如何使用 FOR JSON 以这种格式构建 JSON？

Question

我想使用 FOR JSON 为 HTTP Post 调用构建数据负载。 My Source table 可以用这个片段重新创建：

drop table if exists #jsonData;
drop table if exists #jsonColumns;

select
    'carat' [column] 
into #jsonColumns
union
select 'cut' union
select 'color' union
select 'clarity' union
select 'depth' union
select 'table' union
select 'x' union
select 'y' union
select 'z'

select
    0.23 carat
    ,'Ideal' cut
    ,'E' color
    ,'SI2' clarity
    ,61.5 depth
    ,55.0 [table]
    ,3.95 x
    ,3.98 y
    ,2.43 z
into #jsonData
union
select 0.21,'Premium','E','SI1',59.8,61.0,3.89,3.84,2.31 union
select 0.29,'Premium','I','VS2',62.4,58.0,4.2,4.23,2.63 union
select 0.31,'Good','J','SI2',63.3,58.0,4.34,4.35,2.75
;

数据需要格式化如下：

{
    "columns":["carat","cut","color","clarity","depth","table","x","y","z"],
    "data":[
        [0.23,"Ideal","E","SI2",61.5,55.0,3.95,3.98,2.43],
        [0.21,"Premium","E","SI1",59.8,61.0,3.89,3.84,2.31],
        [0.23,"Good","E","VS1",56.9,65.0,4.05,4.07,2.31],
        [0.29,"Premium","I","VS2",62.4,58.0,4.2,4.23,2.63],
        [0.31,"Good","J","SI2",63.3,58.0,4.34,4.35,2.75]
    ]
}

到目前为止我的尝试如下：

select
    (select * from #jsonColumns for json path) as [columns],
    (select * from #jsonData for json path) as [data]
for json path, without_array_wrapper

然而，对象的returns数组而不是值，像这样：

{
    "columns":[
        {"column":"carat"},
        {"column":"clarity"},
        {"column":"color"},
        {"column":"cut"},
        {"column":"depth"},
        {"column":"table"},
        {"column":"x"},
        {"column":"y"},
        {"column":"z"}
    ]...
}

如何限制数组只显示值？

Answer 1

老实说，使用字符串聚合似乎比使用 JSON 功能更容易。

因为您使用的是 SQL Server 2016，您无法访问 STRING_AGG 或 CONCAT_WS，所以代码要长很多。您必须改用 FOR XML PATH 和 STUFF 并手动插入所有分隔符（为什么 CONCAT 表达式中有这么多 ',' ）。结果如下：

DECLARE @CRLF nchar(2) = NCHAR(13) + NCHAR(10);

SELECT N'{' + @CRLF +
       N'    "columns":[' + STUFF((SELECT ',' + QUOTENAME(c.[name],'"')
                                   FROM tempdb.sys.columns c
                                        JOIN tempdb.sys.tables t ON c.object_id = t.object_id
                                   WHERE t.[name] LIKE N'#jsonData%' --Like isn't needed if not a temporary table. Use the literal name.
                                   ORDER BY c.column_id ASC
                                   FOR XML PATH(N''),TYPE).value('.','nvarchar(MAX)'),1,1,N'') + N'],' + @CRLF +
       N'    "data":[' + @CRLF +
       STUFF((SELECT N',' + @CRLF +
                     N'       ' + CONCAT('[',JD.carat,',',QUOTENAME(JD.cut,'"'),',',QUOTENAME(JD.color,'"'),',',QUOTENAME(JD.clarity,'"'),',',JD.depth,',',JD.[table],',',JD.x,',',JD.y,',',JD.z,']')
              FROM #jsonData JD
              ORDER BY JD.carat ASC
              FOR XML PATH(N''),TYPE).value('.','nvarchar(MAX)'),1,3,N'') + @CRLF +
      N'    ]' + @CRLF +
      N'}';

DB<>Fiddle

如何使用 FOR JSON 以这种格式构建 JSON？

How can I use FOR JSON to build JSON in this format?

json

sql-server

sql-server-2016

for-json