Question

我将XElement对象格式化为：

<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单方法。

Answer 1

代码：

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

Answer 2

这将打印出xml中具有属性的所有节点的属性：

XDocument doc = //your data

var q = from node in doc.Descendants()
        where node.Attributes().Count() > 0
        select new {NodeName = node.Name, Attributes = node.Attributes()};

foreach (var node in q)
{
    Console.WriteLine( node.NodeName );
    foreach (var attribute in node.Attributes)
    {
        Console.WriteLine(attribute.Name + ":" + attribute.Value);
    }
    Console.WriteLine();
}

如果您只希望PatientFieldSetting节点过滤名称：

from node in doc.Descendants("PatientFieldSetting")

如何从嵌套的xml节点获取属性值？

2 个答案: