I have the following:
using(var memoryStream = new MemoryStream())
{
gc.CreatePackage(memoryStream);
}
MemoryStream spits out an excel fil. From the memoryStream, how do I go about actually showing the file that is produced. Note that I do not want to actually save the file to disk but merely display it.
So far I have the following but doesn't seem to work:
using (var memoryStream = new MemoryStream())
{
gc.CreatePackage(memoryStream);
using (var fileStream = File.OpenWrite("Test.xlsx"))
{
memoryStream.WriteTo(fileStream);
}
}
But not sure if I am on the right direction. I get an error saying:
System.Web.Mvc.Controller.File(byte[], string)' is a 'method', which
is not valid in the given context
I am not sure if I am going about this in the right direction.
Making the assumption that you are using ASP.NET MVC, you probably want the File response helper:
using (var memoryStream = new MemoryStream())
{
gc.CreatePackage(memoryStream);
memoryStream.Seek(0, SeekOrigin.Begin);
return File(memoryStream, "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet");
}
If you want to have the file source actually appear in the browser, you can also lie about the MIME type and use text/plain instead. The browser will most likely render this as plain text.
You can also add a third parameter to File in order to specify the filename the download should appear to the end-user to be.
Since the error you post seems to indicate that you're trying to return a generated file from an MVC controller, I think this may be what you're looking for.
public ActionResult MyAction()
{
using (var memoryStream = new MemoryStream())
{
gc.CreatePackage(memoryStream);
//Make sure the position of the stream is at 0
memoryStream.Position = 0;
//Return the contents of the stream with the appropriate MIME type
return File(memoryStream, "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet");
}
}
You could use a MemoryMappedFile to create a virtual file.
example code:
Write:
using (var mmf = MemoryMappedFile.CreateNew("MappedFileName", size, MemoryMappedFileAccess.ReadWriteExecute))
{
using (MemoryMappedViewAccessor accessor = mmf.CreateViewAccessor())
{
accessor.Write(Write your data to mapped file);
}
}
Read:
using (var mmf = MemoryMappedFile.OpenExisting("MappedFileName"))
{
using (MemoryMappedViewAccessor accessor = mmf.CreateViewAccessor())
{
accessor.Read......
}
}
Related
I'm using iText 7, specifically the HtmlConverter.ConvertToDocument method, to convert HTML to PDF. The problem is, I would really rather not create a PDF file on my server, I'd rather do everything in memory and just send it to the users browser so they can download it.
Could anyone show me an example of how to use this library but instead of writing to file write to a MemoryStream so I can send it directly to the browser?
I've been looking for examples and all I can seem to find are those which refer to file output.
I've tried the following, but keep getting an error about cannot access a closed memory stream.
public FileStreamResult pdf() {
using (var workStream = new MemoryStream())
using (var pdfWriter = new PdfWriter(workStream)) {
pdfWriter.SetCloseStream(false);
using (var document = HtmlConverter.ConvertToDocument(html, pdfWriter)) {
//Returns the written-to MemoryStream containing the PDF.
byte[] byteInfo = workStream.ToArray();
workStream.Write(byteInfo, 0, byteInfo.Length);
workStream.Position = 0;
return new FileStreamResult(workStream, "application/pdf");
}
//return new FileStreamResult(workStream, "application/pdf");
}
}
You meddle with the workStream before the document and pdfWriter have finished creating the result in it. Furthermore, the intent of your meddling is unclear, first you retrieve the bytes from the memory stream, then you write them back into it...?
public FileStreamResult pdf()
{
var workStream = new MemoryStream())
using (var pdfWriter = new PdfWriter(workStream))
{
pdfWriter.SetCloseStream(false);
using (var document = HtmlConverter.ConvertToDocument(html, pdfWriter))
{
}
}
workStream.Position = 0;
return new FileStreamResult(workStream, "application/pdf");
}
By the way, as you are essentially doing nothing special with the document returned by HtmlConverter.ConvertToDocument, you probably could use a different HtmlConverter method with less overhead in your code.
Generally this approach works
using (var ms = new MemoryStream())
{
//yourStream.Seek(0, SeekOrigin.Begin)
yourStream.CopyTo(ms);
}
I'm using this code to write an MP3 MemoryStream to file:
using (var nSpeakStreamAsMp3 = new MemoryStream())
using (var nWavFileReader = new WaveFileReader(nSpeakStream))
using (var nMp3Writer = new LameMP3FileWriter(nSpeakStreamAsMp3, nWavFileReader.WaveFormat, LAMEPreset.STANDARD_FAST))
{
nWavFileReader.CopyTo(nMp3Writer);
string sPath = "C:\\inetpub\\wwwroot\\server\\bin\\mymp3.mp3";
using (FileStream nFile = new FileStream(sPath, FileMode.Create, System.IO.FileAccess.Write))
{
nSpeakStreamAsMp3.CopyTo(nFile);
}
sRet = (String.Concat("data:audio/mpeg;base64,", Convert.ToBase64String(nSpeakStreamAsMp3.ToArray())));
}
return sRet;
For some reason which I don't see, this produces a file of 0 bytes.
However, the MP3 stream is valid and does work. I'm passing it as a Base64String to a website, and I do hear it.
Where might be the error here?
nSpeakStreamAsMp3 is currently positioned at the end of the stream; you need to think like a VCR: be kind, rewind (nSpeakStreamAsMp3.Position = 0;) before you copy the value out again
make sure you flush nMp3Writer; if possible, close nMp3Writer completely
I have a function I use for aggregating streams from a zip archive.
private void ExtractMiscellaneousFiles()
{
foreach (var miscellaneousFileName in _fileData.MiscellaneousFileNames)
{
var fileEntry = _archive.GetEntry(miscellaneousFileName);
if (fileEntry == null)
{
throw new ZipArchiveMissingFileException("Couldn't find " + miscellaneousFileName);
}
var stream = fileEntry.Open();
OtherFileStreams.Add(miscellaneousFileName, (DeflateStream) stream);
}
}
This works well in most cases. However, if I have a zip within a zip, I get an excpetion on casting the stream to a DeflateStream:
System.InvalidCastException: Unable to cast object of type 'System.IO.Compression.SubReadStream' to type 'System.IO.Compression.DeflateStream'.
I am unable to find Microsoft documentation for a SubReadStream. I would like my zip within a zip as a DeflateStream. Is this possible? If so how?
UPDATE
Still no success. I attempted #Sunshine's suggestion of copying the stream using the following code:
private void ExtractMiscellaneousFiles()
{
_logger.Log("Extracting misc files...");
foreach (var miscellaneousFileName in _fileData.MiscellaneousFileNames)
{
_logger.Log($"Opening misc file stream for {miscellaneousFileName}");
var fileEntry = _archive.GetEntry(miscellaneousFileName);
if (fileEntry == null)
{
throw new ZipArchiveMissingFileException("Couldn't find " + miscellaneousFileName);
}
var openStream = fileEntry.Open();
var deflateStream = openStream;
if (!(deflateStream is DeflateStream))
{
var memoryStream = new MemoryStream();
deflateStream.CopyTo(memoryStream);
memoryStream.Position = 0;
deflateStream = new DeflateStream(memoryStream, CompressionLevel.NoCompression, true);
}
OtherFileStreams.Add(miscellaneousFileName, (DeflateStream)deflateStream);
}
}
But I get a
System.NotSupportedException: Stream does not support reading.
I inspected deflateStream.CanRead and it is true.
I've discovered this happens not just on zips, but on files that are in the zip but are not compressed (because too small, for example). Surely there's a way to deal with this; surely someone has encountered this before. I'm opening a bounty on this question.
Here's the .NET source for SubReadStream, thanks to #Quantic.
The return type of ZipArchiveEntry.Open() is Stream. An abstract type, in practice it can be a DeflateStream (you'd be happy), a SubReadStream (boo) or a WrappedStream (boo). Woe be you if they decide to improve the class some day and use a ZopfliStream (boo). The workaround is not good, you are trying to deflate data that is not compressed (boo).
Too many boos.
Only good solution is to change the type of your OtherFileStreams member. We can't see it, smells like a List<DeflateStream>. It needs to be a List<Stream>.
So it looks like the when storing a zip file inside another zip it doesn't deflate the zip but rather just inlines the content of the zip with the rest of the files with some information that these entries are part of a sub zip file. Which makes sense because applying compression to something that is already compressed is a waste of time.
This zip file is marked as CompressionMethodValues.Stored in the archive, which causes .NET to just return the original stream it read instead to wrapping it in a DeflateStream.
Source here: https://github.com/dotnet/corefx/blob/master/src/System.IO.Compression/src/System/IO/Compression/ZipArchiveEntry.cs#L670
You could pass the stream into a ZipArchive, if it's not a DeflateStream (if you are interested in the file inside)
var stream = entry.Open();
if (!(stream is DeflateStream))
{
var subArchive = new ZipArchive(stream);
}
Or you can copy the stream to a FileStream (if you want to save it to disk)
var stream = entry.Open();
if (!(stream is DeflateStream))
{
var fs = File.Create(Path.GetTempFileName());
stream.CopyTo(fs);
fs.Close();
}
Or copy to any stream you are interested in using.
Note: This is also how .NET 4.6 behaves
I want to call a web api method and have it allow the user to download a zip file that I create in memory. I also want to create the entries in memory as well.
I'm having trouble getting the server to correctly output the download.
Here is my web api method:
[HttpGet]
[Route("api/downloadstaffdata")]
public HttpResponseMessage DownloadStaffData()
{
var response = new HttpResponseMessage(HttpStatusCode.OK);
using (var stream = new MemoryStream())
{
using (var archive = new ZipArchive(stream, ZipArchiveMode.Create, true))
{
//future for loop to create entries in memory from staff list
var entry = archive.CreateEntry("bob.txt");
using (var writer = new StreamWriter(entry.Open()))
{
writer.WriteLine("Info for: Bob");
}
//future add staff images as well
}
stream.Seek(0, SeekOrigin.Begin);
response.Content = new StreamContent(stream);
}
response.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment")
{
FileName = "staff_1234_1.zip"
};
response.Content.Headers.ContentType = new System.Net.Http.Headers.MediaTypeHeaderValue("application/zip");
return response;
}
Here is my calling js code:
window.open('api/downloadstaffdata');
Here is the response from Chrome:
net::ERR_CONNECTION_RESET
I don't know what I'm doing wrong. I've already searched SO and read the articles about creating the zip file, but I can't get passed the connection reset error when trying to return the zip archive to the client.
Any ideas?
You have your memory stream inside a using block. As such, your memory stream are being disposed before your controller has the chance to write it out (hence the ERR_CONNECTION_RESET).
A MemoryStream does not need to be disposed explicitly (its various derived type may need to be, but not the MemoryStream itself). Garbage Collector can clean it up automatically.
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