如何从 SQL 服务器中的 JSON 获取每个对象的值

Question

DECLARE @json NVARCHAR(1000);
            
    SELECT @json = N'
       {
        "nodeinid": 1345,
        "data": [
          {
            "nodenametext": "Plan for project",
            "nodedesctext": "plan description for execution",
            "nodetypecode": "PLN",
            "statuscode": "IP",
            "prioritycode": "NI",
            "linkeditemid": 0,
            "linkeditemtypecode": "TSK",
            "startdttm": "2021-07-15T09:53:47.9575771",
            "duedttm": "2021-07-15T09:53:47.9575771",
            "hidefromclientindc": "true",
            "draftindc": "no",
            "inworkspaceid": 5678,
            "parentnodeid": 0,
            "categorytext": [
              {
                "categoryid": 111,
                "categoryidname": "college"
              },
              {
                "categoryid": 222,
                "categoryidname": "office"
              }
            ],
            "assigntotext": [
              {
                "workspaceid": 567,
                "roleid": 4545,
                "accesstypecode": "ass",
                "mailurl": "saram@gmail"
              },
              {
                "workspaceid": 977,
                "roleid": 67,
                "accesstypecode": "ass",
                "mailurl": "sarfdam@gmail"
              }
            ]
          }
        ]
      }
    ';

使用的查询：

    SELECT JSON_Value (c.value, '$.nodeinid') as nodein_id 
       ,JSON_Value (P.Value, '$.workspaceid') as workspace_id
     , JSON_Value (p.value, '$.accesstypecode') as accesstype_code
     , JSON_Value (Q.value, '$.categoryid') as category_id
     , JSON_Value (Q.value, '$.categoryidname') as categoryid_name
      
    FROM OPENJSON (@json, '$.data') as C
    CROSS APPLY OPENJSON (C.Value, '$.assigntotext') as P
    CROSS APPLY OPENJSON (C.Value, '$.categorytext') as Q;

预期样本输出：

workspace_id accesstype_code
567          ass
977          ass

结果：查询抛出以下错误

Msg 13609, Level 16, State 4, Line 50
JSON text is not properly formatted. Unexpected character '"' is found at position 998.

如果我保留其中一个对象（assigntotext、categorytext）它工作正常。
请更正查询以获取行和列格式的值。"

Answer 1

好像是nvarchar(1000)不够用，输入的JSON被截断了。将 @json 变量的数据类型更改为 nvarchar(max)。您也可以尝试使用以下方法解析输入 JSON：

DECLARE @json NVARCHAR(max);
SELECT @json = N'... JSON data ...' 

SELECT j2.*
FROM OPENJSON(@json, '$.data') WITH (
   assigntotext nvarchar(max) '$.assigntotext' AS JSON
) j1
OUTER APPLY OPENJSON(j1.assigntotext) WITH (
   workspace_id int '$.workspaceid',
   accesstype_code varchar(10) '$.accesstypecode'
) j2

结果：

workspace_id accesstype_code
----------------------------
567          ass
977          ass

如何从 SQL 服务器中的 JSON 获取每个对象的值

how to get the values of the each object from the JSON in SQL SERVER

sql

sql-server

json

open-json