Actually, I have saved a file in BinaryData in Sql and now I am able to download that file by converting the BinaryData into Bytes.
My code is :
object value = (sender as DevExpress.Web.ASPxGridView.ASPxGridView).GetRowValues(e.VisibleIndex, "ID");
Int64 FileID = Convert.ToInt64(value);
var filedata = (from xx in VDC.SURVEY_QUESTION_REPLIES
where xx.ID == FileID
select xx).FirstOrDefault();
string fileextension = filedata.FILE_EXTENSION.ToString();
string fileName = filedata.ANSWER_TEXT.ToString() + fileextension;
string DocumentName = null;
FileStream FStream = null;
BinaryWriter BWriter = null;
byte[] Binary = null;
const int ChunkSize = 100;
int SizeToWrite = 0;
MemoryStream MStream = null;
DocumentName = fileName;
FStream = new FileStream(#"c:\\" + DocumentName, FileMode.OpenOrCreate, FileAccess.Write);
BWriter = new BinaryWriter(FStream);
Binary = (filedata.FILE_DATA) as byte[];
SizeToWrite = ChunkSize;
MStream = new MemoryStream(Binary);
for (int i = 0; i < Binary.GetUpperBound(0) - 1; i = i + ChunkSize)
{
if (i + ChunkSize >= Binary.Length) SizeToWrite = Binary.Length - i;
byte[] Chunk = new byte[SizeToWrite];
MStream.Read(Chunk, 0, SizeToWrite);
BWriter.Write(Chunk);
BWriter.Flush();
}
BWriter.Close();
FStream.Close();
FStream.Dispose();
System.Diagnostics.Process.Start(#"c:\" + DocumentName);
and it is directly saving the file to the location C Drive.
Now,My Requirement is that,I need to get a Prompt for saving that file and user need to select the location of saving.
Is that Possible ?
You create a filestream with a fixed location here:
FStream = new FileStream(#"c:\\" + DocumentName, FileMode.OpenOrCreate, FileAccess.Write);
What you would have to do is something like this:
var dialog = new SaveFileDialog();
if (dialog.ShowDialog() == DialogResult.OK)
{
FStream = new FileStream(dialog.FileName, FileMode.OpenOrCreate, FileAccess.Write);
// put the rest of your file saving code here
}
Remember to import the Forms namespace
using System.Windows.Forms;
If its Forms app You can use SaveFileDialog class
it is possible,
you can use a saveFileDialog that you create in your designer and call the "show" method to open that dialog :
private void button1_Click(object sender, System.EventArgs e)
{
Stream myStream ;
SaveFileDialog saveFileDialog1 = new SaveFileDialog();
saveFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ;
saveFileDialog1.FilterIndex = 2 ;
saveFileDialog1.RestoreDirectory = true ;
if(saveFileDialog1.ShowDialog() == DialogResult.OK)
{
if((myStream = saveFileDialog1.OpenFile()) != null)
{
// Code to write the stream goes here.
myStream.Close();
}
}
}
Source : http://msdn.microsoft.com/en-gb/library/system.windows.forms.savefiledialog.aspx
I see that you use web components of DevExpress, so assuming that you want to send the stream to response for the client to save the file.
If it is an ASP.NET MVC application, you can return FileContentResult as action result directly. Otherwise, you might use the following sample adapting into your code
Response.ContentType = "application/octet-stream";
Response.AddHeader("Content-Disposition", "attachment; filename=" + DocumentName);
MStream.WriteTo(Response.OutputStream);
Depending on user's browser settings, save file dialog might be shown to the user.
Related
In my C# application, I am trying to make it able to pull a pdf and save it, so that end users can press a button and pull up that pdf while they are in the application. But when I copy the content to the filestream it makes the pdf but it is just blank with nothing from the original pdf. What am I doing wrong?
The pdf's could also have pictures on them, and I don't think the way I'm doing it would allow those to be brought over.
Microsoft.Win32.OpenFileDialog openFileDialog = new Microsoft.Win32.OpenFileDialog();
bool? response = openFileDialog.ShowDialog();
var fileContent = string.Empty;
var filestream = openFileDialog.OpenFile();
using (StreamReader reader = new StreamReader(filestream))
{
fileContent = reader.ReadToEnd();
}
// make folder path
string FolderPath = "ProjectPDFs\\";
string RootPath = "X:\\Vents-US Inventory";
DirectoryInfo FolderDir = new DirectoryInfo(Path.Combine(RootPath, FolderPath));
Directory.CreateDirectory(FolderDir.ToString());
string filePath = "";
string FileName = openFileDialog.SafeFileName;
if (fileContent.Length > 0)
{
filePath = Path.Combine(FolderDir.ToString(), FileName);
using (Stream fileStream = new FileStream(filePath, FileMode.Create, FileAccess.Write))
{
byte[] bytestream = Encoding.UTF8.GetBytes(fileContent);
Stream stream = new MemoryStream(bytestream);
stream.CopyTo(fileStream);
}
}
I have a report and i want to save it directly without a dialog.
Here is my code until now
SaveFileDialog saveFileDialog = new SaveFileDialog();
saveFileDialog.InitialDirectory = #"C:\temp";
saveFileDialog.RestoreDirectory = true;
saveFileDialog.Title = "Browse Text Files";
saveFileDialog.DefaultExt = "doc";
saveFileDialog.Filter = "Word Doc (*.doc)|*.doc|PDF (*.pdf)| *.pdf";
saveFileDialog.CheckFileExists = false;
saveFileDialog.CheckPathExists = true;
Warning[] warnings;
string[] streams;
string mimeType;
string encoding;
string extension;
byte[] bytes = reportTest.LocalReport.Render("Word", null, out mimeType, out encoding, out extension, out streams, out warnings);
//if (saveFileDialog.ShowDialog() == DialogResult.OK)
//{
saveFileDialog.FileName = "123.doc";
var filename = saveFileDialog.FileName;
System.IO.FileStream file = new FileStream(filename, FileMode.Create);
file.Write(bytes, 0, bytes.Length);
file.Close();
//}
Can someone help me how to save the file directly to c:\temp\123.doc (without dialog in the explorer).
Thanks a lot!
You can directly save this with path and file name
filename="PATH" + "test.doc";
System.IO.FileStream file = new FileStream(filename, FileMode.Create);
I need to create a file txt and a file with extension dat. After that I want to compress these files into a zip file and to download it.
My problem is that I don't want to save the files before download. So, can I create and download the zip without creating files on server?
I used MemoryStream. Below you can find the code:
public ZipOutputStream CreateZip(MemoryStream memoryStreamControlFile, MemoryStream memoryStreamDataFile)
{
using (var outputMemStream = new MemoryStream())
{
using (var zipStream = new ZipOutputStream(baseOutputStream))
{
zipStream.SetLevel(3); // 0-9, 9 being the highest level of compression
byte[] bytes = null;
// .dat
var nameFileData = CreateFileName("dat");
var newEntry = new ZipEntry(nameFileData) { DateTime = DateTime.Now };
zipStream.PutNextEntry(newEntry);
bytes = memoryStreamDataFile.ToArray();
var inStream = new MemoryStream(bytes);
StreamUtils.Copy(inStream, zipStream, new byte[4096]);
inStream.Close();
zipStream.CloseEntry();
// .suc
var nameFileControl = CreateFileName("suc");
newEntry = new ZipEntry(nameFileControl) { DateTime = DateTime.Now };
zipStream.PutNextEntry(newEntry);
bytes = memoryStreamControlFile.ToArray();
inStream = new MemoryStream(bytes);
StreamUtils.Copy(inStream, zipStream, new byte[4096]);
inStream.Close();
zipStream.CloseEntry();
// .zip
zipStream.IsStreamOwner = false;
zipStream.Close();
outputMemStream.Position = 0;
return zipStream;
}
}
}
I assume you are working with ASP.Net ?
You can put the file contents (stream, etc...) in a byte array, then output that directly to the user in the response, like :
string fileType = "...zip...";
byte[] fileContent = <your file content> as byte[];
int fileSize = 1000;
string fileName = "filename.zip";
Response.AddHeader("Content-Disposition", "attachment;filename=\"" + fileName + "\"");
Response.ContentType = fileType;
Response.OutputStream.Write(fileContent, 0, fileSize);
I used this example and my code is
private void dialogBtn_Click(object sender, EventArgs e)
{
string file;
DialogResult result = openFileDialog1.ShowDialog(); // Show the dialog.
if (result == DialogResult.OK) // Test result.
{
file = openFileDialog1.FileName;
System.IO.FileInfo fileInfo = new System.IO.FileInfo(file);
TransferServiceClient clientUpload = new TransferServiceClient();
RemoteFileInfo uploadRequestInfo = new RemoteFileInfo();
using (System.IO.FileStream stream = new System.IO.FileStream(file, System.IO.FileMode.Open, System.IO.FileAccess.Read))
{
uploadRequestInfo.FileName = file;
uploadRequestInfo.Length = fileInfo.Length;
uploadRequestInfo.FileByteStream = stream;
//clientUpload.UploadFile(uploadRequestInfo);
clientUpload.UploadFile(fileInfo.Name, fileInfo.Length, stream);
}
}
}
During the clientUpload.UploadFile(fileInfo.Name, fileInfo.Length, stream); I receive an error
An unhandled exception of type 'System.ServiceModel.FaultException`1' occurred in mscorlib.dll
Additional information: Could not find a part of the path 'C:\upload\Book1.xlsx'.
But, the Book1.xlsx is not inside upload folder. It's at the Desktop.
Well, the service implementation of the UploadFile method in the example is as follows:
public void UploadFile(RemoteFileInfo request)
{
FileStream targetStream = null;
Stream sourceStream = request.FileByteStream;
string uploadFolder = #"C:\upload\";
string filePath = Path.Combine(uploadFolder, request.FileName);
using (targetStream = new FileStream(filePath, FileMode.Create,
FileAccess.Write, FileShare.None))
{
//read from the input stream in 65000 byte chunks
const int bufferLen = 65000;
byte[] buffer = new byte[bufferLen];
int count = 0;
while ((count = sourceStream.Read(buffer, 0, bufferLen)) > 0)
{
// save to output stream
targetStream.Write(buffer, 0, count);
}
targetStream.Close();
sourceStream.Close();
}
}
Note the line
string uploadFolder = #"C:\upload\";
You need to change that to some existing folder where you want the uploaded files to be stored.
I am trying to open a Binary file that I plan on converting to hex but I am running into issues with reading the file via FileStream,
private void button1_Click(object sender, EventArgs e)
{
openFD.Title = "Insert a BIN file";
openFD.InitialDirectory = "C:"; // Chooses the default location to open the file
openFD.FileName = " "; // Iniitalizes the File name
openFD.Filter = "Binary File|*.bin|Text File|*.txt"; // FIlters the types of files allowed to by chosen
if (openFD.ShowDialog() != DialogResult.Cancel)
{
chosenFile = openFD.FileName;
string directoryPath = Path.GetDirectoryName(chosenFile); // Returns the directory and the file name to reference the file
string dirName = System.IO.Path.GetDirectoryName(openFD.FileName); // Returns the proper directory with which to refernce the file
richTextBox1.Text += dirName;
richTextBox1.Text += chosenFile;
FileStream InputBin = new FileStream(
directoryPath, FileMode.Open, FileAccess.Read, FileShare.None);
}
}
I am receiving an error saying that the access to the path is denied, any ideas?
Now that I have gotten that error taken care of I have ran into another Issue, I can read the binary file, but I want to display it as a Hex file, I'm not sure what I am doing wrong but I'm not getting an output in HEX, it seems to be Int values...
if (openFD.ShowDialog() != DialogResult.Cancel)
{
chosenFile = openFD.FileName;
string directoryPath = Path.GetDirectoryName(chosenFile);
string dirName = System.IO.Path.GetDirectoryName(openFD.FileName);
using (FileStream stream = new FileStream(chosenFile, FileMode.Open, FileAccess.Read))
{
size = (int)stream.Length;
data = new byte[size];
stream.Read(data, 0, size);
}
while (printCount < size)
{
richTextBox1.Text += data[printCount];
printCount++;
}
Your code is miscommented
string directoryPath = Path.GetDirectoryName(chosenFile); // Returns the directory and the file name to reference the file
is not the filename, it's the directory path. You want:
FileStream InputBin = new FileStream(chosenFile, FileMode.Open,FileAccess.Read, FileShare.None);
Addtionally, if I were to guess based on your intentions, you should update your full function to be:
private void button1_Click(object sender, EventArgs e)
{
openFD.Title = "Insert a BIN file";
openFD.InitialDirectory = "C:"; // Chooses the default location to open the file
openFD.FileName = " "; // Iniitalizes the File name
openFD.Filter = "Binary File|*.bin|Text File|*.txt"; // FIlters the types of files allowed to by chosen
if (openFD.ShowDialog() != DialogResult.Cancel)
{
chosenFile = openFD.FileName;
richTextBox1.Text += chosenFile; //You may want to replace this with = unless you mean to append something that is already there.
FileStream InputBin = new FileStream(chosenFile, FileMode.Open,FileAccess.Read, FileShare.None);
}
}
To answer your second quesiton:
I am receiving an error saying that the access to the path is denied,
any ideas?
Now that I have gotten that error taken care of I have ran into
another Issue, I can read the binary file, but I want to display it as
a Hex file, I'm not sure what I am doing wrong but I'm not getting an
output in HEX, it seems to be Int values...
Modify to use string.Format:
if (openFD.ShowDialog() != DialogResult.Cancel)
{
chosenFile = openFD.FileName;
string directoryPath = Path.GetDirectoryName(chosenFile);
string dirName = System.IO.Path.GetDirectoryName(openFD.FileName);
using (FileStream stream = new FileStream(chosenFile, FileMode.Open, FileAccess.Read))
{
size = (int)stream.Length;
data = new byte[size];
stream.Read(data, 0, size);
}
while (printCount < size)
{
richTextBox1.Text += string.Format( "{0:X} ", data[printCount];
printCount++;
}
}
I've included an ideone example.