无需创建物理文件即可创建流

Question

我需要创建一个包含服务器上存在的文档的zip文件.我正在使用.Net Package类来创建一个新的Package(这是一个zip文件)我必须有一个物理文件或流的路径.我试图不创建一个可能是zip文件的实际文件,而只是创建一个存在于内存或其他东西的流.

我的问题是如何在没有物理文件实例化的情况下实例化新的Stream(即FileStream,MemoryStream等).

Answer 1

MemoryStream有几个构造函数重载,其中没有一个需要文件.

Answer 2

有一个如何在MemoryStream的MSDN页面上执行此操作的示例:

using System;
using System.IO;
using System.Text;

class MemStream
{
    static void Main()
    {
        int count;
        byte[] byteArray;
        char[] charArray;
        UnicodeEncoding uniEncoding = new UnicodeEncoding();

        // Create the data to write to the stream.
        byte[] firstString = uniEncoding.GetBytes(
            "Invalid file path characters are: ");
        byte[] secondString = uniEncoding.GetBytes(
            Path.GetInvalidPathChars());

        using(MemoryStream memStream = new MemoryStream(100))
        {
            // Write the first string to the stream.
            memStream.Write(firstString, 0 , firstString.Length);

            // Write the second string to the stream, byte by byte.
            count = 0;
            while(count < secondString.Length)
            {
                memStream.WriteByte(secondString[count++]);
            }

            // Write the stream properties to the console.
            Console.WriteLine(
                "Capacity = {0}, Length = {1}, Position = {2}\n",
                memStream.Capacity.ToString(),
                memStream.Length.ToString(),
                memStream.Position.ToString());

            // Set the position to the beginning of the stream.
            memStream.Seek(0, SeekOrigin.Begin);

            // Read the first 20 bytes from the stream.
            byteArray = new byte[memStream.Length];
            count = memStream.Read(byteArray, 0, 20);

            // Read the remaining bytes, byte by byte.
            while(count < memStream.Length)
            {
                byteArray[count++] =
                    Convert.ToByte(memStream.ReadByte());
            }

            // Decode the byte array into a char array
            // and write it to the console.
            charArray = new char[uniEncoding.GetCharCount(
                byteArray, 0, count)];
            uniEncoding.GetDecoder().GetChars(
                byteArray, 0, count, charArray, 0);
            Console.WriteLine(charArray);
        }
    }
}

这是你想要的？