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我正在对 SQL Server 2012 中的 XML 数据类型列执行查询。数据示例如下：

<ns:Resume xmlns:ns="http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume">
  <ns:Address>
    <ns:Addr.Type>Home</ns:Addr.Type>
    <ns:Addr.Street>567 3rd Ave</ns:Addr.Street>
    <ns:Addr.Location>
      <ns:Location>
        <ns:Loc.CountryRegion>US </ns:Loc.CountryRegion>
        <ns:Loc.State>MI </ns:Loc.State>
        <ns:Loc.City>Saginaw</ns:Loc.City>
      </ns:Location>
    </ns:Addr.Location>
    <ns:Addr.PostalCode>53900</ns:Addr.PostalCode>
  </ns:Address>
</ns:Resume>

我使用此链接返回 First 一个姓氏，但现在我想返回来自芝加哥和所有简历中的不同州的所有候选人。

对于来自芝加哥的所有候选人，我使用以下代码，但它总是返回列的名称，尽管值存在。你能帮助我吗？

WITH XMLNAMESPACES ('http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume' AS ns)
SELECT 
JobCandidateID,
T.c.value('(ns:Name/ns:Name.First)[1]', 'nvarchar(100)') +N' '+
T.c.value('(ns:Name/ns:Name.Last)[1]', 'nvarchar(100)') as [First & Last Name],
T.c.value('(ns:Address/ns:Addr.Location/ns:Location/ns:Loc.City)[1]', 'nvarchar(100)') as [City Address],
T.c.value('(ns:Employment/ns:Emp.Location/ns:Location/ns:Loc.City)[1]', 'nvarchar(100)') as [City Employment]
FROM   
HumanResources.JobCandidate
CROSS APPLY
[Resume].nodes('/ns:Resume') AS T(c)
where [Resume].exist('/ns:Resume/ns:Address/ns:Addr.Location/ns:Location[ns:Loc.City="Chicago"]')=1;
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1 回答 1

2

对于来自芝加哥的所有候选人：

WITH XMLNAMESPACES ('http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume' AS ns)
SELECT 
JobCandidateID,
T.c.value('(ns:Name/ns:Name.First)[1]', 'nvarchar(100)') +N' '+
T.c.value('(ns:Name/ns:Name.Last)[1]', 'nvarchar(100)') as [First & Last Name],
T.c.value('(ns:Address/ns:Addr.Location/ns:Location/ns:Loc.City)[1]', 'nvarchar(100)') as [City Address],
T.c.value('(ns:Employment/ns:Emp.Location/ns:Location/ns:Loc.City)[1]', 'nvarchar(100)') as [City Employment]
FROM   
HumanResources.JobCandidate
CROSS APPLY
[Resume].nodes('/ns:Resume') AS T(c)
where [Resume].exist('/ns:Resume/ns:Employment/ns:Emp.Location/ns:Location[ns:Loc.City="Chicago"]')=1;

没有交叉应用：

WITH XMLNAMESPACES ('http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume' AS ns)
SELECT 
JobCandidateID,
Resume.value('(/ns:Resume/ns:Name/ns:Name.First)[1]', 'nvarchar(100)') +N' '+
Resume.value('(/ns:Resume/ns:Name/ns:Name.Last)[1]', 'nvarchar(100)') as [First & Last Name],
Resume.value('(/ns:Resume/ns:Address/ns:Addr.Location/ns:Location/ns:Loc.City)[1]', 'nvarchar(100)') as [City Address],
Resume.value('(/ns:Resume/ns:Employment/ns:Emp.Location/ns:Location/ns:Loc.City)[1]', 'nvarchar(100)') as [City Employment]
FROM   
HumanResources.JobCandidate
where Resume.exist('/ns:Resume/ns:Employment/ns:Emp.Location/ns:Location[ns:Loc.City="Chicago"]')=1;

对于所有简历中的不同状态：

WITH XMLNAMESPACES ('http://schemas.microsoft.com/sqlserver/2004/07/adventure-works/Resume' AS ns)
SELECT distinct
Resume.value('(/ns:Resume/ns:Address/ns:Addr.Location/ns:Location/ns:Loc.State)[1]', 'nvarchar(100)') as [State]
FROM   
HumanResources.JobCandidate
于 2015-04-02T13:08:14.447 回答