我已经看到过多个带有此示例的帖子,但由于某种原因,我尝试的所有选项都返回一个空的XmlNodeList。我错过了什么?
我的XML:
<string xmlns='http://ws.jobboard.com/resumes/'>
<Packet>
<Errors />
<PageNumber>1</PageNumber>
<SearchTime>02/25/2014 10:11:47</SearchTime>
<FirstRec>1</FirstRec>
<LastRec>100</LastRec>
<Hits>206816</Hits>
<MaxPage>40</MaxPage>
<Results>
<ResumeResultItem_V3>
<ContactEmail></ContactEmail>
<ContactName>Lisa Yada</ContactName>
<HomeLocation>US-CO-Lakewood</HomeLocation>
<LastUpdate>2014/2/24</LastUpdate>
<ResumeTitle>Web Design & Developement</ResumeTitle>
<JobTitle>Web Design & Developement</JobTitle>
<RecentEmployer>Some Company</RecentEmployer>
<RecentJobTitle>Web Design</RecentJobTitle>
<RecentPay>0</RecentPay>
<ResumeID>RHV47X78Z5CN56</ResumeID>
<UserDID>UHT6R86LL8C581</UserDID>
<ContactEmailMD5>93ce261b843f58962f9</ContactEmailMD5>
<ActionType />
<HighestDegree>Bachelor's Degree</HighestDegree>
<MonthsOfExperience>40</MonthsOfExperience>
<LastActivity>2/24/2014 1:22:19 PM</LastActivity>
</ResumeResultItem_V3> ...
和我的代码:
XmlDocument jobsdoc = new XmlDocument();
jobsdoc.LoadXml(xmlstring);
XmlNodeList xnList = jobsdoc.SelectNodes("/string/Packet/Results/ResumeResultItem_V3");
foreach (XmlNode xn in xnList)
{
DataRow dr = table.NewRow();
dr["ContactName"] = xn["ContactName"].InnerText;
dr["ResumeTitle"] = xn["ResumeTitle"].InnerText;
...
table.Rows.Add(dr);
}
return table;
我已经尝试了许多"/string/Packet/Results/ResumeResultItem_V3"的变体,但无济于事。ws 是第三方,所以我无法控制 XML 的格式。提前感谢。
在 SelectNodes 调用中需要一个命名空间管理器,因为您的 xml 根节点具有命名空间http://ws.jobboard.com/resumes/
请注意,XPath 需要显式使用命名空间别名。
NameTable nt = new NameTable();
XmlNamespaceManager nsmgr = new XmlNamespaceManager(nt);
nsmgr.AddNamespace("r", "http://ws.jobboard.com/resumes/");
XmlDocument jobsdoc = new XmlDocument(nt);
jobsdoc.LoadXml(xmlstring);
var xnList = jobsdoc.SelectNodes(
"/r:string/r:Packet/r:Results/r:ResumeResultItem_V3",
nsmgr);
foreach (XmlNode xn in xnList)
{
// rest of your code
}