60

我正在尝试使用 CSharpCodeProvider 编译下面的代码。该文件已成功编译，但是当我单击生成的 EXE 文件时，出现错误（Windows 正在搜索此问题的解决方案）并且没有任何反应。

当我使用 CSharpCodeProvider 编译下面的代码时，我MySql.Data.dll使用这行代码将其添加为嵌入式资源文件：

if (provider.Supports(GeneratorSupport.Resources))
    cp.EmbeddedResources.Add("MySql.Data.dll");

该文件已成功嵌入（因为我注意到文件大小增加了）。

在下面的代码中，我尝试提取嵌入的 DLL 文件并将其保存到 System32，但由于某种原因，下面的代码不起作用。

namespace ConsoleApplication1
{
    class Program
    {
        public static void ExtractSaveResource(String filename, String location)
        {
            //Assembly assembly = Assembly.GetExecutingAssembly();
            Assembly a = .Assembly.GetExecutingAssembly();
            //Stream stream = assembly.GetManifestResourceStream("Installer.Properties.mydll.dll"); // or whatever
            //string my_namespace = a.GetName().Name.ToString();
            Stream resFilestream = a.GetManifestResourceStream(filename);
            if (resFilestream != null)
            {
                BinaryReader br = new BinaryReader(resFilestream);
                FileStream fs = new FileStream(location, FileMode.Create); // Say
                BinaryWriter bw = new BinaryWriter(fs);
                byte[] ba = new byte[resFilestream.Length];
                resFilestream.Read(ba, 0, ba.Length);
                bw.Write(ba);
                br.Close();
                bw.Close();
                resFilestream.Close();
            }
            // this.Close();
        }

        static void Main(string[] args)
        {
            try
            {
                string systemDir = Environment.SystemDirectory;
                ExtractSaveResource("MySql.Data.dll", systemDir);
            }
            catch (Exception ex)
            {
                Console.WriteLine(ex.Message);
                Console.ReadKey();
            }
        }
    }
}

如何提取作为资源嵌入的 DLL 文件并将其保存到 System32？

4

6 回答 6

82

我建议做起来更轻松。我假设资源存在并且文件是可写的（如果我们谈论系统目录，这可能是一个问题）。

public void WriteResourceToFile(string resourceName, string fileName)
{
    using(var resource = Assembly.GetExecutingAssembly().GetManifestResourceStream(resourceName))
    {
        using(var file = new FileStream(fileName, FileMode.Create, FileAccess.Write))
        {
            resource.CopyTo(file);
        } 
    }
}
于 2013-11-14T15:48:42.107 回答
73

我发现最简单的方法是使用Properties.Resourcesand File。这是我使用的代码（需要using System.IO）...

对于二进制文件： File.WriteAllBytes(fileName, Properties.Resources.file);

对于文本文件： File.WriteAllText(fileName, Properties.Resources.file);

于 2016-01-07T20:29:10.987 回答
31

我一直在使用这种（经过测试的）方法：

OutputDir：要复制资源的位置

ResourceLocation：命名空间（+ dirnames）

文件：要复制的资源位置中的文件列表。

    private static void ExtractEmbeddedResource(string outputDir, string resourceLocation, List<string> files)
    {
        foreach (string file in files)
        {
            using (System.IO.Stream stream = System.Reflection.Assembly.GetExecutingAssembly().GetManifestResourceStream(resourceLocation + @"." + file))
            {
                using (System.IO.FileStream fileStream = new System.IO.FileStream(System.IO.Path.Combine(outputDir, file), System.IO.FileMode.Create))
                {
                    for (int i = 0; i < stream.Length; i++)
                    {
                        fileStream.WriteByte((byte)stream.ReadByte());
                    }
                    fileStream.Close();
                }
            }
        }
    }
于 2012-10-25T07:55:39.640 回答
4

这完美！

public static void Extract(string nameSpace, string outDirectory, string internalFilePath, string resourceName)
{
    //nameSpace = the namespace of your project, located right above your class' name;
    //outDirectory = where the file will be extracted to;
    //internalFilePath = the name of the folder inside visual studio which the files are in;
    //resourceName = the name of the file;
    Assembly assembly = Assembly.GetCallingAssembly();

    using (Stream s = assembly.GetManifestResourceStream(nameSpace + "." + (internalFilePath == "" ? "" : internalFilePath + ".") + resourceName))
    using (BinaryReader r = new BinaryReader(s))
    using (FileStream fs = new FileStream(outDirectory + "\\" + resourcename, FileMode.OpenOrCreate))
    using (BinaryWriter w = new BinaryWriter(fs))
    {
        w.Write(r.ReadBytes((int)s.Length));
    }
}

使用示例：

public static void ExtractFile()
{
    String local = Environment.CurrentDirectory; //gets current path to extract the files

    Extract("Geral", local, "Arquivos", "bloquear_vbs.vbs");
}

如果这仍然没有帮助，请尝试此视频：https ://www.youtube.com/watch?v=_61pLVH2qPk

于 2014-12-17T02:26:17.913 回答
2

或者使用扩展方法...

 /// <summary>
 /// Retrieves the specified [embedded] resource file and saves it to disk.  
 /// If only filename is provided then the file is saved to the default 
 /// directory, otherwise the full filepath will be used.
 /// <para>
 /// Note: if the embedded resource resides in a different assembly use that
 /// assembly instance with this extension method.
 /// </para>
 /// </summary>
 /// <example>
 /// <code>
 ///       Assembly.GetExecutingAssembly().ExtractResource("Ng-setup.cmd");
 ///       OR
 ///       Assembly.GetExecutingAssembly().ExtractResource("Ng-setup.cmd", "C:\temp\MySetup.cmd");
 /// </code>
 /// </example>
 /// <param name="assembly">The assembly.</param>
 /// <param name="resourceName">Name of the resource.</param>
 /// <param name="fileName">Name of the file.</param>
 public static void ExtractResource(this Assembly assembly, string filename, string path=null)
 {
     //Construct the full path name for the output file
     var outputFile = path ?? $@"{Directory.GetCurrentDirectory()}\{filename}";

     // If the project name contains dashes replace with underscores since 
     // namespaces do not permit dashes (underscores will be default to).
     var resourceName = $"{assembly.GetName().Name.Replace("-","_")}.{filename}";

     // Pull the fully qualified resource name from the provided assembly
     using (var resource = assembly.GetManifestResourceStream(resourceName))
     {
         if (resource == null)
             throw new FileNotFoundException($"Could not find [{resourceName}] in {assembly.FullName}!");

         using (var file = new FileStream(outputFile, FileMode.Create, FileAccess.Write))
         {
             resource.CopyTo(file);
         }
     }
 }
于 2018-01-12T15:44:05.247 回答
0

尝试将您的目标程序集读入 aMemoryStream然后保存到FileStream这样的（请记住，此代码未经测试）：

//Imports
using System;
using System.IO;
using System.Reflection;

    Assembly assembly = Assembly.GetExecutingAssembly();
    using (var target = assembly.GetManifestResourceStream("MySql.Data.dll"))
    {
        var size = target.CanSeek ? Convert.ToInt32(target.Length) : 0;

        // read your target assembly into the MemoryStream
        MemoryStream output = null;
        using (output = new MemoryStream(size))
        {
            int len;
            byte[] buffer = new byte[2048];

            do
            {
                len = target.Read(buffer, 0, buffer.Length);
                output.Write(buffer, 0, len);
            } 
            while (len != 0);
        }

        // now save your MemoryStream to a flat file
        using (var fs = File.OpenWrite(@"c:\Windows\System32\MySql.Data.dll"))
        {
            output.WriteTo(fs);
            fs.Flush();
            fs.Close();
        }
    }
于 2012-10-23T13:53:15.590 回答