C# 如何获取XML节点

我试图获得一个节点示例，如下所示： 我怎么能得到父亲的职业我在用密码

oNotificationDoc.Load(sFileName);
oNodeListPerson = oNotificationDoc.GetElementsByTagName("Person");

XmlNode oNodeFather = null;
oNodeFather = oNodeListPerson.Item(2);
 XmlNode oNodeGeneral_temp = oNodeFather.SelectSingleNode("//NmSpace:" + Occupation, nsmgr);

但是得到母亲的职业作为回报

<Person DOB="23121964" Role="Mother" ApproxDateOfMarriage="2" DateOfMarriage="10062015" MaritalStatus="1" Nationality="CN" PPSN="" ApproxDOB="2">

    <PersonName Surname="TEST" Forename1="TEST" OtherSurnames="" BirthSurname="TEST"/>

    <MothersBirthSurname>TEST</MothersBirthSurname>

    <Address Type="Residential" Country="IE" County="D07" Line4="" Line3="TEST" Line2="TEST" Line1="TEST"/>

    <Occupation>BARISTA</Occupation>

    <PrevPregDetails PrevSponAbortions="0" PrevLateFetalDeaths="0" PrevChildrenStillLiving="0" PrevLiveBirths="0" ApproxDateOfLastBirth="" DateOfLastBirth=""/>

    </Person>


    <Person DOB="12101972" Role="Father" Nationality="CN" PPSN="" ApproxDOB="2">

    <PersonName Surname="TEST" Forename1="TEST" OtherSurnames="UNKNOWN" BirthSurname="TEST"/>

    <MothersBirthSurname>TEST</MothersBirthSurname>

    <Address Type="Residential" Country="AA" County="" Line4="" Line3="TEST" Line2="TEST" Line1="TEST"/>

    <Occupation>WAITER</Occupation>

    </Person>

试验
咖啡师
试验
服务员

要在中仅获取父亲角色，请执行以下操作：

<Person DOB="12101972" Role="Father" ...

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XPath

可以解决您的问题。关于XPath：

示例代码：

    var xml = XDocument.Load(sFileName);
    var search = xml.XPathSelectElement("//Person[@DOB='12101972' and @Role='Father']/Occupation");
    Console.WriteLine(search.Value);

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使用linq，因为它很方便，你可以做任何事情

var doc = XDocument.Parse(xml);
var result = from item in doc.Root.Elements("Person")
              select new { Label = (string)item.Element("Occupation") };

---

我已经测试了这段代码，它可以与给定的xml一起使用。

请演示如何访问

onodether

！我已经编辑了问题，请展示如何访问Onodelisperson！请显示什么是

mgroxmlcants.g\u cXML\u Person

问题现在更新更多答案请参阅：

var doc = XDocument.Parse(xml);
var result = from item in doc.Root.Elements("Person")
              select new { Label = (string)item.Element("Occupation") };