问题
am trying to add a file to an existing archive using the following code. When run no errors or exceptions are shown but no files are added to the archive either. Any ideas why?
using (FileStream fileStream = File.Open(archivePath, FileMode.Open, FileAccess.ReadWrite))
using (ZipOutputStream zipToWrite = new ZipOutputStream(fileStream))
{
zipToWrite.SetLevel(9);
using (FileStream newFileStream = File.OpenRead(sourceFiles[0]))
{
byte[] byteBuffer = new byte[newFileStream.Length - 1];
newFileStream.Read(byteBuffer, 0, byteBuffer.Length);
ZipEntry entry = new ZipEntry(sourceFiles[0]);
zipToWrite.PutNextEntry(entry);
zipToWrite.Write(byteBuffer, 0, byteBuffer.Length);
zipToWrite.CloseEntry();
zipToWrite.Close();
zipToWrite.Finish();
}
}
回答1:
In DotNetZip, adding files to an existing zip is really simple and reliable.
using (var zip = ZipFile.Read(nameOfExistingZip))
{
zip.CompressionLevel = Ionic.Zlib.CompressionLevel.BestCompression;
zip.AddFile(additionalFileToAdd);
zip.Save();
}
If you want to specify a directory path for that new file, then use a different overload for AddFile().
using (var zip = ZipFile.Read(nameOfExistingZip))
{
zip.CompressionLevel = Ionic.Zlib.CompressionLevel.BestCompression;
zip.AddFile(additionalFileToAdd, "directory\\For\\The\\Added\\File");
zip.Save();
}
If you want to add a set of files, use AddFiles().
using (var zip = ZipFile.Read(nameOfExistingZip))
{
zip.CompressionLevel = Ionic.Zlib.CompressionLevel.BestCompression;
zip.AddFiles(listOfFilesToAdd, "directory\\For\\The\\Added\\Files");
zip.Save();
}
You don't have to worry about Close(), CloseEntry(), CommitUpdate(), Finish() or any of that other gunk.
回答2:
From Codeproject someone used this code. Only difference is close and finish otherway around and the write part:
using (ZipOutputStream s = new
ZipOutputStream(File.Create(txtSaveTo.Text + "\\" +
sZipFileName + ".zip")))
{
s.SetLevel(9); // 0-9, 9 being the highest compression
byte[] buffer = new byte[4096];
foreach (string file in filenames)
{
ZipEntry entry = new
ZipEntry(Path.GetFileName(file));
entry.DateTime = DateTime.Now;
s.PutNextEntry(entry);
using (FileStream fs = File.OpenRead(file))
{
int sourceBytes;
do
{
sourceBytes = fs.Read(buffer, 0,
buffer.Length);
s.Write(buffer, 0, sourceBytes);
} while (sourceBytes > 0);
}
}
s.Finish();
s.Close();
}
BTW:
byte[] byteBuffer = new byte[newFileStream.Length - 1];
newFileStream.Read(byteBuffer, 0, byteBuffer.Length);
This is incorrect, the size is newFileStream.length else the Read goes wrong. You have an array and you make it for example 10-1 is 9 bytes long, from 0 to 8.
But your reading from 0 to 9...
回答3:
I think your Finish call should be before your Close call.
Update: This looks like a known bug. It's possible it may already have been fixed - you'll need to check your SharpZipLib version to see if it incorporates any fix. If not, you can work around it by copying all files to a new archive, adding the new file, then moving the new archive to the old archive name.
回答4:
/// <summary>
/// 添加压缩文件 p 为客户端传回来的文件/夹列表,用分号隔开，不包括主路径， zipfile压缩包的名称
/// </summary>
/// <param name="p"></param>
/// <param name="zipfile"></param>
public void AddZipFile(string p, string zipfile)
{
if (ServerDir.LastIndexOf(@"\") != ServerDir.Length - 1)
{
ServerDir += @"\";
}
string[] tmp = p.Split(new char[] { ';' }); //分离文件列表
if (zipfile != "") //压缩包名称不为空
{
string zipfilepath=ServerDir + zipfile;
if (_ZipOutputStream == null)
{
_ZipOutputStream = new ZipOutputStream(File.Create(zipfilepath));
}
for (int i = 0; i < tmp.Length; i++)
{
if (tmp[i] != "") //分离出来的文件名不为空
{
this.AddZipEntry(tmp[i], _ZipOutputStream, out _ZipOutputStream); //向压缩文件流加入内容
}
}
}
}
private static ZipOutputStream _ZipOutputStream;
public void Close()
{
_ZipOutputStream.Finish();
_ZipOutputStream.Close();
}
回答5:
there is a folder ZippedFolder in site's root directory , inside it we have a archive MyZipFiles.
There is a folder with name siteImages which consists of all image files. The following is the code to zip the images
string zipPath = Server.MapPath("~/ZippedFolder/MyZipFiles.zip");
using (ZipFile zip = new ZipFile())
{
zip.AddFile(Server.MapPath("~/siteImages/img1.jpg"),string.Empty);
zip.AddFile(Server.MapPath("~/siteImages/img2.jpg"),string.Empty);
zip.AddFile(Server.MapPath("~/siteImages/img2.jpg"),string.Empty);
zip.Save(zipPath);
}
if we have different file formats and we want your files to be saved in respective folders,you can specify the code as follows.
string zipPath = Server.MapPath("~/ZippedFolder/MyZipFiles.zip");
using (ZipFile zip = new ZipFile())
{
zip.AddFile(Server.MapPath("~/siteimages/img1.jpg"), "images");
zip.AddFile(Server.MapPath("~/siteimages/img2.jpg"), "images");
zip.AddFile(Server.MapPath("~/documents/customer.pdf"), "files");
zip.AddFile(Server.MapPath("~/documents/sample.doc"), "files");
zip.Save(zipPath);
}
now the archive contains two folders images ---- > img1.jpg , img2,.jpg and another folder files --> customer.pdf, sample.doc
回答6:
The ZipOutputStream class does not update existing ZIP files. Use the ZipFile class instead.
回答7:
I have found a simple solution keeping it to ZipFile and ZipEntry only
ZipFile zipExisting = ZipFile.Read(Server.MapPath("/_Layouts/includes/Template.zip"));
ICollection<ZipEntry> entries = _zipFileNew.Entries;
foreach (ZipEntry zipfile in entries)
{
zipExisting.AddEntry(zipfile.FileName, zipfile.InputStream);
}
zipExisting.Save(Response.OutputStream);
Response.End();
来源：https://stackoverflow.com/questions/1356003/c-sharp-sharpziplib-adding-file-to-existing-archive