I have been developing an web application with Asp.Net and I'm using SharpZipLib to work with odt files (from Open Office) and in the future docx files (for ms office). I need to open an odt file (like a zip file) change a xml file inside it, zip again and give it to the browser send to my client.
I can do this in file system but it will get a space in my disk temporarily and we don't want it. I would like to do this in memory (with a MemoryStream class), but I don't know how to unzip folders/files in a memory stream with SharpZipLib, change and use it to zip again. Is there any sample about how to do this?
Thank you
You can use something like
Stream inputStream = //... File.OpenRead(...);
//for read file
ZipInputStream zipInputStream = new ZipInputStream(inputStream));
//for output
MemoryStream memoryStream = new MemoryStream();
using ( ZipOutputStream zipStream = new ZipOutputStream(memoryStream))
{
ZipEntry entry = new ZipEntry("...");
//...
zipStream.PutNextEntry(entry);
zipStream.Write(data, 0, data.Length);
//...
zipStream.Finish();
zipStream.Close();
}
Edit ::
You need in general unZip your file, get ZipEntry , change , and write in ZipOutputStream with MemoryStream.
Use this article http://www.codeproject.com/KB/cs/Zip_UnZip.aspx
Related
I am calling REST API which is accepting Stream to upload file from local device, so for that right now I am using following code to get Stream from a file and than closing that stream after it get's uploaded:
var stream = new FileStream(file, FileMode.Open, FileAccess.ReadWrite);
The problem with the above approach is that, until entire file gets uploaded to server user don't have any chance to delete that file because stream of that file is open, what would be the solution to resolve this issue?
If your typical file is reasonably sized (and I'm hoping you won't be uploading 2GB+ files to a REST API), you could always just read the stream into memory and before feeding it to your API, like so:
using (MemoryStream memoryStream = new MemoryStream())
{
using (FileStream fileStream = new FileStream(file, FileMode.Open, FileAccess.ReadWrite)) {
fileStream.CopyTo(memoryStream);
}
memoryStream.Position = 0; // Reset to origin.
// Now use the MemoryStream as you would a FileStream:
api.Upload(memoryStream);
}
Another alternative is to create a temp copy of the file on your hard drive and feed that to the API - but then dealing with cleanup can become a bit cumbersome. FileOptions.DeleteOnClose is your friend and may very well suffice for your purposes, but it still offers no bulletproof guarantees.
I have use ICSharpCode.SharpZipLib.Zip library for creating Zip file it's working fine, but when i have attach this Zip file to mail attachment, mail is not sent due to attached zip file,
here is code for attaching zip file to mail
System.Net.Mail.Attachment attachment = null;
try
{
MemoryStream memoryStream = new MemoryStream();
attachment = new System.Net.Mail.Attachment(memoryStream, "test.zip");
}
catch (Exception e)
{
return false;
}
please know me how i can send zip file through mail?.
Your code
MemoryStream memoryStream = new MemoryStream();
attachment = new System.Net.Mail.Attachment(memoryStream, "test.zip");
passes a Stream, but that Stream is empty (there's nothing in memoryStream).
If you want to use a MemoryStream, you must read the contents of the ZIP file into memory. You can also use a FileStream if the ZIP is already on disk.
If using a MemoryStream, be sure and set its position 0.
memoryStream.Position = 0;
Depending on how you are using SharpZipLib, you may have access to a ZipOutputStream. If you do, I think you could use that.
There could be lot of reasons for the issue,
Check file date & time
Size of the file ( Based on server settings)
Try different formats (rar,7z, etc)
I'm needing to create a zipped document containing files that exist on the server. I am using the Ionic.Zip to do so, and to create a new Package (which is the zip file) I have to have either a path to a physical file or a stream. I am trying to not create an actual file that would be the zip file, instead just create a stream that would exist in memory or something. how can i do this?
Create the package using a MemoryStream then.
You can try the save method in the ZipFile Class. It can save to a stream
try this.
using (MemoryStream ms = new MemoryStream())
{
using (Ionic.Zip.ZipFile zipFile = new Ionic.Zip.ZipFile())
{
zipFile.AddFiles(filesToBeZipped, false, "NewFolder");//filesTobeZipped is a List<string>
zipFile.Save(ms);
}
}
You'll want to use the .AddEntry method on the ZipFile you've created specifying a name and the byte[] containing the actual file data.
ex.
ZipFile zipFile = new ZipFile();
zipFile.AddEntry(file.FileName, file.FileData);
where file.FileName will be the entry name (in the zip file) and file.FileData is the byte array.
I have see this term branded around but I don't really understand how you open a file in memory.
I have the files written to disk in a temp location but this needs cleaning when a certain form closes and I can't do it when it's open. It's a must that this folder gets emptied. I was wondering if I opened files in memory instead whether it would make a difference?
MemoryStream inMemoryCopy = new MemoryStream();
using (FileStream fs = File.OpenRead(path))
{
fs.CopyTo(inMemoryCopy);
}
// Now you can delete the file at 'path' and still have an in memory copy
I think you want to work with Memory Mapped files added recently to .NET 4.
http://blogs.msdn.com/b/salvapatuel/archive/2009/06/08/working-with-memory-mapped-files-in-net-4.aspx
Memory Mapped Files .NET
I think it means to read the content of that file into memory as a whole and then close the connection to the file. Assuming it's a file that's not too big you could just read it into a byte[]:
byte[] fileContent = File.ReadAllBytes(fileName);
If it's a text file read it into a string using
string fileContent = File.ReadAllText(fileName);
Once you've done that use a StreamReader to read it later as you would a file on disk.
You can use DeleteOnClose parameter of FileStream constructor:
FileStream fs = new FileStream("<Path Here>", FileMode.Create,
FileAccess.ReadWrite, FileShare.None, 1024, FileOptions.DeleteOnClose);
and the file will be deleted when closed.
I am using the following C# code to compress a file:
// Open the stream we want to compress
FileStream fs = File.Create(#"C:\Projects\Samples\test\compressed.zip", 0);
// Creates the GZipStream
GZipStream gzip = new GZipStream(fs, CompressionMode.Compress);
// Reading the content to compress
byte[] bytes = File.ReadAllBytes(#"C:\Projects\Samples\samplefile.xml");
// Writing compressed content
gzip.Write(bytes, 0, bytes.Length);
gzip.Close(); // This also closes the FileStream (the underlying stream)
However, when I extract the file from windows explorer the file loses it's extension so instead of samplefile.xml it just becomes samplefile. Same thing happened with .txt file not just .xml file.
Can you help me see what I'm doing wrong?
ok found the problem:
Line 2 has to be as follows:
FileStream fs = File.Create(#"C:\Projects\Samples\test\compressed.xml.zip", 0);
GZipStream doesn't create zip archives. It creates a gzip file, which contains only one file, and doesn't necessarily store a filename at all. Normally you should use the .gz extension to identify a gzip file, and it's conventional to use the entire name of the original file with .gz appended on the end. See also here for more information about gzip format: http://en.wikipedia.org/wiki/Gzip#File_format
If you actually want to create zip archives, you might want to use a library like SharpZipLib: http://www.icsharpcode.net/opensource/sharpziplib/