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社区首页 >问答首页 >根据c# LINQ中的子节点XML获取特定父节点

问根据c# LINQ中的子节点XML获取特定父节点
EN

Stack Overflow用户
提问于 2018-05-04 13:12:25
回答 3查看 1.2K关注 0票数 3

我有一个很长的XML，它的父节点是sdnEntry，每个父节点都有它的子sdnType，它定义了条目的类型。我正在尝试仅获取包含 sdnType to Individual**.**的节点

下面是我的xml的简短示例；

代码语言：javascript
复制
<sdnEntry>
    <uid>6905</uid>
    <lastName>abc</lastName>
    <sdnType>Entity</sdnType> // type is entity

    <akaList>
    <aka>
        <uid>4741</uid>
        <type>a.k.a.</type>
        <category>strong</category>
        <lastName>ABC</lastName>
        <firstName>ABCCCC</firstName>
    </aka>
    <aka>
        <uid>4742</uid>
        <type>a.k.a.</type>
        <category>weak</category>
        <lastName>ADCS</lastName>
    </aka>
    </akaList>

    <nationalityList>
        <nationality>
        <uid>5416</uid>
        <country>XYZ</country>
        <mainEntry>true</mainEntry>
        </nationality>
    </nationalityList>
</sdnEntry>

<sdnEntry>
    <uid>6905</uid>
    <lastName>abc</lastName>
    <sdnType>Individual</sdnType> // type is individual

    <akaList>
    <aka>
        <uid>4741</uid>
        <type>a.k.a.</type>
        <category>strong</category>
        <lastName>ABC</lastName>
        <firstName>ABCCCC</firstName>
    </aka>
    <aka>
        <uid>4742</uid>
        <type>a.k.a.</type>
        <category>weak</category>
        <lastName>ADCS</lastName>
    </aka>
    </akaList>

    <nationalityList>
        <nationality>
        <uid>5416</uid>
        <country>XYZ</country>
        <mainEntry>true</mainEntry>
        </nationality>
    </nationalityList>
</sdnEntry>

<sdnEntry>
    <uid>6905</uid>
    <lastName>abc</lastName>
    <sdnType>Individual</sdnType>

    <akaList>
    <aka>
        <uid>4741</uid>
        <type>a.k.a.</type>
        <category>strong</category>
        <lastName>ABC</lastName>
        <firstName>ABCCCC</firstName>
    </aka>
    <aka>
        <uid>4742</uid>
        <type>a.k.a.</type>
        <category>weak</category>
        <lastName>ADCS</lastName>
    </aka>
    </akaList>

    <nationalityList>
        <nationality>
        <uid>5416</uid>
        <country>XYZ</country>
        <mainEntry>true</mainEntry>
        </nationality>
    </nationalityList>
</sdnEntry>

<sdnEntry>
    <uid>6905</uid>
    <lastName>abc</lastName>
    <sdnType>Entity</sdnType>

    <akaList>
    <aka>
        <uid>4741</uid>
        <type>a.k.a.</type>
        <category>strong</category>
        <lastName>ABC</lastName>
        <firstName>ABCCCC</firstName>
    </aka>
    <aka>
        <uid>4742</uid>
        <type>a.k.a.</type>
        <category>weak</category>
        <lastName>ADCS</lastName>
    </aka>
    </akaList>

    <nationalityList>
        <nationality>
        <uid>5416</uid>
        <country>XYZ</country>
        <mainEntry>true</mainEntry>
        </nationality>
    </nationalityList>
</sdnEntry>

我的代码是这样的，但我得到了错误；

代码语言：javascript
复制
 var lXelements = XElement.Parse(xml);
 var lParentNode = "sdnEntry";
  if (lParentNode == "sdnEntry")
            {
 //lXelements = (XElement)lXelements.Descendants("sdnType").Where(x => x.Name.LocalName == "Individual");
                lXelements = (XElement)lXelements.Descendants("sdnType").Where(x => (string)x.Value == "Individual");
            }

我得到了转换错误，我不知道我的这段代码会给我我想要的结果或不。

错误：

附加信息:无法将类型为'WhereEnumerableIterator`1System.Xml.Linq.XElement‘的对象强制转换为类型“”System.Xml.Linq.XElement“”。

xml-parsing
c#
xml
EN

回答 3

Stack Overflow用户

回答已采纳

发布于 2018-05-08 18:40:09

错误是因为您正在尝试将Linq Where结果重新分配给XElement。

除此之外，您基本上希望获得具有子<sdnType>Individual</sdnType>的所有<sdnEntry>节点

代码语言：javascript
复制
XElement elements = XElement.Parse(xml);
var parentNode = "sdnEntry";
var childNode = "sdnType";
var childNodeValue = "Individual";
List<XElement> entries = elements
    .Descendants(parentNode)
    .Where(parent => parent.Descendants(childNode)
        .Any(child => child.Value == childNodeValue)
    ).ToList();

entries应仅包含与所提供的子元素筛选器匹配的所需父元素。

上面的方法基于父节点搜索子节点。

以下方法首先查找子节点，然后在树中查找父节点

代码语言：javascript
复制
List<XElement> entries = elements
    .Descendants(childNode)
    .Where(child => child.Value == childNodeValue)
    .SelectMany(child => child.Ancestors(parentNode))
    .ToList();

