I am attempting to create a zipfile in ASP.NET MVC a single PDF file within it. However, using the code below, an empty zipfile is created. Can someone please advise what I am doing incorrectly?
public FileResult DownloadZipfile(string html)
{
MemoryStream memoryStream = new MemoryStream();
ZipArchive archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true);
byte[] rawDownload = PDFConverterUtils.PdfSharpConvert(html);
ZipArchiveEntry entry = archive.CreateEntry("MyPDF.pdf");
using (Stream entryStream = entry.Open())
using (StreamWriter streamWriter = new StreamWriter(entryStream))
{
streamWriter.BaseStream.Write(rawDownload, 0, rawDownload.Length);
}
return new FileStreamResult(memoryStream, System.Net.Mime.MediaTypeNames.Application.Zip) { FileDownloadName = "test.zip" };
}
When using a ZipArchive with a MemoryStream, I would suggest resetting the position of the stream after writing to it so that the content of the stream can be read by the response.
public FileResult DownloadZipfile(string html) {
byte[] rawDownload = PDFConverterUtils.PdfSharpConvert(html);
MemoryStream memoryStream = new MemoryStream();
using(ZipArchive archive = new ZipArchive(
stream: memoryStream,
mode: ZipArchiveMode.Create,
leaveOpen: true //To leave the memory stream open after disposal
)){
ZipArchiveEntry entry = archive.CreateEntry("MyPDF.pdf");
using (Stream entryStream = entry.Open()) {
entryStream.Write(rawDownload, 0, rawDownload.Length);
}
}
memoryStream.Position = 0;//reset memory stream position for read
return new FileStreamResult(memoryStream, System.Net.Mime.MediaTypeNames.Application.Zip) {
FileDownloadName = "test.zip"
};
}
As suggested in another answer you should dispose of the archive to force it to write its content to its underlying memory stream, but take note of the following
ZipArchive.Dispose()
Unless you construct the object by using the ZipArchive(Stream, ZipArchiveMode, Boolean) constructor overload and set its leaveOpen parameter to true, all underlying streams are closed and no longer available for subsequent write operations.
When you are finished using this instance of ZipArchive, call Dispose() to release all resources used by this instance. You should eliminate further references to this ZipArchive instance so that the garbage collector can reclaim the memory of the instance instead of keeping it alive for finalization.
Because you want to make use of the memory stream after writing to it, you need to make sure that it remains open, and that the position of the stream is reset to the beginning so that the content of the stream can be read from the start.
Related
I'm trying to serve a file to the users that is packed inside a zip archive on the server.
The project is ASP.NET Core 5.0 MVC project.
I managed to do it by using ZipArchiveEntry.Open() and copying that to a memory stream.
string zipFile = #"D:\all_installs.zip";
using (FileStream fs = new FileStream(zipFile, FileMode.Open))
{
using (ZipArchive zip = new ZipArchive(fs))
{
ZipArchiveEntry entry = zip.Entries.FirstOrDefault(x => x.FullName == "downloadable file.iso");
string name = entry.FullName;
string baseName = Path.GetFileName(name);
//open a stream to the zip entry
Stream stream = entry.Open();
//copy stream to memory
MemoryStream memoryStream = new MemoryStream();
stream.CopyTo(memoryStream); //big memory usage?
memoryStream.Position = 0;
return this.File(memoryStream, "application/octet-stream", baseName);
}
}
This would require a lot of RAM if there are many simultaneous downloads, so instead I would like to serve it directly from the archive, which I know will require the CPU while unpacking it, but that's fine since the download speed will be very limited anyways.
I tried serving stream directly, but I get the following error:
NotSupportedException: Stream does not support reading.
How can I serve the entry-stream directly?
Problem is both FileStream fs and ZipArchive zip are disposed here, so when it's time to write response and asp.net tries to read your zip entry (stream) - it's not available any more, since everything has been disposed.
You need to not dispose them right away but instead tell asp.net to dispose them when it's done writing the response. For that, HttpResponse has method RegisterForDispose, so you need to do something like that:
string zipFile = #"C:\tmp\record.zip";
FileStream fs = null;
ZipArchive zip = null;
Stream stream = null;
try {
fs = new FileStream(zipFile, FileMode.Open);
zip = new ZipArchive(fs);
ZipArchiveEntry entry = zip.Entries.First(x => x.FullName == "24fa535b-2fc9-4ce5-96f4-2ff1ef0d9b64.json");
string name = entry.FullName;
string baseName = Path.GetFileName(name);
//open a stream to the zip entry
stream = entry.Open();
return this.File(stream, "application/octet-stream", baseName);
}
finally {
if (stream != null)
this.Response.RegisterForDispose(stream);
if (zip != null)
this.Response.RegisterForDispose(zip);
if (fs != null)
this.Response.RegisterForDispose(fs);
}
Now asp.net will first write the response, then dispose all your disposables for you.
Here is my code.
using (FileStream zipToOpen = new FileStream(#"D:\test\1.txt", 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("========================");
}
}
// I reused this stream again and below code just a sample.
// It would get exception
using (ZipArchive archive = new ZipArchive(zipToOpen, ZipArchiveMode.Update))
{
//System.ArgumentException: 'Update mode requires a stream with read, write, and seek capabilities.'
}
}
Is this exception caused by 'stream cannot be resued in C#' ?
If the answer is yes, any office documents can let me take a reference ?
Thanks in advance, it's appreciate if your guys can provide some help.
