Develop Reference

C# ASP.NET MVC application with file type input on a Mac [duplicate]

I am trying to read an IFormFile received from a HTTP POST request like this:
public async Task<ActionResult> UploadDocument([FromForm]DataWrapper data)
{
IFormFile file = data.File;
string fileName = file.FileName;
long length = file.Length;
if (length < 0)
return BadRequest();
using FileStream fileStream = new FileStream(fileName, FileMode.OpenOrCreate);
byte[] bytes = new byte[length];
fileStream.Read(bytes, 0, (int)file.Length);
...
}
but something is wrong, after this line executes:
fileStream.Read(bytes, 0, (int)file.Length);
all of the elements of bytes are zero.
Also, the file with the same name is created in my Visual Studio project, which I would prefer not to happen.

You can't open an IFormFile the same way you would a file on disk. You'll have to use IFormFile.OpenReadStream() instead. Docs here
public async Task<ActionResult> UploadDocument([FromForm]DataWrapper data)
{
IFormFile file = data.File;
long length = file.Length;
if (length < 0)
return BadRequest();
using var fileStream = file.OpenReadStream();
byte[] bytes = new byte[length];
fileStream.Read(bytes, 0, (int)file.Length);
}
The reason that fileStream.Read(bytes, 0, (int)file.Length); appears to be empty is, because it is. The IFormFile.Filename is the name of the file given by the request and doesn't exist on disk.

Your code's intent seems to be to write to a FileStream, not a byte buffer. What it actually does though, is create a new empty file and read from it into an already cleared buffer. The uploaded file is never used.
Writing to a file
If you really want to save the file, you can use CopyTo :
using(var stream = File.Create(Path.Combine(folder_I_Really_Want,file.FileName))
{
file.CopyTo(stream);
}
If you want to read from the uploaded file into a buffer without saving to disk, use a MemoryStream. That's just a Stream API buffer over a byte[] buffer. You don't have to specify the size but that reduces reallocations as the internal buffer grows.
Reading into byte[]
Reading into a byte[] through MemoryStream is essentially the same :
var stream = new MemoryStream(file.Length);
file.CopyTo(stream);
var bytes=stream.ToArray();

The problem is that you are opening a new filestream based on the file name in your model which will be the name of the file that the user selected when uploading. Your code will create a new empty file with that name, which is why you are seeing the file in your file system. Your code is then reading the bytes from that file which is empty.
You need to use IFormFile.OpenReadStream method or one of the CopyTo methods to get the actual data from the stream.
You then write that data to your file on your file system with name you want.
var filename ="[Enter or create name for your file here]";
using (FileStream fs = new FileStream(filename, FileMode.OpenOrCreate))
//Create the file in your file system with the name you want.
{
using (MemoryStream ms = new MemoryStream())
{
//Copy the uploaded file data to a memory stream
file.CopyTo(ms);
//Now write the data in the memory stream to the new file
fs.Write(ms.ToArray());
}
}

C#: Barcode to byte array to valid zip file

I'm scaning a Pdf417 barcode which returns me a byte[] array. The DataString itself is a cryptic value like face to keyboard multiple times and fast. So I'm guessing it could be a zip file which is stored in the barcode. In the zip file there should be a xml file.
I had different attempts so far to convert my byte[] array to a valid zip file. In the end I was never able to open said zip file.
The barcodes are created by certified software solutions, so the barcode is not the problem for sure.
I can't be the only one who has had this problem, right?
Output when reading a barcode with dummy data in it:
"\0\0\0\vz\0\u0002B\u0001\u0002PK\u0003\u0004\n\0\0\0\b\0 E\u0081Q|\u0015\u00163Î\u0001\0\0\u0004\u0004\0\0\u0004\0\0\0txab}SMo£0\u0010½ï¯°|\a\u001bª¤Ý\u0015Påc\u0093Fê&\b²9ôÆÂ$ \r¦²M\u0093üûNø\b$Që\u008bí7ï½\u0019{lçù\u0098ïÉ\aH\u0095\u0015Â¥\u0096É)\u0001\u0011\u0017I&v.ý»\u009e\u0019OôÙsÖ\u0004iB¹4Õúý\u0017c\u0087ÃÁT\u0087L©\u0004b3N\u0099\u008aSÈ#¦\u0012fs>ä6·X\u0018í#y\u009aB\u008cS¤Ñ}}\u001c\u008d)\t\u0017S\u0097ÎV+\\u009dÔ¦^z?\b\u000egRäï\u00918\u0091\u0097À\b&Æ2ÊÁ¥/\u0091Pä\u000fd ÉhNÉÛÂwé\u0013ç\u009c²Fäcé\u0085XLÉk¤´¨4\u0015\u009d\u0092Y&[äìÒ\u0017×ÚJ\u001fn\u008cQh,¥÷8\u0018\u009a\u0016\u000eÓÆa>ütØ%Tgbmªf\u001fö\0\u0094\u0015I\aTàV\u0016¹gãe\u0018|`pËa\u0015pÍ)\u0085Îö5ɲ\u008d\a$ÕHgÍn½\u009d\u0005ö'\ao\u0080'á&ç\u000ek\u0080\u008e1\u0093Ø>\u0018\u0083\u0080m¦\u0015ëEV:\u0005Ù\u0006njYÃQ{ãB\u0094ÊaÕú:\u001c\u0096¹g]r\u009ew½"¿ðuæ²Pª©ox\u0011÷Ñ\u008e;ÞÌ\u008dWß7&\u0085Ð2ûW\u009e\u001fÍM\r\u0001ìJ|Oxa\u008dS\v\ÓüRÆi¤ ª·â]\u0090^Íßçs\u0096 Û\u009b~lm:¬ãM!)ã³v¤Ã\u0002Óô²Þ\u0087:Ù$\u008dä\u000eTPî\u0081ÝÃ7\aú½Ý\u0002\u001a}\QÙ\u001d·nï×Ý\u000fð\u0093Êÿí×aø\u0082±ÓÞ'PK\u0001\u0002\u0014\0\n\0\0\0\b\0 E\u0081Q|\u0015\u00163Î\u0001\0\0\u0004\u0004\0\0\u0004\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0txabPK\u0005\u0006\0\0\0\0\u0001\0\u0001\02\0\0\0ð\u0001\0\0\0\0"
Dont pay too much attention to this function, at this stage I'm probably trying too much to get to a solution. This is just one of many test voids of mine.
public void Test(byte[] bytes)
{
byte[] zipBytes;
using (var memoryStream = new MemoryStream())
{
using (var zipArchive = new ZipArchive(memoryStream, ZipArchiveMode.Create, leaveOpen: true))
{
var zipEntry = zipArchive.CreateEntry("test");
using (Stream entryStream = zipEntry.Open())
{
entryStream.Write(bytes, 0, bytes.Length);
}
}
zipBytes = memoryStream.ToArray();
}
using (var fileStream = new FileStream(#"C:\BarcodeReaderTesting\test.zip", FileMode.OpenOrCreate))
{
fileStream.Write(zipBytes, 0, zipBytes.Length);
}
}
Any tips on this topic?

C# read IFormFile into byte[]

I am trying to read an IFormFile received from a HTTP POST request like this:
public async Task<ActionResult> UploadDocument([FromForm]DataWrapper data)
{
IFormFile file = data.File;
string fileName = file.FileName;
long length = file.Length;
if (length < 0)
return BadRequest();
using FileStream fileStream = new FileStream(fileName, FileMode.OpenOrCreate);
byte[] bytes = new byte[length];
fileStream.Read(bytes, 0, (int)file.Length);
...
}
but something is wrong, after this line executes:
fileStream.Read(bytes, 0, (int)file.Length);
all of the elements of bytes are zero.
Also, the file with the same name is created in my Visual Studio project, which I would prefer not to happen.

You can't open an IFormFile the same way you would a file on disk. You'll have to use IFormFile.OpenReadStream() instead. Docs here
public async Task<ActionResult> UploadDocument([FromForm]DataWrapper data)
{
IFormFile file = data.File;
long length = file.Length;
if (length < 0)
return BadRequest();
using var fileStream = file.OpenReadStream();
byte[] bytes = new byte[length];
fileStream.Read(bytes, 0, (int)file.Length);
}
The reason that fileStream.Read(bytes, 0, (int)file.Length); appears to be empty is, because it is. The IFormFile.Filename is the name of the file given by the request and doesn't exist on disk.

Your code's intent seems to be to write to a FileStream, not a byte buffer. What it actually does though, is create a new empty file and read from it into an already cleared buffer. The uploaded file is never used.
Writing to a file
If you really want to save the file, you can use CopyTo :
using(var stream = File.Create(Path.Combine(folder_I_Really_Want,file.FileName))
{
file.CopyTo(stream);
}
If you want to read from the uploaded file into a buffer without saving to disk, use a MemoryStream. That's just a Stream API buffer over a byte[] buffer. You don't have to specify the size but that reduces reallocations as the internal buffer grows.
Reading into byte[]
Reading into a byte[] through MemoryStream is essentially the same :
var stream = new MemoryStream(file.Length);
file.CopyTo(stream);
var bytes=stream.ToArray();

The problem is that you are opening a new filestream based on the file name in your model which will be the name of the file that the user selected when uploading. Your code will create a new empty file with that name, which is why you are seeing the file in your file system. Your code is then reading the bytes from that file which is empty.
You need to use IFormFile.OpenReadStream method or one of the CopyTo methods to get the actual data from the stream.
You then write that data to your file on your file system with name you want.
var filename ="[Enter or create name for your file here]";
using (FileStream fs = new FileStream(filename, FileMode.OpenOrCreate))
//Create the file in your file system with the name you want.
{
using (MemoryStream ms = new MemoryStream())
{
//Copy the uploaded file data to a memory stream
file.CopyTo(ms);
//Now write the data in the memory stream to the new file
fs.Write(ms.ToArray());
}
}

Extract tgz file in memory and access files in C#

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
}
}
}

SharpZipLib Examine and select contents of a ZIP file

I am using SharpZipLib in a project and am wondering if it is possible to use it to look inside a zip file, and if one of the files within has a data modified in a range I am searching for then to pick that file out and copy it to a new directory? Does anybody know id this is possible?

Yes, it is possible to enumerate the files of a zip file using SharpZipLib. You can also pick files out of the zip file and copy those files to a directory on your disk.
Here is a small example:
using (var fs = new FileStream(#"c:\temp\test.zip", FileMode.Open, FileAccess.Read))
{
using (var zf = new ZipFile(fs))
{
foreach (ZipEntry ze in zf)
{
if (ze.IsDirectory)
continue;
Console.Out.WriteLine(ze.Name);
using (Stream s = zf.GetInputStream(ze))
{
byte[] buf = new byte[4096];
// Analyze file in memory using MemoryStream.
using (MemoryStream ms = new MemoryStream())
{
StreamUtils.Copy(s, ms, buf);
}
// Uncomment the following lines to store the file
// on disk.
/*using (FileStream fs = File.Create(#"c:\temp\uncompress_" + ze.Name))
{
StreamUtils.Copy(s, fs, buf);
}*/
}
}
}
}
In the example above I use a MemoryStream to store the ZipEntry in memory (for further analysis). You could also store the ZipEntry (if it meets certain criteria) on disk.
Hope, this helps.