这两种方法都基于以下XML生成相同的两个匹配元素结果

代码语言：javascript
复制
var xml = @"
<sdnList>
    <sdnEntry>
        <uid>6905</uid>
        <lastName>abc</lastName>
        <sdnType>Entity</sdnType>

        <akaList>
        <aka>
            <uid>4741</uid>
            <type>a.k.a.</type>
            <category>strong</category>
            <lastName>ABC</lastName>
            <firstName>ABCCCC</firstName>
        </aka>
        <aka>
            <uid>4742</uid>
            <type>a.k.a.</type>
            <category>weak</category>
            <lastName>ADCS</lastName>
        </aka>
        </akaList>

        <nationalityList>
            <nationality>
            <uid>5416</uid>
            <country>XYZ</country>
            <mainEntry>true</mainEntry>
            </nationality>
        </nationalityList>
    </sdnEntry>

    <sdnEntry>
        <uid>6905</uid>
        <lastName>abc</lastName>
        <sdnType>Individual</sdnType> 

        <akaList>
        <aka>
            <uid>4741</uid>
            <type>a.k.a.</type>
            <category>strong</category>
            <lastName>ABC</lastName>
            <firstName>ABCCCC</firstName>
        </aka>
        <aka>
            <uid>4742</uid>
            <type>a.k.a.</type>
            <category>weak</category>
            <lastName>ADCS</lastName>
        </aka>
        </akaList>

        <nationalityList>
            <nationality>
            <uid>5416</uid>
            <country>XYZ</country>
            <mainEntry>true</mainEntry>
            </nationality>
        </nationalityList>
    </sdnEntry>

    <sdnEntry>
        <uid>6905</uid>
        <lastName>abc</lastName>
        <sdnType>Individual</sdnType>

        <akaList>
        <aka>
            <uid>4741</uid>
            <type>a.k.a.</type>
            <category>strong</category>
            <lastName>ABC</lastName>
            <firstName>ABCCCC</firstName>
        </aka>
        <aka>
            <uid>4742</uid>
            <type>a.k.a.</type>
            <category>weak</category>
            <lastName>ADCS</lastName>
        </aka>
        </akaList>

        <nationalityList>
            <nationality>
            <uid>5416</uid>
            <country>XYZ</country>
            <mainEntry>true</mainEntry>
            </nationality>
        </nationalityList>
    </sdnEntry>

    <sdnEntry>
        <uid>6905</uid>
        <lastName>abc</lastName>
        <sdnType>Entity</sdnType>

        <akaList>
        <aka>
            <uid>4741</uid>
            <type>a.k.a.</type>
            <category>strong</category>
            <lastName>ABC</lastName>
            <firstName>ABCCCC</firstName>
        </aka>
        <aka>
            <uid>4742</uid>
            <type>a.k.a.</type>
            <category>weak</category>
            <lastName>ADCS</lastName>
        </aka>
        </akaList>

        <nationalityList>
            <nationality>
            <uid>5416</uid>
            <country>XYZ</country>
            <mainEntry>true</mainEntry>
            </nationality>
        </nationalityList>
    </sdnEntry>
</sdnList>
";
票数 5
EN

Stack Overflow用户

发布于 2018-05-09 12:16:49

您可以尝试执行以下操作，看看它是否对您有帮助:这不使用lambda表达式，但与上面的Nkosi代码执行相同的操作

代码语言：javascript
复制
 XElement xe = XElement.Parse(xml);
        IEnumerable<XElement> newlist = (from x in xe.Elements("sdnEntry")
                                         where x.Element("sdnType").Value == "Individual"
                                         select x);
       ///Then at this point newlist contains all xelement where the sdnType=Individual
票数 1
EN

Stack Overflow用户

发布于 2018-05-04 14:07:41

您正在尝试将IEnumerable<XElement>转换为XElement。去掉演员模型，它应该会起作用：

代码语言：javascript
复制
lXelements = lXelements.Descendants("sdnType")
          .Where(x => (string)x.Value == "Individual");

...

foreach(var element in lXelements)
{
   DoSomething(element);
}
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：

https://stackoverflow.com/questions/50167560

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