获取祖先节点+的XQuery-SQL

Question

的数据这是我的XML看起来像：

<Product sequence_number="1" number="1543448904" id="S1" unit_number="1"> 

    <consumer_narrative name="GLENN,GREGORY" date_filed="02/13/2009"> 
     <message type="Consumer Comments">THE CONSUMER STATES THIS WAS NOT </message> 
     <message type="Consumer Comments">THE PRODUCT REQUESTED.</message> 
    </consumer_narrative></Product> <Product sequence_number="2" number="1543448905" id="S1" unit_number="1"> 

    <consumer_narrative name="JOHN,GORDON" date_filed="08/23/2009"> 
     <message type="Consumer Comments">THE CONSUMER STATES THAT</message> 
     <message type="Consumer Comments">WRONG PRODUCT WAS SENT.</message> 
    </consumer_narrative> </Product> 

我的查询：

SELECT 

    tab.col.value('../@number', 'varchar(30)') [Claim Number], 
    tab.col.value('../@name', 'varchar(30)') [Name], 
    tab.col.value('../@date_filed', 'varchar(30)') [DateField], 
    tab.col.value('@type', 'varchar(50)') [Type], 
    tab.col.value('.', 'varchar(250)') [CustomerComments] 

FROM XMLTABLE AS B 
CROSS APPLY xmldocument.nodes('//Product/consumer_narrative/message') tab(col) 
WHERE B.XMLId = 123 

给我 “空” 的索赔数目。我应该用什么来代替../@number来获取索赔号。

Answer 1

另一个 “../”

第一../只收了你的高达元素consumer_narrative