By default, ZipArchive assumes lifetime control of the Stream it is handed, and disposes it when it is disposed. To avoid this, set leaveOpen to true in the constructor overloads. However: you will also probably need to rewind the stream between usages, so you'll need zipToOpen.Position = 0; between the two using blocks. Anecdotally, it does seem unlikely that 1.txt is a zip file, but... I guess it could be!
I want to create a zip-file and return it to the browser so that it downloads the zip to the downloads-folder.
var images = imageRepository.GetAll(allCountryId);
using (FileStream f2 = new FileStream("SudaAmerica", FileMode.Create))
using (GZipStream gz = new GZipStream(f2, CompressionMode.Compress, false))
{
foreach (var image in images)
{
gz.Write(image.ImageData, 0, image.ImageData.Length);
}
return base.File(gz, "application/zip", "SudaAmerica");
}
i have tried the above but then i get an error saying the stream is disposed.
Is this possible or should i use another library then gzipstream?
The problem here is exactly what it says: you are handing it something based on gz, but gz gets disposed the moment you leave the using.
One option would be to wait until outside the using block, then tell it to use the filename of the thing you just wrote ("SudaAmerica"). However, IMO you shouldn't actually be writing a file here at all. If you use a MemoryStream instead, you can use .ToArray() to get a byte[] of the contents, which you can use in the File method. This requires no IO access, which is a win in about 20 different ways. Well, maybe 3 ways. But...
var images = imageRepository.GetAll(allCountryId);
using (MemoryStream ms = new MemoryStream())
{
using (GZipStream gz = new GZipStream(ms, CompressionMode.Compress, false))
{
foreach (var image in images)
{
gz.Write(image.ImageData, 0, image.ImageData.Length);
}
}
return base.File(ms.ToArray(), "application/zip", "SudaAmerica");
}
Note that a gzip stream is not the same as a .zip archive, so I very much doubt this will have the result you want. Zip archive creation is available elsewhere in the .NET framework, but it is not via GZipStream.
You probably want ZipArchive
I am reading a file using StreamReader fileReader = File.OpenText(filePath). I would like to modify one line in the file in memory and push the modified stream to another method.
What I would like to avoid is reading the whole file into a string and modifying the string (doesn't scale). I would also like to avoid modifying the actual file.
Is there a straightforward way of doing this?
There is no built-in way to do that in .Net framework.
Stream and StreamReader/StreamWriter classes are designed to be chained if necessary (like GZipStream wraps stream to compress it). So you can create wrapper StreamReader and update data as you need for every operation after calling wrapped reader.
You can open two stream -one for read, one for write- at the same time. I tested simple code that works, but not sure that's what you want:
// "2.bar\r\n" will be replaced by "!!!!!\r\n"
File.WriteAllText("test.txt",
#"1.foo
2.bar
3.fake");
// open inputStream for StreamReader, and open outputStream for StreamWriter
using (var inputStream = File.Open("test.txt", FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
using (var reader = new StreamReader(inputStream))
using (var outputStream = File.Open("test.txt", FileMode.Open, FileAccess.Write, FileShare.Read))
using (var writer = new StreamWriter(outputStream))
{
var position = 0L; // track the reading position
var newLineLength = Environment.NewLine.Length;
while (!reader.EndOfStream)
{
var line = reader.ReadLine();
// your particular conditions here.
if (line.StartsWith("2."))
{
// seek line start position
outputStream.Seek(position, SeekOrigin.Begin);
// replace by something,
// but the length should be equal to original in this case.
writer.WriteLine(new String('!', line.Length));
}
position += line.Length + newLineLength;
}
}
/* as a result, test.txt will be:
1.foo
!!!!!
3.fake
*/
As you can see, both streams can be accessed by StreamReader and StreamWriter at the same time. And you can also manipulate both read/write position as well.
I don't understand what I'm doing wrong here. I generate couple of memory streams and in debug-mode I see that they are populated. But when I try to copy MemoryStream to FileStream in order to save the file fileStream is not populated and file is 0bytes long (empty).
Here is my code
if (file.ContentLength > 0)
{
var bytes = ImageUploader.FilestreamToBytes(file); // bytes is populated
using (var inStream = new MemoryStream(bytes)) // inStream is populated
{
using (var outStream = new MemoryStream())
{
using (var imageFactory = new ImageFactory())
{
imageFactory.Load(inStream)
.Resize(new Size(320, 0))
.Format(ImageFormat.Jpeg)
.Quality(70)
.Save(outStream);
}
// outStream is populated here
var fileName = "test.jpg";
using (var fileStream = new FileStream(Server.MapPath("~/content/u/") + fileName, FileMode.CreateNew, FileAccess.ReadWrite))
{
outStream.CopyTo(fileStream); // fileStream is not populated
}
}
}
}
You need to reset the position of the stream before copying.
outStream.Position = 0;
outStream.CopyTo(fileStream);
You used the outStream when saving the file using the imageFactory. That function populated the outStream. While populating the outStream the position is set to the end of the populated area. That is so that when you keep on writing bytes to the steam, it doesn't override existing bytes. But then to read it (for copy purposes) you need to set the position to the start so you can start reading at the start.
If your objective is simply to dump the memory stream to a physical file (e.g. to look at the contents) - it can be done in one move:
System.IO.File.WriteAllBytes(#"C:\\filename", memoryStream.ToArray());
No need to set the stream position first either, since the .ToArray() operation explicitly ignores that, as per #BaconBits comment below https://learn.microsoft.com/en-us/dotnet/api/system.io.memorystream.toarray?view=netframework-4.7.2.
Another alternative to CopyTo is WriteTo.
Advantage:
No need to reset Position.
Usage:
outStream.WriteTo(fileStream);
Function Description:
Writes the entire contents of this memory stream to another stream.