Question

我试图从我的SQL数据库构建一个JSON序列化的键/值对项列表（compat level 140）。诀窍是值可以是任何值：数字，字符串，null或其他JSON对象。

应该能够看起来像这样：

[{"key":"key1","value":"A String"},{"key":"key2","value":{"InnerKey":"InnerValue"}}]

然而，SQL似乎迫使我选择字符串或一个对象。

SELECT
       [key] = kvp.[key],
       [value] = CASE
              WHEN ISJSON(kvp.[value]) = 1 THEN JSON_QUERY(kvp.[value])
              ELSE '"' + kvp.[value] + '"'  -- See note below
              END
FROM (VALUES
        ('key1', 'This value is a string')
       ,('key2', '{"description":"This value is an object"}')
       ,('key3', '["This","value","is","an","array","of","strings"]')
       ,('key4', NULL)

       -- Without these lines, the above 4 work fine; with either of them, even those 4 are broken
       --,('key5', (SELECT [description] = 'This value is a dynamic object' FOR JSON PATH, WITHOUT_ARRAY_WRAPPER))
       --,('key6', JSON_QUERY((SELECT [description] = 'This value is a dynamic object' FOR JSON PATH, WITHOUT_ARRAY_WRAPPER)))
) AS kvp([key], [value])
FOR JSON PATH

我是否尝试做一些SQL无法支持的事情，或者我只是缺少使其正常工作的正确语法？

*请注意，添加双引号似乎不应该是必要的。但没有这些，SQL无法包装字符串并生成错误的JSON：

[{"key":"key1","value":This value is a string},...

Answer 1

如果您的查询被修改为此，则可以：

SELECT
   [key] = kvp.[key],
   [value] = ISNULL(
          JSON_QUERY(CASE WHEN ISJSON(kvp.[value]) = 1 THEN kvp.[value] END),
          '"' + STRING_ESCAPE(kvp.[value], 'json') + '"'
   )
FROM (VALUES
   ('key1', 'This value is a "string"')
   ,('key2', '{"description":"This value is an object"}')
   ,('key3', '["This","value","is","an","array","of","strings"]')
   ,('key4', NULL)

   -- These now work
   ,('key5', (SELECT [description] = 'This value is a dynamic object' FOR JSON PATH, WITHOUT_ARRAY_WRAPPER))
   ,('key6', JSON_QUERY((SELECT [description] = 'This value is a dynamic object' FOR JSON PATH, WITHOUT_ARRAY_WRAPPER)))
) AS kvp([key], [value])
FOR JSON PATH, INCLUDE_NULL_VALUES

当然，如果value是int，这还不够。另外，我无法解释为什么你的工作无效。

你可以在SQL JSON中混合字符串和对象输出吗？

1 个答案: