sql-server-2005 - 从 XML 文件中选择数据作为 SQL 中的表

Question

有人可以向我展示一些用于查询 xml 文件的 TSQL，就好像它是一个表一样吗？

该文件位于服务器上，“C:\xmlfile.xml”

并且包含

<ArrayOfSpangemansFilter xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <SpangemansFilter>
        <FilterID>1219</FilterID>
        <Name>Fred</Name>
        <Code>510</Code>
        <Department>N</Department>
        <Number>305327</Number>
    </SpangemansFilter>
    <SpangemansFilter>
        <FilterID>3578</FilterID>
        <Name>Gary</Name>
        <Code>001</Code>
        <Department>B</Department>
        <Number>0692690</Number>
    </SpangemansFilter>
    <SpangemansFilter>
        <FilterID>3579</FilterID>
        <Name>George</Name>
        <Code>001</Code>
        <Department>X</Department>
        <Number>35933</Number>
    </SpangemansFilter>
</ArrayOfSpangemansFilter>

我想要的示例输出

FilterID    |Name       |Code       |Department             |Number
-------------------------------------------------------------------
1219        |Fred       |510        |N                      |305327
3578        |Gary       |001        |B                      |0692690
3579        |George     |001        |X                      |35933

Answer 1

set @xmlData='<?xml version="1.0"?>
<ArrayOfSpangemansFilter xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<SpangemansFilter>
<FilterID>1219</FilterID>
<Name>Fred</Name>
<Code>510</Code>
<Department>N</Department>
<Number>305327</Number>
</SpangemansFilter>
<SpangemansFilter>
<FilterID>3578</FilterID>
<Name>Gary</Name>
<Code>001</Code>
<Department>B</Department>
<Number>0692690</Number>
</SpangemansFilter>
<SpangemansFilter>
<FilterID>3579</FilterID>
<Name>George</Name>
<Code>001</Code>
<Department>X</Department>
<Number>35933</Number>
</SpangemansFilter>
</ArrayOfSpangemansFilter>'


SELECT 
  ref.value('FilterID[1]', 'int') AS FilterID ,
  ref.value('Name[1]', 'NVARCHAR (10)') AS Name ,
  ref.value('Code[1]', 'NVARCHAR (10)') AS Code ,
  ref.value('Department[1]', 'NVARCHAR (3)') AS Department,
  ref.value('Number[1]', 'int') AS Number      
FROM @xmlData.nodes('/ArrayOfSpangemansFilter/SpangemansFilter') 
xmlData( ref )

产品:

FilterID    Name       Code       Department Number
----------- ---------- ---------- ---------- -----------
1219        Fred       510        N          305327
3578        Gary       001        B          692690
3579        George     001        X          35933

注意:需要 [1] 来指示您要选择序列的第一个值，因为查询可能会每行返回多个匹配值(想象一下您的 XML 包含多个 FilterID)根据 SpangemansFilter)。

我认为了解这一点很有用，所以我在 Google 上搜索并阅读了很多帖子，直到找到 this one.

更新从文件加载:

DECLARE @xmlData XML
SET @xmlData = (
  SELECT * FROM OPENROWSET (
    BULK 'C:\yourfile.xml', SINGLE_CLOB
  ) AS xmlData
)

选择@xmlData