I want to load a file from the Asset, I found the solution but with Java. How can I convert following Java code to c#.
public String loadKMLFromAsset() {
String kmlData = null;
try {
InputStream is = getAssets().open("yourKMLFile");
int size = is.available();
byte[] buffer = new byte[size];
is.read(buffer);
is.close();
kmlData = new String(buffer, "UTF-8");
} catch (IOException ex) {
ex.printStackTrace();
return null;
}
return kmlData;
}
Use AssetManager
// Read the contents of our asset
string content;
AssetManager assets = this.Assets;
using (StreamReader sr = new StreamReader (assets.Open ("read_asset.txt")))
{
content = sr.ReadToEnd ();
}
Use BinaryReader instead of streamReader, if u are working with files such as db, kml, shapefiles,, video formats, etc. StreamReader reads only strings or just plain text, so when reading binary file some of the content may be skipped, since streamreader doesnt read byte by byte
This code writes the asset file to a file in your mobile file system:
if (!System.IO.File.Exists("yourKMLFile_mobile"))
{
var s = Resources.OpenRawResource(Resource.Raw.yourKMLFile);
FileStream writeStream = new FileStream("yourKMLFile_mobile", FileMode.OpenOrCreate, FileAccess.Write);
ReadWriteStream(s, writeStream);
}
Related
I have files (from 3rd parties) that are being FTP'd to a directory on our server. I download them and process them even 'x' minutes. Works great.
Now, some of the files are .zip files. Which means I can't process them. I need to unzip them first.
FTP has no concept of zip/unzipping - so I'll need to grab the zip file, unzip it, then process it.
Looking at the MSDN zip api, there seems to be no way i can unzip to a memory stream?
So is the only way to do this...
Unzip to a file (what directory? need some -very- temp location ...)
Read the file contents
Delete file.
NOTE: The contents of the file are small - say 4k <-> 1000k.
Zip compression support is built in:
using System.IO;
using System.IO.Compression;
// ^^^ requires a reference to System.IO.Compression.dll
static class Program
{
const string path = ...
static void Main()
{
using(var file = File.OpenRead(path))
using(var zip = new ZipArchive(file, ZipArchiveMode.Read))
{
foreach(var entry in zip.Entries)
{
using(var stream = entry.Open())
{
// do whatever we want with stream
// ...
}
}
}
}
}
Normally you should avoid copying it into another stream - just use it "as is", however, if you absolutely need it in a MemoryStream, you could do:
using(var ms = new MemoryStream())
{
stream.CopyTo(ms);
ms.Position = 0; // rewind
// do something with ms
}
You can use ZipArchiveEntry.Open to get a stream.
This code assumes the zip archive has one text file.
using (FileStream fs = new FileStream(path, FileMode.Open))
using (ZipArchive zip = new ZipArchive(fs) )
{
var entry = zip.Entries.First();
using (StreamReader sr = new StreamReader(entry.Open()))
{
Console.WriteLine(sr.ReadToEnd());
}
}
using (ZipArchive archive = new ZipArchive(webResponse.GetResponseStream()))
{
foreach (ZipArchiveEntry entry in archive.Entries)
{
Stream s = entry.Open();
var sr = new StreamReader(s);
var myStr = sr.ReadToEnd();
}
}
Looks like here is what you need:
using (var za = ZipFile.OpenRead(path))
{
foreach (var entry in za.Entries)
{
using (var r = new StreamReader(entry.Open()))
{
//your code here
}
}
}
You can use SharpZipLib among a variety of other libraries to achieve this.
You can use the following code example to unzip to a MemoryStream, as shown on their wiki:
using ICSharpCode.SharpZipLib.Zip;
// Compresses the supplied memory stream, naming it as zipEntryName, into a zip,
// which is returned as a memory stream or a byte array.
//
public MemoryStream CreateToMemoryStream(MemoryStream memStreamIn, string zipEntryName) {
MemoryStream outputMemStream = new MemoryStream();
ZipOutputStream zipStream = new ZipOutputStream(outputMemStream);
zipStream.SetLevel(3); //0-9, 9 being the highest level of compression
ZipEntry newEntry = new ZipEntry(zipEntryName);
newEntry.DateTime = DateTime.Now;
zipStream.PutNextEntry(newEntry);
StreamUtils.Copy(memStreamIn, zipStream, new byte[4096]);
zipStream.CloseEntry();
zipStream.IsStreamOwner = false; // False stops the Close also Closing the underlying stream.
zipStream.Close(); // Must finish the ZipOutputStream before using outputMemStream.
outputMemStream.Position = 0;
return outputMemStream;
// Alternative outputs:
// ToArray is the cleaner and easiest to use correctly with the penalty of duplicating allocated memory.
byte[] byteArrayOut = outputMemStream.ToArray();
// GetBuffer returns a raw buffer raw and so you need to account for the true length yourself.
byte[] byteArrayOut = outputMemStream.GetBuffer();
long len = outputMemStream.Length;
}
Ok so combining all of the above, suppose you want to in a very simple way take a zip file called
"file.zip" and extract it to "C:\temp" folder. (Note: This example was only tested for compress text files) You may need to do some modifications for binary files.
using System.IO;
using System.IO.Compression;
static void Main(string[] args)
{
//Call it like this:
Unzip("file.zip",#"C:\temp");
}
static void Unzip(string sourceZip, string targetPath)
{
using (var z = ZipFile.OpenRead(sourceZip))
{
foreach (var entry in z.Entries)
{
using (var r = new StreamReader(entry.Open()))
{
string uncompressedFile = Path.Combine(targetPath, entry.Name);
File.WriteAllText(uncompressedFile,r.ReadToEnd());
}
}
}
}
I have files (from 3rd parties) that are being FTP'd to a directory on our server. I download them and process them even 'x' minutes. Works great.
Now, some of the files are .zip files. Which means I can't process them. I need to unzip them first.
FTP has no concept of zip/unzipping - so I'll need to grab the zip file, unzip it, then process it.
Looking at the MSDN zip api, there seems to be no way i can unzip to a memory stream?
So is the only way to do this...
Unzip to a file (what directory? need some -very- temp location ...)
Read the file contents
Delete file.
NOTE: The contents of the file are small - say 4k <-> 1000k.
Zip compression support is built in:
using System.IO;
using System.IO.Compression;
// ^^^ requires a reference to System.IO.Compression.dll
static class Program
{
const string path = ...
static void Main()
{
using(var file = File.OpenRead(path))
using(var zip = new ZipArchive(file, ZipArchiveMode.Read))
{
foreach(var entry in zip.Entries)
{
using(var stream = entry.Open())
{
// do whatever we want with stream
// ...
}
}
}
}
}
Normally you should avoid copying it into another stream - just use it "as is", however, if you absolutely need it in a MemoryStream, you could do:
using(var ms = new MemoryStream())
{
stream.CopyTo(ms);
ms.Position = 0; // rewind
// do something with ms
}
You can use ZipArchiveEntry.Open to get a stream.
This code assumes the zip archive has one text file.
using (FileStream fs = new FileStream(path, FileMode.Open))
using (ZipArchive zip = new ZipArchive(fs) )
{
var entry = zip.Entries.First();
using (StreamReader sr = new StreamReader(entry.Open()))
{
Console.WriteLine(sr.ReadToEnd());
}
}
using (ZipArchive archive = new ZipArchive(webResponse.GetResponseStream()))
{
foreach (ZipArchiveEntry entry in archive.Entries)
{
Stream s = entry.Open();
var sr = new StreamReader(s);
var myStr = sr.ReadToEnd();
}
}
Looks like here is what you need:
using (var za = ZipFile.OpenRead(path))
{
foreach (var entry in za.Entries)
{
using (var r = new StreamReader(entry.Open()))
{
//your code here
}
}
}
You can use SharpZipLib among a variety of other libraries to achieve this.
You can use the following code example to unzip to a MemoryStream, as shown on their wiki:
using ICSharpCode.SharpZipLib.Zip;
// Compresses the supplied memory stream, naming it as zipEntryName, into a zip,
// which is returned as a memory stream or a byte array.
//
public MemoryStream CreateToMemoryStream(MemoryStream memStreamIn, string zipEntryName) {
MemoryStream outputMemStream = new MemoryStream();
ZipOutputStream zipStream = new ZipOutputStream(outputMemStream);
zipStream.SetLevel(3); //0-9, 9 being the highest level of compression
ZipEntry newEntry = new ZipEntry(zipEntryName);
newEntry.DateTime = DateTime.Now;
zipStream.PutNextEntry(newEntry);
StreamUtils.Copy(memStreamIn, zipStream, new byte[4096]);
zipStream.CloseEntry();
zipStream.IsStreamOwner = false; // False stops the Close also Closing the underlying stream.
zipStream.Close(); // Must finish the ZipOutputStream before using outputMemStream.
outputMemStream.Position = 0;
return outputMemStream;
// Alternative outputs:
// ToArray is the cleaner and easiest to use correctly with the penalty of duplicating allocated memory.
byte[] byteArrayOut = outputMemStream.ToArray();
// GetBuffer returns a raw buffer raw and so you need to account for the true length yourself.
byte[] byteArrayOut = outputMemStream.GetBuffer();
long len = outputMemStream.Length;
}
Ok so combining all of the above, suppose you want to in a very simple way take a zip file called
"file.zip" and extract it to "C:\temp" folder. (Note: This example was only tested for compress text files) You may need to do some modifications for binary files.
using System.IO;
using System.IO.Compression;
static void Main(string[] args)
{
//Call it like this:
Unzip("file.zip",#"C:\temp");
}
static void Unzip(string sourceZip, string targetPath)
{
using (var z = ZipFile.OpenRead(sourceZip))
{
foreach (var entry in z.Entries)
{
using (var r = new StreamReader(entry.Open()))
{
string uncompressedFile = Path.Combine(targetPath, entry.Name);
File.WriteAllText(uncompressedFile,r.ReadToEnd());
}
}
}
}
I would like to know if it is possible to gzip a powerpoint file. the reason for me asking this question is because all of the article which I have found are gzipping a .txt and would like to know if its possible to gzip a .pptx.. through the use of c#.. the code below is what i am using
static void Main()
{
try
{
string anyString = File.ReadAllText("presentation.pptx");
CompressStringToFile("new.gz", anyString);
}
catch
{
// Couldn't compress.
}
}
public static void CompressStringToFile(string fileName, string value)
{
// A.
// Write string to temporary file.
string temp = Path.GetTempFileName();
File.WriteAllText(temp, value);
// B.
// Read file into byte array buffer.
byte[] b;
using (FileStream f = new FileStream(temp, FileMode.Open))
{
b = new byte[f.Length];
f.Read(b, 0, (int)f.Length);
}
using (FileStream f2 = new FileStream(fileName, FileMode.Create))
using (GZipStream gz = new GZipStream(f2, CompressionMode.Compress, false))
{
gz.Write(b, 0, b.Length);
}
}
A .pptx file (also .docx and .xlsx) is already zipped. If you change the file extension to .zip and open the file you'll see the contents.
So, while you should be able to gzip one of these files, it's unlikely you'd see a great deal of further compression.
I'm working in C#, and I'm downloading for the internet a zip file with one XML file in it. and I wish to load this XML file. This is what I have so far:
byte[] data;
WebClient webClient = new WebClient();
try {
data = webClient.DownloadData(downloadUrl);
}
catch (Exception ex) {
Console.WriteLine("Error in DownloadData (Ex:{0})", ex.Message);
throw;
}
if (data == null) {
Console.WriteLine("Bulk data is null");
throw new Exception("Bulk data is null");
}
//Create the stream
MemoryStream stream = new MemoryStream(data);
XmlDocument document = new XmlDocument();
//Gzip
GZipStream gzipStream = new GZipStream(stream, CompressionMode.Decompress);
//Load report straight from the gzip stream
try {
document.Load(gzipStream);
}
catch (Exception ex) {
Console.WriteLine("Error in Load (Ex:{0})", ex.Message);
throw;
}
in document.Load I'm always getting the following exception:
The magic number in GZip header is not correct. Make sure you are passing in a GZip stream.
What I'm doing wrong?
Apparently SharpZipLib is now unmaintained and you probably want to avoid it:
https://stackoverflow.com/a/593030
In .NET 4.5 there is now built in support for zip files, so for your example it would be:
var data = new WebClient().DownloadData(downloadUrl);
//Create the stream
var stream = new MemoryStream(data);
var document = new XmlDocument();
//zip
var zipArchive = new ZipArchive(stream);
//Load report straight from the zip stream
document.Load(zipArchive.Entries[0].Open());
If you have a byte array that contains a zip archive with a single file, you can use the ZipArchive class to get an unzipped byte array with the file's data.
ZipArchive is contained in .NET 4.5, in the assembly System.IO.Compression.FileSystem (you need to reference it explicitly).
The following function, adapted from this answer, works for me:
public static byte[] UnzipSingleEntry(byte[] zipped)
{
using (var memoryStream = new MemoryStream(zipped))
{
using (var archive = new ZipArchive(memoryStream))
{
foreach (ZipArchiveEntry entry in archive.Entries)
{
using (var entryStream = entry.Open())
{
using (var reader = new BinaryReader(entryStream))
{
return reader.ReadBytes((int)entry.Length);
}
}
}
}
}
return null; // To quiet my compiler
}
I am using SharpZipLib and it's working great !
Below is a function that encapsulate the library
public static void Compress(FileInfo sourceFile, string destinationFileName,string destinationTempFileName)
{
Crc32 crc = new Crc32();
string zipFile = Path.Combine(sourceFile.Directory.FullName, destinationTempFileName);
zipFile = Path.ChangeExtension(zipFile, ZIP_EXTENSION);
using (FileStream fs = File.Create(zipFile))
{
using (ZipOutputStream zOut = new ZipOutputStream(fs))
{
zOut.SetLevel(9);
ZipEntry entry = new ZipEntry(ZipEntry.CleanName(destinationFileName));
entry.DateTime = DateTime.Now;
entry.ZipFileIndex = 1;
entry.Size = sourceFile.Length;
using (FileStream sourceStream = sourceFile.OpenRead())
{
crc.Reset();
long len = sourceFile.Length;
byte[] buffer = new byte[bufferSize];
while (len > 0)
{
int readSoFar = sourceStream.Read(buffer, 0, buffer.Length);
crc.Update(buffer, 0, readSoFar);
len -= readSoFar;
}
entry.Crc = crc.Value;
zOut.PutNextEntry(entry);
len = sourceStream.Length;
sourceStream.Seek(0, SeekOrigin.Begin);
while (len > 0)
{
int readSoFar = sourceStream.Read(buffer, 0, buffer.Length);
zOut.Write(buffer, 0, readSoFar);
len -= readSoFar;
}
}
zOut.Finish();
zOut.Close();
}
fs.Close();
}
}
As the others have mentioned GZip and Zip are not the same so you might need to use a zip library. I use a library called: DotNetZip - available from the below site:
http://dotnetzip.codeplex.com/
From GZipStream Class description:
Compressed GZipStream objects written to a file with an extension of .gz can be decompressed using many common compression tools; however, this class does not inherently provide functionality for adding files to or extracting files from .zip archives
So unless you control server-side files, I'd suggest looking for specific zip-targeted library (SharpZipLib for example).
I am attempting to create a new FileStream object from a byte array. I'm sure that made no sense at all so I will try to explain in further detail below.
Tasks I am completing:
1) Reading the source file which was previously compressed
2) Decompressing the data using GZipStream
3) copying the decompressed data into a byte array.
What I would like to change:
1) I would like to be able to use File.ReadAllBytes to read the decompressed data.
2) I would then like to create a new filestream object usingg this byte array.
In short, I want to do this entire operating using byte arrays. One of the parameters for GZipStream is a stream of some sort, so I figured I was stuck using a filestream. But, if some method exists where I can create a new instance of a FileStream from a byte array - then I should be fine.
Here is what I have so far:
FolderBrowserDialog fbd = new FolderBrowserDialog(); // Shows a browser dialog
fbd.ShowDialog();
// Path to directory of files to compress and decompress.
string dirpath = fbd.SelectedPath;
DirectoryInfo di = new DirectoryInfo(dirpath);
foreach (FileInfo fi in di.GetFiles())
{
zip.Program.Decompress(fi);
}
// Get the stream of the source file.
using (FileStream inFile = fi.OpenRead())
{
//Create the decompressed file.
string outfile = #"C:\Decompressed.exe";
{
using (GZipStream Decompress = new GZipStream(inFile,
CompressionMode.Decompress))
{
byte[] b = new byte[blen.Length];
Decompress.Read(b,0,b.Length);
File.WriteAllBytes(outfile, b);
}
}
}
Thanks for any help!
Regards,
Evan
It sounds like you need to use a MemoryStream.
Since you don't know how many bytes you'll be reading from the GZipStream, you can't really allocate an array for it. You need to read it all into a byte array and then use a MemoryStream to decompress.
const int BufferSize = 65536;
byte[] compressedBytes = File.ReadAllBytes("compressedFilename");
// create memory stream
using (var mstrm = new MemoryStream(compressedBytes))
{
using(var inStream = new GzipStream(mstrm, CompressionMode.Decompress))
{
using (var outStream = File.Create("outputfilename"))
{
var buffer = new byte[BufferSize];
int bytesRead;
while ((bytesRead = inStream.Read(buffer, 0, BufferSize)) != 0)
{
outStream.Write(buffer, 0, bytesRead);
}
}
}
}
Here is what I ended up doing. I realize that I did not give sufficient information in my question - and I apologize for that - but I do know the size of the file I need to decompress as I am using it earlier in my program. This buffer is referred to as "blen".
string fi = #"C:\Path To Compressed File";
// Get the stream of the source file.
// using (FileStream inFile = fi.OpenRead())
using (MemoryStream infile1 = new MemoryStream(File.ReadAllBytes(fi)))
{
//Create the decompressed file.
string outfile = #"C:\Decompressed.exe";
{
using (GZipStream Decompress = new GZipStream(infile1,
CompressionMode.Decompress))
{
byte[] b = new byte[blen.Length];
Decompress.Read(b,0,b.Length);
File.WriteAllBytes(outfile, b);
}
}
}