1

我有像下面这样的xml文件和Device类,我想List<Device>从我的xml文件中获取我如何通过linq做到这一点

            XDocument loaded = XDocument.Load(SharedData.CONFIGURATION_FULL_PATH);
            var q = loaded.Descendants("device").Select(c => c);

但当然这段代码不起作用

<?xml version="1.0" encoding="utf-8"?>
<settings>
<device>
  <username>aa</username>
  <AgentName>aa</AgentName>
  <password>aa</password>
  <domain>aa</domain>
</device>
<device>
  <username>bb</username>
  <AgentName>bb</AgentName>
  <password>bb</password>
  <domain>bb</domain>
</device>
<device>
  <username>cc</username>
  <AgentName>cc</AgentName>
  <password>cc</password>
  <domain>cc</domain>
</device>

</settings>
4

2 回答 2

4
List<Device> devices = new List<Device>(loaded.Descendants("Device")
                                             .Select(e =>
                                               new Device(e.Element("username").Value,
                                                          e.Element("AgentName").Value,
                                                          e.Element("password").Value,
                                                          e.Element("domain").Value
                                                         )));
于 2012-08-15T02:41:37.013 回答
0

你可以这样做:

var devices =
    loaded
        .Descendants("Device")
        .Select(e => new Device(
            e.Element("username").Value,
            e.Element("AgentName").Value,
            e.Element("password").Value,
            e.Element("domain").Value))
        .ToList();
于 2012-08-15T03:09:36.690 回答