Question

我正在尝试读取XML文件并解析其内容，但我无法从文件中提取参数。

我正在尝试解析的XML文件如下所示：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<root>
    <register_map>
        <Register ID="1" Index="0x100000" DataType="0x0007" ObjectType="0x07" Name="Device Type"/>
        <Register ID="2" Index="0x100100" DataType="0x0005" ObjectType="0x07" Name="Error Register"/>
    </register_map>
</root>

到目前为止我的代码看起来像这样

namespace Test_XML
{
    class Program
    {
        struct RegisterEntry
        {
            public UInt32 index;
            public UInt16 dataType;
            public UInt16 objectType;
            public string name;
        };

        static void Main(string[] args)
        {

            XDocument doc = XDocument.Load("registers.xml");

            var registers = doc.Descendants("register_map");

            foreach (var register in registers)
            {
                // Fill up a list of RegisterEntrys with contents of XML
            }
            Console.ReadLine();
        }
    }
}

如何从“寄存器”中提取参数并将它们放在RegisterEntry对象中？

Answer 1

您可以使用

var registers = doc.XPathSelectElements("/root/register_map/Register");

它将为您提供Register个节点的集合，因此您将能够访问其属性并填充RegisterEntry对象，如：

foreach (var register in registers)
{
    var dataType = register.Attribute("DataType").Value;
    //the rest of the code
}

注意XPathSelectElements是System.Xml.XPath命名空间中的扩展方法。确保您已引用System.Xml.Linq程序集以便使用它。

Answer 2

您应该使用.Attributes["name"].Value。我认为你想要将这些值转换为Int，所以我们还需要一个额外的Convert.ToInt(string, base);

var RegisteryEntryList = new List<RegistryEntry>();

foreach (var register in registers)
{
    //create a new RegistryEntry
    var obj = new RegistryEntry();
    //convert your string to an int value and save it
    obj.index = Convert.ToInt32(register.Attributes["Index"].Value.Split('x')[1], 8);
    obj.datatype = Convert.ToInt32(register.Attributes["DataType"].Value.Split('x')[1], 8);

    //... your remaining properties
    RegisteryEntryList.Add(obj);
}

请注意：如果您的索引是二进制（基数2），则需要相应地调整转换。有关详细信息，请参阅https://msdn.microsoft.com/en-us/library/1k20k614(v=vs.110).aspx

Answer 3

您的查询将为您提供名称为register_map的所有元素 - 您想要所有Register个元素。将其更改为：

var registers = doc.Descendants("Registers");

然后遍历它们并获取所需的值，将它们转换为所需的类型。

foreach (var register in registers)
{
    var indexHex = (string)register.Attribute("Index");
    var index = Convert.ToUInt32(indexHex, 16);

    var dataTypeHex = (string)register.Attribute("DataType");
    var dataType = Convert.ToUInt16(dataTypeHex, 16);

    var objectTypeHex = (string)register.Attribute("ObjectType");
    var objectType = Convert.ToUInt16(objectTypeHex, 16);

    var name = (string)register.Attribute("Name");

    var entry = new RegisterEntry
    {
        index = index,
        dataType = dataType,
        objectType = objectType,
        name = name,
    };

    // do something with entry
}

Answer 4

使用xml Linq

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
using System.Globalization;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            string xml =
                "<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?>" +
                "<root>" +
                    "<register_map>" +
                        "<Register ID=\"1\" Index=\"0x100000\" DataType=\"0x0007\" ObjectType=\"0x07\" Name=\"Device Type\"/>" +
                        "<Register ID=\"2\" Index=\"0x100100\" DataType=\"0x0005\" ObjectType=\"0x07\" Name=\"Error Register\"/>" +
                    "</register_map>" +
                "</root>";

            XDocument doc = XDocument.Parse(xml);

            var results = doc.Descendants("Register").Select(x => new {
                id = (int)x.Attribute("ID"),
                index = int.Parse(x.Attribute("Index").Value.Substring(2), NumberStyles.HexNumber,  CultureInfo.CurrentCulture),
                dataType = int.Parse(x.Attribute("DataType").Value.Substring(2), NumberStyles.HexNumber, CultureInfo.CurrentCulture),
                objectType = int.Parse(x.Attribute("ObjectType").Value.Substring(2), NumberStyles.HexNumber, CultureInfo.CurrentCulture),
                name = (string)x.Attribute("Name")
            }).ToList();
        }
    }
}

解析XML文件并创建其内容列表

4 个答案: