如何从嵌套的xml节点获取属性值？

Question

我有这样格式化的XElement对象：

<Setting guid="3bcedf55-b75f-456b-b90a-a92cbbb022ga">
    <PatientFieldList>
        <PatientFieldSetting PatientName="UserDecision" PatentFieldLength="64" />
        <PatientFieldSetting PatientName="prohibited" PatentFieldLength="128" />
    </PatientFieldList>
</Setting>

我必须获取所有节点中所有属性的值，但我不知道如何：/我尝试过

xml.Elements("PatientFieldList")

xml.Descendants("PatientsSettingsFieldsList").Where(x => x.Attribute("PatentFieldLength").Value == 64)`

我有很多这样的节点，所以我想知道是否有通过“ []”或某种方式访问​​这些属性的简便方法。

Answer 1

码：

using System;
using System.Linq;
using System.Xml.Linq

var xml = "<Setting ...";
var doc = XElement.Parse(xml);
int i; // for int parse
var q = from node in doc.Descendants("PatientFieldSetting")
        let name = node.Attribute("PatientName")
        let length = node.Attribute("PatentFieldLength")
        select new { Name = (name != null) ? name.Value : "", Length = (length != null && Int32.TryParse(length.Value, out i)) ? i : 0 };

foreach (var node in q)
{
    Console.WriteLine("Name={0}, Length={1}", node.Name, node.Length);
}

输出：

Name=UserDecision, Length=64
Name=prohibited, Length=128