如何从 C# 中的 xml 读取特定节点？
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How to read a particular node from xml in C#?
提问by
I have following XML:
我有以下 XML：
<Loop Name="MasterData">
<Loop Name="SlaveData">
<Segment Name="AAA">
<Node1>hello</Node1>
<Node2>john</Node2>
<Node3>hi</Node3>
<Node4>marry</Node4>
</Segment>
<Segment Name="BBB">
<Node1>00</Node1>
<Node2> </Node2>
<Node3>00</Node3>
<Node4> </Node4>
</Segment>
</Loop>
</Loop>
I have to read value of each Node i.e, Node1, Node2, Node3, Node4 which are under Segment node whose attribute ie Name = "AAA". How can I do this. I am referring following link from stackoverflow but thats not working for me.
我必须读取每个节点的值，即 Node1、Node2、Node3、Node4，它们在属性 ie 的 Segment 节点下Name = "AAA"。我怎样才能做到这一点。我指的是来自 stackoverflow 的以下链接，但这对我不起作用。
How to read attribute value from XmlNode in C#?
I need output like this
我需要这样的输出
Lets I have four sting variables strNode1, strNode2, strNode3, strNode4. I want to store values in above four variables like following
让我有四个刺痛变量strNode1, strNode2, strNode3, strNode4。我想将值存储在以上四个变量中，如下所示
strNode1 = "hello"
strNode2 = "john"
strNode3 = "hi"
strNode4 = "marry"
采纳答案by Casperah
I found a simple solution for my problem
我为我的问题找到了一个简单的解决方案
XmlNodeList xnList = doc.SelectNodes("/Loop/Loop/Segment[@Name='AAA']");
foreach (XmlNode xn in xnList)
{
if (xn.HasChildNodes)
{
foreach (XmlNode item in xn.ChildNodes)
{
Console.WriteLine(item.InnerText);
}
}
}
回答by Rich O'Kelly
I'd recommend using an XDocument (NB specific filtering based on parent nodes etc omitted):
我建议使用 XDocument（基于父节点等的 NB 特定过滤被省略）：
var document = XDocument.Load(path);
var nodes = document.Descendents().Where(e => e.Name.LocalName.StartsWith("Node"));
Update to include filtering of parent element
更新以包括对父元素的过滤
var nodes = document.Descendents()
.Where(e => e.Atrributes().Any(a => a.Name.localName == "Name" && a.Value == "AAA"))
.SelectMany(e => e.Descendents().Where(e => e.Name.LocalName.StartsWith("Node"));
var values = nodes.Select(n => n.Value).ToList(); // This will be a list containing "hello", "john, "hi", "marry"
回答by Erwin
If you ever used linq-to-entities or linq-to-sql, there is a linq-to-xmlyou can use.
如果您曾经使用过 linq-to-entities 或 linq-to-sql，则可以使用linq-to-xml。
回答by Thorsten Dittmar
Assuming you have an XmlDocumentyou can use XPath like:
假设你有一个XmlDocument你可以使用 XPath 像：
XmlNode node = doc.SelectSingleNode("//Segment[@Name='AAA']");
to get the Segmentnode and then iterate all its children in a loop.
获取Segment节点，然后在循环中迭代其所有子节点。
回答by Hossein Narimani Rad
Try this:
尝试这个：
System.Xml.Linq.XDocument doc = XDocument.Load(your file);
var nodes =
doc.Element("Loop").Element("Loop").Elements("Segment")
.Where(input => (string)input.Attribute("Name") == "AAA")
.Select(input => input.Elements()).ToList();
Then:
然后：
List<string> result = new List<string>();
foreach (List<XElement> item in nodes)
{
result.AddRange(item.Select(i => i.Value));
}
回答by phadaphunk
You could use XmlDocumentto load your xml as an object and then query the specific nodes you want using XPath. Your xpath query (which I can't test right now) would probably look like this.
您可以使用XmlDocument将您的 xml 作为对象加载，然后使用XPath查询您想要的特定节点。您的 xpath 查询（我现在无法测试）可能看起来像这样。
XmlNodeList xNodes = xmlDocument.SelectNodes("//Segment[@Name = 'AAA']");
回答by Mehdi Bugnard
Do you can try this ?
你可以试试这个吗？
XDocument xmlFile = XDocument.Load("myFile.xml");
foreach (var nodeSegment in xmlFile.Descendants("Loop"))
{
foreach (var nodes in nodeSegment.Descendants().Where(e => e.Name.LocalName.StartsWith("Node")))
{
}
}
回答by Casperah
You could use XmlDocument and XPath:
您可以使用 XmlDocument 和 XPath：
XmlDocument doc = new XmlDocument();
doc.Load(xml);
foreach(XmlNode node in doc.SelectNodes("//Segment[@Name='AAA']/node()"))
{
string name = node.Name;
string value = node.innerText;
// ...
}
回答by Asif Iqbal
Try this out..
试试这个..
List<string> nodeDetails = new List<string>();
var nodes = from n in XDocument.Load(@"D:\pfXml.xml")
.Element("Loop")
.Element("loop")
.Elements("Segment")
where (string)n.Attribute("Name") == "AAAA"
select new
{
n1 = n.Element("Node1").Value,
n2 = n.Element("Node2").Value,
n3 = n.Element("Node3").Value,
n4 = n.Element("Node4").Value
};
foreach (var node in nodes)
{
nodeDetails.Add(node.n1);
nodeDetails.Add(node.n2);
nodeDetails.Add(node.n3);
nodeDetails.Add(node.n4);
}
回答by Kiran Jadhav
This might help u,
<br/>
My XML File Is like:
<br/>
< People>
<br/>
< GroupName Name="NY"><br/>
< Address1>1540 Broadway< /Address1> <br/>
< State>NY< /State> <br/>
< City>New York< /City> <br/>
< ZipCode>10036< /ZipCode> <br/>
< /GroupName><br/><br/>
< GroupName Name="TX"> <br/>
< Address1>1755 Wittington Place< /Address1> <br/>
< State>TX< /State> <br/>
< City>Dallas< /City> <br/>
< ZipCode>75234< /ZipCode> <br/>
< /GroupName><br/><br/>
< GroupName Name="Gr"> <br/>
< Address1>124 Platinum< /Address1> <br/>
< State>MH< /State> <br/>
< City>Mum< /City> <br/>
< ZipCode>400221< /ZipCode> <br/>
< /GroupName><br/><br/>
< /People><br/>
In Code behind file :<br/>
System.Xml.Linq.XDocument doc = XDocument.Load(Path of XML);
var xmlData = (from p in doc.Descendants("People").Descendants("GroupName")
let Name = (p.Attribute("Name").Value)
where Name.Equals("NY")
select new {Name = Name, Add = p.Element("Address1").Value, State = p.Element("State").Value}).Single();
<br/><br/><br/>
strName = xmlData.Name;<br/>
strAdd = xmlData.Add;<br/>
strState = xmlData.State;<br/>
<br/>
Hope this will be perfect answer for your question.