Question

根据答案here我想将多个文件流写入一个流，如下所示：

为每个流的长度编号保留4个字节每个流内容在其长度编号后写入（4字节后）在结束流将是这样的

流= File1 len + File1 stream content + File2 len + File2 stream content + ....

示例代码：

result = new ExportResult_C()
            {
                PackedStudy = packed.ToArray() ,
                Stream = new MemoryStream()
            };
            string[] zipFiles = Directory.GetFiles(zipRoot);
            foreach (string fileN in zipFiles)
            {
                MemoryStream outFile = new MemoryStream(File.ReadAllBytes(fileN));
                MemoryStream len = new MemoryStream(4);
                //initiate outFile len to 4 byte push it to main stream
                //Then push outFile stream to main stream
                //Continue and do this for another file 
            } 
            //For test Save stream to file(s)

这是个好主意吗？我真的不知道这些评论如何成为代码行。

提前致谢。

Answer 1

试试这个

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

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            byte[] testMessage = Encoding.UTF8.GetBytes("The quick brown fox jumped over the lazy dog");
            MemoryStream outFile = new MemoryStream();
            BinaryWriter writer = new BinaryWriter(outFile);
            for (int i = 0; i < 10; i++ )
            {
                writer.Write(BitConverter.GetBytes(testMessage.Length), 0, 4);
                writer.Write(testMessage, 0, testMessage.Length);
            }
            writer.Flush();

            outFile.Position = 0;
            BinaryReader reader = new BinaryReader(outFile, Encoding.UTF8);
            while (outFile.Position < outFile.Length)
            {
                int size = reader.ReadInt32();
                byte[] data = reader.ReadBytes(size);
            }
        }
    }
}

Answer 2

我认为我发布了一个更好的解决方案来回答我的问题here 多个文件字节将被序列化为一个流，而在客户端，它将被反序列化为一个字节数组。

请参阅here，它可能很有用。

但我接受了@jdweng解决方案，感谢他的关注和帮助。

一个Stream中的多个文件，自定义流

2 个答案: