使用System.IO.Compression在内存中创建ZIP存档
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我正在尝试使用MemoryStream创建一个带有简单演示文本文件的ZIP存档，如下所示：
using (var memoryStream = new MemoryStream())
using (var archive = new ZipArchive(memoryStream , ZipArchiveMode.Create))
{
var demoFile = archive.CreateEntry("foo.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write("Bar!");
}
using (var fileStream = new FileStream(@"C:Temp est.zip", FileMode.Create))
{
stream.CopyTo(fileStream);
}
}
如果我运行此代码，则会创建存档文件本身，但不会创建foo.txt。
但是，如果我直接用文件流替换MemoryStream，则会正确创建存档：
using (var fileStream = new FileStream(@"C:Temp est.zip", FileMode.Create))
using (var archive = new ZipArchive(fileStream, FileMode.Create))
{
// ...
}
是否可以使用MemoryStream创建没有FileStream的ZIP存档？
答案
感谢https://stackoverflow.com/a/12350106/222748，我得到了：
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var demoFile = archive.CreateEntry("foo.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write("Bar!");
}
}
using (var fileStream = new FileStream(@"C:Temp est.zip", FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
因此我们需要在使用ZipArchive之前调用dispose，这意味着将'true'作为第三个参数传递给ZipArchive，这样我们仍然可以在处理它之后访问它。
另一答案
只是另一个版本的压缩而不写任何文件。
string fileName = "export_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".xlsx";
byte[] fileBytes = here is your file in bytes
byte[] compressedBytes;
string fileNameZip = "Export_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";
using (var outStream = new MemoryStream())
{
using (var archive = new ZipArchive(outStream, ZipArchiveMode.Create, true))
{
var fileInArchive = archive.CreateEntry(fileName, CompressionLevel.Optimal);
using (var entryStream = fileInArchive.Open())
using (var fileToCompressStream = new MemoryStream(fileBytes))
{
fileToCompressStream.CopyTo(entryStream);
}
}
compressedBytes = outStream.ToArray();
}
另一答案
在将流复制到zip流之前，将流的位置设置为0。
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var demoFile = archive.CreateEntry("foo.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write("Bar!");
}
}
using (var fileStream = new FileStream(@"C:Temp est.zip", FileMode.Create))
{
memoryStream.Position=0;
memoryStream.WriteTo(fileStream);
}
}
另一答案
您需要完成写入内存流然后再读取缓冲区。
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create))
{
var demoFile = archive.CreateEntry("foo.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream))
{
streamWriter.Write("Bar!");
}
}
using (var fileStream = new FileStream(@"C:Temp est.zip", FileMode.Create))
{
var bytes = memoryStream.GetBuffer();
fileStream.Write(bytes,0,bytes.Length );
}
}
另一答案
MVC的工作解决方案
public ActionResult Index()
{
string fileName = "test.pdf";
string fileName1 = "test.vsix";
string fileNameZip = "Export_" + DateTime.Now.ToString("yyyyMMddhhmmss") + ".zip";
byte[] fileBytes = System.IO.File.ReadAllBytes(@"C: est est.pdf");
byte[] fileBytes1 = System.IO.File.ReadAllBytes(@"C: est est.vsix");
byte[] compressedBytes;
using (var outStream = new MemoryStream())
{
using (var archive = new ZipArchive(outStream, ZipArchiveMode.Create, true))
{
var fileInArchive = archive.CreateEntry(fileName, CompressionLevel.Optimal);
using (var entryStream = fileInArchive.Open())
using (var fileToCompressStream = new MemoryStream(fileBytes))
{
fileToCompressStream.CopyTo(entryStream);
}
var fileInArchive1 = archive.CreateEntry(fileName1, CompressionLevel.Optimal);
using (var entryStream = fileInArchive1.Open())
using (var fileToCompressStream = new MemoryStream(fileBytes1))
{
fileToCompressStream.CopyTo(entryStream);
}
}
compressedBytes = outStream.ToArray();
}
return File(compressedBytes, "application/zip", fileNameZip);
}
另一答案
using System;
using System.IO;
using System.IO.Compression;
namespace ConsoleApplication
{
class Program`enter code here`
{
static void Main(string[] args)
{
using (FileStream zipToOpen = new FileStream(@"c:usersexampleuser
elease.zip", FileMode.Open))
{
using (ZipArchive archive = new ZipArchive(zipToOpen, ZipArchiveMode.Update))
{
ZipArchiveEntry readmeEntry = archive.CreateEntry("Readme.txt");
using (StreamWriter writer = new StreamWriter(readmeEntry.Open()))
{
writer.WriteLine("Information about this package.");
writer.WriteLine("========================");
}
}
}
}
}
}
另一答案
这是将实体转换为XML文件然后压缩它的方法：
private void downloadFile(EntityXML xml) {
string nameDownloadXml = "File_1.xml";
string nameDownloadZip = "File_1.zip";
var serializer = new XmlSerializer(typeof(EntityXML));
Response.Clear();
Response.ClearContent();
Response.ClearHeaders();
Response.AddHeader("content-disposition", "attachment;filename=" + nameDownloadZip);
using (var memoryStream = new MemoryStream())
{
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
var demoFile = archive.CreateEntry(nameDownloadXml);
using (var entryStream = demoFile.Open())
using (StreamWriter writer = new StreamWriter(entryStream, System.Text.Encoding.UTF8))
{
serializer.Serialize(writer, xml);
}
}
using (var fileStream = Response.OutputStream)
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
Response.End();
}
另一答案
private void button6_Click(object sender, EventArgs e)
{
//create With Input FileNames
AddFileToArchive_InputByte(new ZipItem[]{ new ZipItem( @"E:1.jpg",@"images1.jpg"),
new ZipItem(@"E:2.txt",@"text2.txt")}, @"C: est.zip");
//create with input stream
AddFileToArchive_InputByte(new ZipItem[]{ new ZipItem(File.ReadAllBytes( @"E:1.jpg"),@"images1.jpg"),
new ZipItem(File.ReadAllBytes(@"E:2.txt"),@"text2.txt")}, @"C: est.zip");
//Create Archive And Return StreamZipFile
MemoryStream GetStreamZipFile = AddFileToArchive(new ZipItem[]{ new ZipItem( @"E:1.jpg",@"images1.jpg"),
new ZipItem(@"E:2.txt",@"text2.txt")});
//Extract in memory
ZipItem[] ListitemsWithBytes = ExtractItems(@"C: est.zip");
//Choese Files For Extract To memory
List<string> ListFileNameForExtract = new List<string>(new string[] { @"images1.jpg", @"text2.txt" });
ListitemsWithBytes = ExtractItems(@"C: est.zip", ListFileNameForExtract);
// Choese Files For Extract To Directory
ExtractItems(@"C: est.zip", ListFileNameForExtract, "c:\extractFiles");
}
public struct ZipItem
{
string _FileNameSource;
string _PathinArchive;
byte[] _Bytes;
public ZipItem(string __FileNameSource, string __PathinArchive)
{
_Bytes=null ;
_FileNameSource = __FileNameSo以上是关于使用System.IO.Compression在内存中创建ZIP存档的主要内容，如果未能解决你的问题，请参考以下文章
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