SQL Server - OPENXML 如何获取属性值 [英] SQL Server - OPENXML how to get attribute value

问题描述

我有以下 XML:

<Field FieldRowId="1000">
    <Items>
        <Item Name="CODE"/>
        <Item Name="DATE"/>
    </Items>
</Field>

我需要使用 OPENXML 获取 FieldRowId.到目前为止我拥有的 SQL:

I need to get the FieldRowId using OPENXML. The SQL i have so far:

INSERT INTO @tmpField
      ([name], [fieldRowId])
SELECT [Name], --Need to get row id of the parent node
 FROM OPENXML (@idoc, '/Field/Items/Item', 1) 

推荐答案

我向 xml 添加了一个根节点.并演示了如何抓取 ID.我假设您在 xml 中有多个字段元素.这假设您有起始 XML；你是否给了一个 Item 并且必须向上遍历?

I added a root node to the xml. and demonstrated grabbing the ID. I'm assuming you have more than one field element in the xml. This is assuming you have the starting XML; are you given an Item and have to traverse upwards?

DECLARE @T varchar(max) 
SET @T = 
'<root>
    <Field FieldRowId="1000">
        <Items>
            <Item Name="CODE"/>
            <Item Name="DATE"/>
        </Items>
    </Field>
    <Field FieldRowId="2000">
        <Items>
            <Item Name="CODE"/>
            <Item Name="DATE"/>
        </Items>
    </Field>
</root>'

DECLARE @X xml

SET @X = CAST(@T as xml)
SELECT Y.ID.value('../../@FieldRowId', 'int') as FieldID, 
       Y.ID.value('@Name', 'varchar(max)') as "Name"
FROM @X.nodes('/root/Field/Items/Item') as Y(ID)

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