如何在c#中读取XML文件？
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How to read XML file in c#?
提问by Hemant Kothiyal
I have following XML file, i want to know best way to read this XML file
我有以下 XML 文件，我想知道阅读此 XML 文件的最佳方法
<MyFile>
<Companies>
<Company>123</Company>
<Company>456</Company>
<Company>789</Company>
</Companies>
</MyFile>
As an output i need collection of values like "123,456,789" or it could be array of string[]
作为输出，我需要收集诸如“123,456,789”之类的值，或者它可以是字符串数组[]
Can we use Linq to xml? How?
我们可以使用Linq to xml吗？如何？
回答by spender
var xmlStr=@"<MyFile>
<Companies>
<Company>123</Company>
<Company>456</Company>
<Company>789</Company>
</Companies>
</MyFile>";
var xDoc = XDocument.Parse(xmlStr);
var companyIds = xDoc.Descendants("Company").Select(e => (int)e);
回答by Daren Thomas
var xdoc = XDocument.Load(PATH_TO_FILE);
var companies = xdoc.Descendants("Company").Select(c => (string)c).ToArray();
This will give you a string[].
这会给你一个string[].
回答by Habib
Use LINQ to XML, Include using System.Xml.Linq;
使用 LINQ to XML，包括 using System.Xml.Linq;
XDocument xmlDoc = XDocument.Load("yourfile.xml");
var test = xmlDoc.Descendants("Companies").Elements("Company").Select(r => r.Value).ToArray();
string result = string.Join(",", test);
Output would be:
输出将是：
123,456,789
123,456,789
回答by Habib
In dataset you can read xml file
在数据集中，您可以读取 xml 文件
Following are lines of code to read XML file in DataSet
以下是在 DataSet 中读取 XML 文件的代码行
DataSet dsMenu = new DataSet(); //Create Dataset Object
dsMenu.ReadXml("XMLFILENAME.XML"); // Read XML file in Dataset
DataTable dtXMLFILE// Create DatyaTable object
dtXMLFILE= dsMenu.Tables[0]; // Store XML Data in Data Table
回答by Keplah
In the past, I have used an XmlReaderand had no difficulties.
过去，我使用过XmlReader，并且没有任何困难。
MSDN Documentation: http://msdn.microsoft.com/en-us/library/system.xml.xmlreader(v=vs.110).aspx
MSDN 文档：http: //msdn.microsoft.com/en-us/library/system.xml.xmlreader(v= vs.110).aspx
It is very straightforward and the documentation is pretty well written. A quick demonstration of how to use it:
它非常简单，文档也写得很好。如何使用它的快速演示：
XmlReader reader = XmlReader.Create(targetFile);
while (reader.Read())
{
switch (reader.NodeType)
{
case XmlNodeType.Element:
if (reader.Name.Equals("Company")
{
// Read the XML Node's attributes and add to string
}
break;
}
}
回答by thezar
string pathToXmlFile = @"C:\test.xml";
XElement patternDoc = XElement.Load(pathToXmlFile);
List<string> values = new List<string>();
foreach (var element in patternDoc.Elements("Companies").Elements("Company"))
{
values.Add(element.Value);
}
