Question

嗨，我有一个像下面的xml。

<Services>
      <Service Name="ReportWriter" AssemplyName="ReportService.ReportWriters" ClassName="ExcelWriter">
        <ServiceConfigurations>
         <Configuration key="key1" value="value1" />
 <Configuration key="key2" value="value2" />

        </ServiceConfigurations>
      </Service>
      <Service Name="LogtWriter" AssemplyName="" ClassName="">
        <ServiceConfigurations>
          <Configuration key="" value="" />
        </ServiceConfigurations>
      </Service>
      <Service Name="OutputHandler" AssemplyName="" ClassName="">
        <ServiceConfigurations>
          <Configuration key="key1" value="value1" />
 <Configuration key="key2" value="value2" />
        </ServiceConfigurations>
      </Service>
</Services>

我想获取服务名称=＆＃34; ReportWriter＆＃34;的配置键和值属性。

FOR ex-输出应为key1，value1，key2，value2为服务名称=＆＃39; ReportWriter＆＃39;。

任何人都可以帮我解决如何取悦

Answer 1

您可以使用Linq2Xml和XPath

var xDoc = XDocument.Load(filename);
var conf = xDoc.XPathSelectElements("//Service[@Name='ReportWriter']/ServiceConfigurations/Configuration")
           .Select(c => new { Key = (string)c.Attribute("key"), Value = (string)c.Attribute("value") })
           .ToList();

或没有XPath

var conf = xDoc.Descendants("Service")
                .First(srv => srv.Attribute("Name").Value == "ReportWriter")
                .Descendants("Configuration")
                .Select(c => new { Key = (string)c.Attribute("key"), Value = (string)c.Attribute("value") })
                .ToList();

Answer 2

试试这个： -

 Dictionary<string,string> result = xdoc.Descendants("Service")
                  .First(x => (string)x.Attribute("Name") == "ReportWriter")
                  .Descendants("Configuration")
                  .Select(x => new
                             {
                                 Key = x.Attribute("key").Value,
                                 Value = x.Attribute("value").Value
                             }).ToDictionary(x => x.Key, y => y.Value) ;

<强>更新

以上查询返回Dictionary<string,string>，您可以使用foreach来迭代它的元素并将其添加到现有字典中。

在xml中按父级获取子节点属性

3 个答案: