Question

我试图获得的是一个简单的SQL语句来构建：

 {"status":{"code":404,"message":"Not found"},"otherthing":20}

如果我设为：

DECLARE @ReturnJSON nvarchar(max)

SET @ReturnJSON = (
    SELECT ( 
        SELECT 404 as [code]
              ,'Not found' as [message] 
               FOR JSON PATH ) as [status]
       , 20 as [otherthing]
   FOR JSON PATH, WITHOUT_ARRAY_WRAPPER) ;

SELECT @ReturnJSON

我在数组包装器下获得第二级，如下所示：

{"status":[{"code":404,"message":"Not found"}],"otherthing":20}

但如果我在第二级添加WITHOUT_ARRAY_WRAPPER ......

DECLARE @ReturnJSON nvarchar(max)

SET @ReturnJSON = (
    SELECT ( 
        SELECT 404 as [code]
              ,'Not found' as [message] 
               FOR JSON PATH, WITHOUT_ARRAY_WRAPPER ) as [status]
       , 20 as [otherthing]
   FOR JSON PATH, WITHOUT_ARRAY_WRAPPER) ;

SELECT @ReturnJSON

发生了一些有趣的事情：

{"status":"{\"code\":404,\"message\":\"Not found\"}","otherthing":20}

我遗失了一些东西，我知道，当然，但是我不能看到

Answer 1

我认为Matheno（在评论中）是对的：显然问题是FOR JSON逃脱了你的文本。为了防止这种不必要的内部JSON转义，你可以用JSON_QUERY()包裹它：

DECLARE @ReturnJSON nvarchar(max)
DECLARE @innerJSON nvarchar(max)

set @innerJSON =(        SELECT 404 as [code]
              ,'Not found' as [message] 
               FOR JSON PATH, WITHOUT_ARRAY_WRAPPER )

SET @ReturnJSON = (
    SELECT ( 
        JSON_QUERY(@innerJSON)
               ) as [status]
       , 20 as [otherthing]
   FOR JSON PATH, WITHOUT_ARRAY_WRAPPER) ;

SELECT @ReturnJSON

输出：

{"status":{"code":404,"message":"Not found"},"otherthing":20}

Answer 2

这不是您问题的确切答案，但我希望它能为您解决问题。

您可以构建不带嵌套查询的预期输出，只需使用属性名称定义层次结构即可，

DECLARE @ReturnJSON nvarchar(max)

SET @ReturnJSON = (
    SELECT 
        404 as [status.code]
        ,'Not found' as [status.message]
        , 20 as [otherthing]
    FOR JSON PATH, WITHOUT_ARRAY_WRAPPER) ;

SELECT @ReturnJSON

JSON没有较低级别的数组包装器

2 个答案: