    ],
    "2021-12-28": [
        {
            "date": "2021-12-28T00:00:00",
            "title": "Coloring",
            "start_time": "2021-12-27T15:20:00",
            "end_time": "2021-12-27T16:00:00"
        },
        {
            "date": "2021-12-28T00:00:00",
            "title": "Coloring",
            "start_time": "2021-12-27T12:20:00",
            "end_time": "2021-12-27T14:00:00"
        }
    ]
}

Answer 2

在 SQL 服务器的现代版本中，您可以为此使用 STRING_AGG，一次按日期聚合，然后再次聚合整个事件。

您还可以获得仅查询一次 table 的好处（APPLY 仅指回外部 table）。

Note that if you are using arbitrary text as the keys, you should use STRING_ESCAPE to escape them

SELECT
  JsonOutput = N'{' + STRING_AGG(N'"' + CAST(t.date AS nvarchar(30)) + N'":' + t.[Json], N',') WITHIN GROUP (ORDER BY [date] DESC) + N'}'
FROM (
    SELECT
      date = CAST(t.date AS date),
      Json = STRING_AGG(N'[' + c.Json + N']', ',') WITHIN GROUP (ORDER BY [date] DESC)
    FROM @tbl t
    CROSS APPLY (
        SELECT t.date, t.title, t.start_time, t.end_time
        FOR JSON PATH, WITHOUT_ARRAY_WRAPPER
    ) c(Json)
    GROUP BY CAST(t.date AS date)
) t;

db<>fiddle

Answer 3

分组日table，按id排序，按日期字段的日期部分排序，日期部分作为字段头。每个字段的值是对应组的table个序列，然后我们将它们转化为JSON格式。 SQL 只能assemble 将JSON 字符串按指定格式手动。该声明冗长且难以阅读。一般的替代方法是从数据库中提取原始数据并在 Python 或 SPL 中对其进行处理。 SPL，open-source Java 包，更容易集成到 Java 程序中并生成简单的代码。它只用两行代码就完成了任务：

	A
1	=MSSQL.query("select date,title,start_time,end_time from json order by id")
2	=json(transpose(A1.group@o(date(date)).run(~=[date(date)]

JSON 每个选定日期的根

JSON with roots for every selected day

sql

tsql

sql-server

json