Question

我有一个以下格式的xml，当我传递ParentEntityId时需要获取所有SubEntity的id和名称。这需要在silverlight中完成。

<Treeview>
    <ParentEntity ParentEntityId="1" ParentEntityName="P1">
        <Facility FacilityId="F1" FacilityName="F1">
            <Category CategoryId="C1" CategoryName="C1"/>             
            <Category CategoryId="C2" CategoryName="C2" >
                <Activity ActivityId="A1" ActivityName="A1" /> 
            </Category>
        </Facility>
        <Facility FacilityId="F2" FacilityName="F2">
            <Category CategoryId="C1" CategoryName="C1">
                <Activity ActivityId="A2" ActivityName="A2" /> 
                <Activity ActivityId="A3" ActivityName="A3" />   
            </Category>
        </Facility>
        <SubEntity SubEntityId="S1" SubEntityName="S1">
            <SubEntity SubEntityId="S2" SubEntityName="S2"/> 
        </SubEntity>
        <SubEntity SubEntityId="S3" SubEntityName="S3">
            <SubEntity SubEntityId="S4" SubEntityName="S4">
                <Facility FacilityId="F3" FacilityName="F3">
                    <Category CategoryId="C1" CategoryName="C1 >
                        <Activity ActivityId="A1" ActivityName="A1" />   
                    </Category>
                </Facility>
            </SubEntity>
            <SubEntity SubEntityId="S5" SubEntityName="S5">
                <Facility FacilityId="F5" FacilityName="F5">
                    <Category CategoryId="C1" CategoryName="C1" />
                    <Category CategoryId="C2" CategoryName="C2"/>                
                </Facility>
            </SubEntity>
            <SubEntity SubEntityId="S6" SubEntityName="S6" />
        </SubEntity>
    </ParentEntity>
    <ParentEntity ParentEntityId="2" ParentEntityName="P2">
        <Facility FacilityId="F1" FacilityName="F1">
            <Category CategoryId="C1" CategoryName="C1"/>                   
        </Facility>
        <SubEntity SubEntityId="S7" SubEntityName="S7" />
    </ParentEntity>
</Treeview>

假设我传递了parentEntityId = 1，我需要输出S1，S2，S3，S4，S5，S6。

提前致谢， Sunitha

Answer 1

使用LINQ to XML

  foreach(XElement subEntity in doc.Root.Elements("ParentEntity")
       .First(elem => elem.Attribute("ParentEntityID").Value == "1")
       .Descendents("SubEntity") )
  {
     // Do stuff with each SubEntity e.g.:-
     string name = subEntity.Attribute("SubEntityName");
  }

如果您使用Silverlight 4，则另一种方法是使用XPath。您需要将System.Xml.XPath.dll添加到您的引用。在这种情况下，上述循环变为： -

  foreach(XElement subEntity in doc.Root
      .XPathSelectElements("ParentEntity[ParentEntityId='1']//SubEntity" )
  {
     // Do stuff with each SubEntity e.g.:-
     string name = subEntity.Attribute("SubEntityName");
  }

顺便说一下，你的真实XML前缀中没有你所放置元素名称的所有属性名称吗？例如，为什么“SubEntityName”而不仅仅是“Name”？

如果我将父节点id传递给silverlight中的xml，如何获取子节点？

1 个答案: