Question

我有一个FileInfo，我知道它是XML，我想转换为XmlDocument。有什么办法可以做到吗？这是我的代码

string name = "filename";
XmlDocument xml = new XmlDocument();
FileInfo xmlFileInfo = directoryInfo.GetFiles(filename + extension).FirstOrDefault();

//Im trying to do something like: xml = xmlFileInfo as XmlDocument

Answer 1

您不想将FileInfo对象转换为XmlDocument，而是想要将FileInfo指向的文件的内容加载到此类对象中。

尝试这样做：

var doc = new XmlDocument();
doc.Load(xmlFileInfo.FullName);

Answer 2

这是我一年多以前写的代码。如果您没有读取文件夹的权限，则“获取文件”将失败。因此，解决方案是递归获取每个文件夹。：

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

namespace WriteFileNamesXml
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        const string FOLDER = @"c:\temp";
        static XmlWriter writer = null;
        static void Main(string[] args)
        {
            XmlWriterSettings settings = new XmlWriterSettings();
            settings.Indent = true;

            writer = XmlWriter.Create(FILENAME, settings);
            writer.WriteStartDocument(true);

            DirectoryInfo info = new DirectoryInfo(FOLDER);
            WriteTree(info);

            writer.WriteEndDocument();
            writer.Flush();
            writer.Close();


        }
        static long WriteTree(DirectoryInfo info)
        {
            long size = 0;
            writer.WriteStartElement("Folder");
            try
            {
                writer.WriteAttributeString("name", info.Name);
                writer.WriteAttributeString("numberSubFolders", info.GetDirectories().Count().ToString());
                writer.WriteAttributeString("numberFiles", info.GetFiles().Count().ToString());
                writer.WriteAttributeString("date", info.LastWriteTime.ToString());


                foreach (DirectoryInfo childInfo in info.GetDirectories())
                {
                    size += WriteTree(childInfo);
                }

            }
            catch (Exception ex)
            {
                string errorMsg = string.Format("Exception Folder : {0}, Error : {1}", info.FullName, ex.Message);
                Console.WriteLine(errorMsg);
                writer.WriteElementString("Error", errorMsg);
            }

            FileInfo[] fileInfo = null;
            try
            {
                fileInfo = info.GetFiles();
            }
            catch (Exception ex)
            {
                string errorMsg = string.Format("Exception FileInfo : {0}, Error : {1}", info.FullName, ex.Message);
                Console.WriteLine(errorMsg);
                writer.WriteElementString("Error", errorMsg);
            }

            if (fileInfo != null)
            {
                foreach (FileInfo finfo in fileInfo)
                {
                    try
                    {
                        writer.WriteStartElement("File");
                        writer.WriteAttributeString("name", finfo.Name);
                        writer.WriteAttributeString("size", finfo.Length.ToString());
                        writer.WriteAttributeString("date", info.LastWriteTime.ToString());
                        writer.WriteEndElement();
                        size += finfo.Length;
                    }
                    catch (Exception ex)
                    {
                        string errorMsg = string.Format("Exception File : {0}, Error : {1}", finfo.FullName, ex.Message);
                        Console.WriteLine(errorMsg);
                        writer.WriteElementString("Error", errorMsg);
                    }
                }
            }

            writer.WriteElementString("size", size.ToString());
            writer.WriteEndElement();
            return size;

        }
    }
}

有什么方法可以将FileInfo转换为XmlDocument吗？

2 个答案: