I have taken a IFormFile as an input and I need to read its content.
And, I am going to read the file content using the following code. However, File.OpenText requires a file path as input.
TextReader fileReader = File.OpenText(pathToFile)
Since I am not going to save the file in a physical location I have no file path with me at hand to pass to File.OpenText as an input.
How can I solve this ?
public async Task<string> PdfFileSave(IFormFile file, string nameOfThePerson)
{
TextReader fileReader = File.OpenText(pathToFile);
}
Try this:
using (var memoryStream = new MemoryStream())
{
await file.CopyToAsync(memoryStream);
byte[] fileBytes = memoryStream.ToArray();
string text = Encoding.UTF8.GetString(fileBytes);
}
you can use File.OpenReadStream():
public async Task<string> PdfFileSave(IFormFile file, string nameOfThePerson)
{
TextReader fileReader = new StreamReader( file.OpenReadStream() );
}
Related
I have a service that downloads a *.tgz file from a remote endpoint. I use SharpZipLib to extract and write the content of that compressed archive to disk. But now I want to prevent writing the files to disk (because that process doesn't have write permissions on that disk) and keep them in memory.
How can I access the decompressed files from memory? (Let's assume the archive holds simple text files)
Here is what I have so far:
public void Decompress(byte[] byteArray)
{
Stream inStream = new MemoryStream(byteArray);
Stream gzipStream = new GZipInputStream(inStream);
TarArchive tarArchive = TarArchive.CreateInputTarArchive(gzipStream);
tarArchive.ExtractContents(#".");
tarArchive.Close();
gzipStream.Close();
inStream.Close();
}
Check this and this out.
Turns out, ExtractContents() works by iterating over TarInputStream. When you create your TarArchive like this:
TarArchive.CreateInputTarArchive(gzipStream);
it actually wraps the stream you're passing into a TarInputStream. Thus, if you want more fine-grained control over how you extract files, you must use TarInputStream directly.
See, if you can iterate over files, directories and actual file contents like this:
Stream inStream = new MemoryStream(byteArray);
Stream gzipStream = new GZipInputStream(inStream);
using (var tarInputStream = new TarInputStream(gzipStream))
{
TarEntry entry;
while ((entry = tarInputStream.GetNextEntry()) != null)
{
var fileName = entry.Name;
using (var fileContents = new MemoryStream())
{
tarInputStream.CopyEntryContents(fileContents);
// use entry, fileName or fileContents here
}
}
}
I am writing a REST API which will take in a JSON request object. The request object will have to be serialized to a file in JSON format; the file has to be compressed into a zip file and the ZIP file has to be posted to another service, for which I would have to deserialize the ZIP file. All this because the service I have to call expects me to post data as ZIP file. I am trying to see if I can avoid disk IO. Is there a way to directly convert the object into a byte array representing ZIP content in-memory instead of all the above steps?
Note : I'd prefer accomplishing this using .net framework libraries (as against external libraries)
Yes, it is possible to create a zip file completely on memory, here is an example using SharpZip Library (Update: A sample using ZipArchive added at the end):
public static void Main()
{
var fileContent = Encoding.UTF8.GetBytes(
#"{
""fruit"":""apple"",
""taste"":""yummy""
}"
);
var zipStream = new MemoryStream();
var zip = new ZipOutputStream(zipStream);
AddEntry("file0.json", fileContent, zip); //first file
AddEntry("file1.json", fileContent, zip); //second file (with same content)
zip.Close();
//only for testing to see if the zip file is valid!
File.WriteAllBytes("test.zip", zipStream.ToArray());
}
private static void AddEntry(string fileName, byte[] fileContent, ZipOutputStream zip)
{
var zipEntry = new ZipEntry(fileName) {DateTime = DateTime.Now, Size = fileContent.Length};
zip.PutNextEntry(zipEntry);
zip.Write(fileContent, 0, fileContent.Length);
zip.CloseEntry();
}
You can obtain SharpZip using Nuget command PM> Install-Package SharpZipLib
Update:
Note : I'd prefer accomplishing this using .net framework libraries (as against external libraries)
Here is an example using Built-in ZipArchive from System.IO.Compression.Dll
public static void Main()
{
var fileContent = Encoding.UTF8.GetBytes(
#"{
""fruit"":""apple"",
""taste"":""yummy""
}"
);
var zipContent = new MemoryStream();
var archive = new ZipArchive(zipContent, ZipArchiveMode.Create);
AddEntry("file1.json",fileContent,archive);
AddEntry("file2.json",fileContent,archive); //second file (same content)
archive.Dispose();
File.WriteAllBytes("testa.zip",zipContent.ToArray());
}
private static void AddEntry(string fileName, byte[] fileContent,ZipArchive archive)
{
var entry = archive.CreateEntry(fileName);
using (var stream = entry.Open())
stream.Write(fileContent, 0, fileContent.Length);
}
You could use the GZipStream class along with MemoryStream.
A quick example:
using System.IO;
using System.IO.Compression;
//Put JSON into a MemoryStream
var theJson = "Your JSON Here";
var jsonStream = new MemoryStream();
var jsonStreamWriter = new StreamWriter(jsonStream);
jsonStreamWriter.Write(theJson);
jsonStreamWriter.Flush();
//Reset stream so it points to the beginning of the JSON
jsonStream.Seek(0, System.IO.SeekOrigin.Begin);
//Create stream to hold your zipped JSON
var zippedStream = new MemoryStream();
//Zip JSON and put it in zippedStream via compressionStream.
var compressionStream = new GZipStream(zippedStream, CompressionLevel.Optimal);
jsonStream.CopyTo(compressionStream);
//Reset zipped stream to point at the beginning of data
zippedStream.Seek(0, SeekOrigin.Begin);
//Get ByteArray with zipped JSON
var zippedJsonBytes = zippedStream.ToArray();
You should try the ZipArchive Class streaming to a MemoryStream Class
Yes. You can return it as a binary stream. Depending on the language, you can use special libraries. You will also need libraries on the client.
I'm trying to convert a .db file to binary so I can stream it across a web server. I'm pretty new to C#. I've gotten as far as looking at code snippets online but I'm not really sure if the code below puts me on the right track. How I can write the data once I read it? Does BinaryReader automatically open up and read the entire file so I can then just write it out in binary format?
class Program
{
static void Main(string[] args)
{
using (FileStream fs = new FileStream("output.bin", FileMode.Create))
{
using (BinaryWriter bw = new BinaryWriter(fs))
{
long totalBytes = new System.IO.FileInfo("input.db").Length;
byte[] buffer = null;
BinaryReader binReader = new BinaryReader(File.Open("input.db", FileMode.Open));
}
}
}
}
Edit: Code to stream the database:
[WebGet(UriTemplate = "GetDatabase/{databaseName}")]
public Stream GetDatabase(string databaseName)
{
string fileName = "\\\\computer\\" + databaseName + ".db";
if (File.Exists(fileName))
{
FileStream stream = File.OpenRead(fileName);
if (WebOperationContext.Current != null)
{
WebOperationContext.Current.OutgoingResponse.ContentType = "binary/.bin";
}
return stream;
}
return null;
}
When I call my server, I get nothing back. When I use this same type of method for a content-type of image/.png, it works fine.
All the code you posted will actually do is copy the file input.db to the file output.bin. You could accomplish the same using File.Copy.
BinaryReader will just read in all of the bytes of the file. It is a suitable start to streaming the bytes to an output stream that expects binary data.
Once you have the bytes corresponding to your file, you can write them to the web server's response like this:
using (BinaryReader binReader = new BinaryReader(File.Open("input.db",
FileMode.Open)))
{
byte[] bytes = binReader.ReadBytes(int.MaxValue); // See note below
Response.BinaryWrite(bytes);
Response.Flush();
Response.Close();
Response.End();
}
Note: The code binReader.ReadBytes(int.MaxValue) is for demonstrating the concept only. Don't use it in production code as loading a large file can quickly lead to an OutOfMemoryException. Instead, you should read in the file in chunks, writing to the response stream in chunks.
See this answer for guidance on how to do that
https://stackoverflow.com/a/8613300/141172
Basically all I want is to load a Gziped file into a rich text box. I found some code on the MS .NET site for decompressing the file. Now I want to point that stream to a rich text box, but I keep getting the error "An object reference is required for the non-static field, method, or property 'WindowsFormsApplication1.Form1.richTextBox1' "
Code is here. What am I doing wrong? Thanks in advance.
public static void Decompress(FileInfo fi)
{
// Get the stream of the source file.
using (FileStream inFile = fi.OpenRead())
{
// Get original file extension, for example
// "doc" from report.doc.gz.
string curFile = fi.FullName;
string origName = curFile.Remove(curFile.Length -
fi.Extension.Length);
//Create the decompressed file.
using (FileStream outFile = File.Create(origName))
{
using (GZipStream Decompress = new GZipStream(inFile,
CompressionMode.Decompress))
{
// Copy the decompression stream
// into the output file.
Decompress.CopyTo(outFile);
richTextBox1.LoadFile(Decompress.CopyTo(outFile), RichTextBoxStreamType.PlainText);
// problem right here ^^^^
}//using
}//using
}//using
}//DeCompress
Just a hunch, but try this instead:
richTextBox1.LoadFile(outFile, RichTextBoxStreamType.PlainText);
The Decompress.CopyTo(outFile) is a method and doesn't return anything, which is probably why the LoadFile method is coughing on that line.
Also, change your function to this (you can't have your control referenced in a static method):
public void Decompress(FileInfo fi)
What I ended up doing is a hack, but basically I dump the uncompressed data to a file then load that file in the RTF. I'm sure it's much slower than streaming it directly to RTF but I could not get that piece working. It's functional, but not great. I pass in the fi variable to Decompress based on what the program arguement is, and I then assign that program to be run when a user double clicks on a gz file in windows. So the code looks like this:
public void Decompress(FileInfo fi)
{
// Get the stream of the source file.
using (FileStream inFile = fi.OpenRead())
{
// Get original file extension, for example
// "doc" from report.doc.gz.
string curFile = fi.FullName;
string origName = curFile.Remove(curFile.Length -
fi.Extension.Length);
//Create the decompressed file.
using (FileStream outFile = File.Create(origName))
{
using (GZipStream Decompress = new GZipStream(inFile,
CompressionMode.Decompress))
{
// Copy the decompression stream
// into the output file.
Decompress.CopyTo(outFile);
Decompress.Close();
outFile.Close();
inFile.Close();
rtbOut.LoadFile(origName, RichTextBoxStreamType.PlainText);
string tmp = rtbOut.Text;
}//using
}//using
}//using
} //Decompress
Is there a way to create a Zip archive that contains multiple files, when the files are currently in memory? The files I want to save are really just text only and are stored in a string class in my application. But I would like to save multiple files in a single self-contained archive. They can all be in the root of the archive.
It would be nice to be able to do this using SharpZipLib.
Use ZipEntry and PutNextEntry() for this. The following shows how to do it for a file, but for an in-memory object just use a MemoryStream
FileStream fZip = File.Create(compressedOutputFile);
ZipOutputStream zipOStream = new ZipOutputStream(fZip);
foreach (FileInfo fi in allfiles)
{
ZipEntry entry = new ZipEntry((fi.Name));
zipOStream.PutNextEntry(entry);
FileStream fs = File.OpenRead(fi.FullName);
try
{
byte[] transferBuffer[1024];
do
{
bytesRead = fs.Read(transferBuffer, 0, transferBuffer.Length);
zipOStream.Write(transferBuffer, 0, bytesRead);
}
while (bytesRead > 0);
}
finally
{
fs.Close();
}
}
zipOStream.Finish();
zipOStream.Close();
Using SharpZipLib for this seems pretty complicated. This is so much easier in DotNetZip. In v1.9, the code looks like this:
using (ZipFile zip = new ZipFile())
{
zip.AddEntry("Readme.txt", stringContent1);
zip.AddEntry("readings/Data.csv", stringContent2);
zip.AddEntry("readings/Index.xml", stringContent3);
zip.Save("Archive1.zip");
}
The code above assumes stringContent{1,2,3} contains the data to be stored in the files (or entries) in the zip archive. The first entry is "Readme.txt" and it is stored in the top level "Directory" in the zip archive. The next two entries are stored in the "readings" directory in the zip archive.
The strings are encoded in the default encoding. There is an overload of AddEntry(), not shown here, that allows you to explicitly specify the encoding to use.
If you have the content in a stream or byte array, not a string, there are overloads for AddEntry() that accept those types. There are also overloads that accept a Write delegate, a method of yours that is invoked to write data into the zip. This works for easily saving a DataSet into a zip file, for example.
DotNetZip is free and open source.
This function should create a byte array from a stream of data: I've created a simple interface for handling files for simplicity
public interface IHasDocumentProperties
{
byte[] Content { get; set; }
string Name { get; set; }
}
public void CreateZipFileContent(string filePath, IEnumerable<IHasDocumentProperties> fileInfos)
{
using (var memoryStream = new MemoryStream())
{
using (var zipArchive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
foreach(var fileInfo in fileInfos)
{
var entry = zipArchive.CreateEntry(fileInfo.Name);
using (var entryStream = entry.Open())
{
entryStream.Write(fileInfo.Content, 0, fileInfo.Content.Length);
}
}
}
using (var fileStream = new FileStream(filePath, FileMode.OpenOrCreate, System.IO.FileAccess.Write))
{
memoryStream.Position = 0;
memoryStream.CopyTo(fileStream);
}
}
}
Yes, you can use SharpZipLib to do this - when you need to supply a stream to write to, use a MemoryStream.
I come across this problem, using the MSDN example I created this class:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO.Packaging;
using System.IO;
public class ZipSticle
{
Package package;
public ZipSticle(Stream s)
{
package = ZipPackage.Open(s, FileMode.Create);
}
public void Add(Stream stream, string Name)
{
Uri partUriDocument = PackUriHelper.CreatePartUri(new Uri(Name, UriKind.Relative));
PackagePart packagePartDocument = package.CreatePart(partUriDocument, "");
CopyStream(stream, packagePartDocument.GetStream());
stream.Close();
}
private static void CopyStream(Stream source, Stream target)
{
const int bufSize = 0x1000;
byte[] buf = new byte[bufSize];
int bytesRead = 0;
while ((bytesRead = source.Read(buf, 0, bufSize)) > 0)
target.Write(buf, 0, bytesRead);
}
public void Close()
{
package.Close();
}
}
You can then use it like this:
FileStream str = File.Open("MyAwesomeZip.zip", FileMode.Create);
ZipSticle zip = new ZipSticle(str);
zip.Add(File.OpenRead("C:/Users/C0BRA/SimpleFile.txt"), "Some directory/SimpleFile.txt");
zip.Add(File.OpenRead("C:/Users/C0BRA/Hurp.derp"), "hurp.Derp");
zip.Close();
str.Close();
You can pass a MemoryStream (or any Stream) to ZipSticle.Add such as:
FileStream str = File.Open("MyAwesomeZip.zip", FileMode.Create);
ZipSticle zip = new ZipSticle(str);
byte[] fileinmem = new byte[1000];
// Do stuff to FileInMemory
MemoryStream memstr = new MemoryStream(fileinmem);
zip.Add(memstr, "Some directory/SimpleFile.txt");
memstr.Close();
zip.Close();
str.Close();
Note this answer is outdated; since .Net 4.5, the ZipArchive class allows zipping files in-memory. See johnny 5's answer below for how to use it.
You could also do it a bit differently, using a Serializable object to store all strings
[Serializable]
public class MyStrings {
public string Foo { get; set; }
public string Bar { get; set; }
}
Then, you could serialize it into a stream to save it.
To save on space you could use GZipStream (From System.IO.Compression) to compress it. (note: GZip is stream compression, not an archive of multiple files).
That is, of course if what you need is actually to save data, and not zip a few files in a specific format for other software.
Also, this would allow you to save many more types of data except strings.
I was utilizing Cheeso's answer by adding MemoryStreams as the source of the different Excel files. When I downloaded the zip, the files had nothing in them. This could be the way we were getting around trying to create and download a file over AJAX.
To get the contents of the different Excel files to be included in the Zip, I had to add each of the files as a byte[].
using (var memoryStream = new MemoryStream())
using (var zip = new ZipFile())
{
zip.AddEntry("Excel File 1.xlsx", excelFileStream1.ToArray());
zip.AddEntry("Excel File 2.xlsx", excelFileStream2.ToArray());
// Keep the file off of disk, and in memory.
zip.Save(memoryStream);
}
Use a StringReader to read from your string objects and expose them as Stream s.
That should make it easy to feed them to your zip-building code.