如何使用 json sql server 获取此值？

Question

如何在此 json 文件中获取question_id字段？我试过了，但它返回null。

DECLARE @json NVARCHAR(MAX)
SET @json =   
  N'{ 
            "solution": "xxxxxxxxxxxxxxxxxxxxx",
            "options": [
                {
                    "choice_id": 205073,
                    "choice": "aaaa"
                },
                {
                    "choice_id": 205074,
                    "choice": "bbbb"
                },
                {
                    "choice_id": 205075,
                    "choice": "cccc"
                },
                {
                    "choice_id": 205076,
                    "choice": "dddd"
                }
            ],
            "question_id": 12345
        }'  

SELECT * FROM  
 OPENJSON ( @json, '$.options')
WITH (      
              choice_id varchar(8000) '$.choice_id',
              question_id int '$.question_id'
 ) 

结果

Answer 1

你把question_id它当成一个孩子，options而实际上它是一个兄弟姐妹options。

您可以json_value在列列表中使用来获取 的值question_id。

select O.choice_id,
       json_value(@json, '$.question_id') as question_id
from openjson(@json, '$.options')
     with (
            choice_id varchar(8000) '$.choice_id'
          ) as O;

如果根上有一组对象，则首先需要切碎$以获取每个对象的一行，然后使用 across apply来获取对象。

declare @json nvarchar(max)
set @json = N'
[
   {
      "solution":"xxxxxxxxxxxxxxxxxxxxx",
      "options":[
         {
            "choice_id":205073,
            "choice":"aaaa"
         },
         {
            "choice_id":205074,
            "choice":"bbbb"
         },
         {
            "choice_id":205075,
            "choice":"cccc"
         },
         {
            "choice_id":205076,
            "choice":"dddd"
         }
      ],
      "question_id":12345
   },
   {
      "solution":"xxxxxxxxxxxxxxxxxxxxx",
      "options":[
         {
            "choice_id":205073,
            "choice":"aaaa"
         },
         {
            "choice_id":205074,
            "choice":"bbbb"
         },
         {
            "choice_id":205075,
            "choice":"cccc"
         },
         {
            "choice_id":205076,
            "choice":"dddd"
         }
      ],
      "question_id":22345
   }
]'  

select O.choice_id,
       json_value(@json, '$.question_id') as question_id
from openjson(@json, '$.options')
     with (
            choice_id varchar(8000) '$.choice_id'
          ) as O;