Question

虽然这适用于检索显示here的local-name(.)节点名称，但我无法获得一个包含简历名字和恢复每个求职者姓氏的表格。

    USE AdventureWorks2012;

    WITH XMLNAMESPACES(
     'http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume'
      AS ns)

      SELECT 
        T.rows.value('(ns:Name.First)[1]', 'nvarchar(100)') AS firstName, 
        T.rows.value('(ns:Name.Last)[1]', 'nvarchar(100)') AS lastName
      FROM HumanResources.JobCandidate
      CROSS APPLY
        Resume.nodes('//ns:Name/*') AS T(rows);

错误消息列出了上下文中的节点，包括似乎是目标节点的节点。

XQuery [HumanResources.JobCandidate.Resume.value（）]：没有命名的元素＆＃34; {http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume} Name.First ＆＃34; 在类型中＆＃34; element（ns {http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume}：Name.Prefix，xs：string）| 元素（ns {http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume}： Name.First ，xs：string）

...

element（ns {http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume}：Name.Last，xs：string）| 元件（NS {http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume}：Name.Suffix，XS：字符串）。＆＃34;

问题element是否达到了水平？

Answer 1

问题是上下文太深了。通过从/*路径移除nodes来解决此问题。

WITH XMLNAMESPACES(
 'http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume'
  AS ns)

  SELECT 
    T.rows.value('(ns:Name.First)[1]', 'nvarchar(100)') AS firstName, 
    T.rows.value('(ns:Name.Last)[1]', 'nvarchar(100)') AS lastName
  FROM HumanResources.JobCandidate
  CROSS APPLY
    Resume.nodes('//ns:Name/*') AS T(rows);

T-SQL XQuery节点值路径不匹配

1 个答案: