如何从 XML 内容中拆分字符串并获取所需的值

Question

大家好，我正在转换一个 xml 内容并将其插入到一个 table 变量中，如下所示

DECLARE @iDoc int

SET @XMLData = '<NewDataSet>
  <Table>
    <DataId>2324205.3933251.7336404</DataId>
    <IsVisible>true</IsVisible>
    <Notes />
    <Marks>85.5</Marks>
   </Table>
   </NewDataSet>' 

EXEC sp_xml_preparedocument @idoc OUTPUT, @XMLData 

SELECT DataId FROM OPENXML(@idoc, 'NewDataSet/Table', 1)
WITH   (DataId NVARCHAR(250) 'DataId')```

I would like to split the dot value and retrieve the the first value, can some one help me how to can I do that with in XML

IsVisible is a bit filed, Marks is deicmal like that I will have

Answer 1

从SQLServer 2005开始，在处理XML数据类型时，最好使用基于w3c标准的XQuery语言。保留 Microsoft 专有 OPENXML 及其同伴 sp_xml_preparedocument 和 sp_xml_removedocument 只是为了与过时的 SQL 向后兼容 Server 2000。它们的使用减少到极少数边缘情况。强烈建议重新编写 SQL 并将其切换为 XQuery。

SQL

-- DDL and sample data population, start
DECLARE @tbl TABLE (ID INT IDENTITY PRIMARY KEY, Notes NVARCHAR(MAX));
INSERT INTO @tbl (Notes) VALUES
(N'<NewDataSet>
    <Table>
        <DataId>2324205.3933251.7336404</DataId>
    </Table>
</NewDataSet>');
-- DDL and sample data population, end

WITH rs AS
(
    SELECT * 
        , TRY_CAST(Notes AS XML).value('(/NewDataSet/Table/DataId/text())[1]', 'VARCHAR(MAX)') AS x
    FROM @tbl
)
SELECT * 
    , LEFT(x, CHARINDEX('.', x) - 1) AS [After]
    , PARSENAME(x, 3) AS [After2]
FROM rs;

输出

+-------------------------+---------+
|         Before          |  After  |
+-------------------------+---------+
| 2324205.3933251.7336404 | 2324205 |
+-------------------------+---------+

如何从 XML 内容中拆分字符串并获取所需的值

How to split string from XML content and get the required value

sql-server

openxml