【发布时间】:2022-01-12 12:50:10
【问题描述】:
我有一个修改一些 JSON 的 SQL 查询。我正在遍历数据并根据迭代修改部分 JSON。
为此,我需要将可变参数传递给 JSON_MODIFY,但由于某种原因,它不起作用!
SET @json = JSON_MODIFY(@ProData, '$.' + @ProKey + '.hasAnswer', CAST(1 as BIT))
我也尝试将整个表达式作为变量传递:
DECLARE @hasAnswerPath VARCHAR(100);
SET @hasAnswerPath = '$.' + @ProKey + '.hasAnswer';
SET @json = JSON_MODIFY(@ProData, @hasAnswerPath, CAST(1 as BIT))
但是它有相同的输出,hasAnswer被添加到JSON的根而不是@ProKey指定的元素中。
这很好用:
SET @json = JSON_MODIFY(@ProData, '$.SomeName1.hasAnswer', CAST(1 as BIT))
就像@ProKey 被忽略了。
完整查询:
BEGIN TRAN
DECLARE @ProID as uniqueidentifier;
DECLARE @ProData as nvarchar(max);
DECLARE @ProKey as varchar(200);
DECLARE ProCursor CURSOR FOR
SELECT Id, [Data] FROM [dbo].[PRO]
OPEN ProCursor;
FETCH NEXT FROM ProCursor INTO @ProID, @ProData;
WHILE @@FETCH_STATUS = 0
BEGIN
DECLARE @json NVARCHAR(max);
DECLARE DataCursor CURSOR FOR
SELECT [key] FROM OPENJSON(@ProData) WHERE type = 5; --5 is object data
OPEN DataCursor;
FETCH NEXT FROM DataCursor INTO @ProKey;
WHILE @@FETCH_STATUS = 0
BEGIN
SET @json=JSON_MODIFY(@ProData, '$.' + @ProKey + '.hasAnswer', CAST(1 as BIT))
SET @json=JSON_MODIFY(@json,'$.' + @ProKey + '.questionType','intro')
FETCH NEXT FROM DataCursor INTO @ProKey;
END;
UPDATE [dbo].[PRO]
SET [Data] = @json
WHERE Id = @ProID
PRINT @json
CLOSE DataCursor;
DEALLOCATE DataCursor;
FETCH NEXT FROM ProCursor INTO @ProID, @ProData;
END
CLOSE ProCursor;
DEALLOCATE ProCursor;
ROLLBACK
示例 JSON:
{
"SomeName1": {
"header": "Some text",
"answer": {
"type": "specified",
"numberValue": 1.0
}
},
"SomeName2": {
"header": "Some text",
"answer": {
"type": "specified",
"numberValue": 4.0
}
},
"SomeName3": {
"header": "Some text",
"answer": {
"type": "specified",
"numberValue": 2.0
}
}
}
预期结果:
},
"SomeName1": {
"header": "Some text",
"answer": {
"type": "specified",
"numberValue": 1.0
}
"hasAnswer": true,
"questionType": "intro",
}
}
实际结果:
},
"SomeName1": {
"header": "Some text",
"answer": {
"type": "specified",
"numberValue": 1.0
}
}
},
"hasAnswer":true,
"questionType":"intro"
}
我在这里做错了什么?
【问题讨论】:
-
MS SQL Server 2019。
-
您未发布的代码可能有问题。有什么有用的可以补充吗?像 json 的样子和意想不到的结果?
-
问题更新了更多信息。
-
它可以很容易地通过创建一个包含列 Id 和 Data 的新表并插入 NewGuid 和 JSON 样本来测试
-
哦,那是复制/粘贴错误!我已经编辑了问题。
标签: sql json sql-server tsql sql-server-2019