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)。
答案 0 :(得分:31)
MemoryStream有几个constructor overloads,其中没有一个需要文件。
答案 1 :(得分:8)
有一个如何在MSDN page for MemoryStream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);
}
}
}
这是你在找什么?
答案 2 :(得分:3)