Question

我的XML具有以下结构：

<?xml version="1.0"?>
<NidanArchiveDS>
<xs:schema id="NidanArchiveDS" targetNamespace="http://tempuri.org/NidanArchiveDS.xsd" xmlns:mstns="http://tempuri.org/NidanArchiveDS.xsd" xmlns="http://tempuri.org/NidanArchiveDS.xsd" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" attributeFormDefault="qualified" elementFormDefault="qualified">
<xs:element name="NidanArchiveDS" msdata:IsDataSet="true" msdata:Locale="ru-RU">
  <xs:complexType>
    <xs:choice minOccurs="0" maxOccurs="unbounded">
      <xs:element name="Settings">
        <xs:complexType>
          <xs:attribute name="ContentTypeName" form="unqualified" type="xs:string" use="required" />
          <xs:attribute name="ListName" form="unqualified" type="xs:string" use="required" />
          <xs:attribute name="DocTypeFieldName" form="unqualified" type="xs:string" />
          <xs:attribute name="DocNumberFieldName" form="unqualified" type="xs:string" />
          <xs:attribute name="DOcDateFieldName" form="unqualified" type="xs:string" />
        </xs:complexType>
      </xs:element>
    </xs:choice>
  </xs:complexType>
  <xs:unique name="Constraint1" msdata:PrimaryKey="true">
    <xs:selector xpath=".//mstns:Settings" />
    <xs:field xpath="@ContentTypeName" />
    <xs:field xpath="@ListName" />
  </xs:unique>
</xs:element>
</xs:schema>
<diffgr:diffgram xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1">
<NidanArchiveDS xmlns="http://tempuri.org/NidanArchiveDS.xsd">
  <Settings diffgr:id="Settings1" msdata:rowOrder="0" ContentTypeName="CONTENTTYPE 1" ListName="Заявки на оплату" DocTypeFieldName="Имя11" DocNumberFieldName="Имя22" DOcDateFieldName="Имя33" />
  <Settings diffgr:id="Settings2" msdata:rowOrder="1" ContentTypeName="CONTENTTYPE 2" ListName="LST" DocTypeFieldName="Имя44" DocNumberFieldName="Имя" DOcDateFieldName="Имя5555" />

</NidanArchiveDS>

下面是我使用ContentTypeName =“CONTENTTYPE 2”获取设置的查询

docNameFieldName = (from xn in nidanarchiveSettings.Descendants(System.Xml.Linq.XName.Get("Settings"))
                                                        where xn.Attribute(System.Xml.Linq.XName.Get("ContentTypeName")).Value == "CONTENTTYPE 2"
                                                        && xn.Attribute(System.Xml.Linq.XName.Get("ListName")).Value == "LST"
                                                        select xn.Attribute(System.Xml.Linq.XName.Get("DocTypeFieldName")).Value).SingleOrDefault();

但它给了我一个空字符串。我想知道如何以正确的方式编写查询以获得此属性。

Answer 1

你忘记了命名空间：

XDocument xdoc = XDocument.Load(path_to_xml);
XNamespace ns = "http://tempuri.org/NidanArchiveDS.xsd"; 

var docNameFieldName = 
    (from s in xdoc.Descendants(ns + "Settings")
     where (string)s.Attribute("ContentTypeName") == "CONTENTTYPE 2" &&
           (string)s.Attribute("ListName") == "LST"
     select (string)s.Attribute("DocTypeFieldName")).SingleOrDefault();

也不要使用节点的Value属性。简单地将它转换为字符串（或其他类型） - 如果某个节点丢失，它不会引发异常。

Answer 2

你可以用lambda路线：

var docNameFieldName = xdoc.Descendants("Settings")
.First(sett => sett.Attribute("ContentTypeName") == "CONTENTTYPE 2")
.Attribute("DocTypeFieldName").Value;

此外，您可能刚刚发布了部分XML片段，但您的<diffgr:diffgram>代码未关闭...

如何通过LINQ to XML获取间接后代节点？

2 个答案: