Develop Reference

How to get a specific file from directory in C#

Using Directory.GetFiles() returns all the files in the directory as follows:
var sqlFiles = Directory.GetFiles($"{AppDomain.CurrentDomain.BaseDirectory}Content\\DbScripts\\","*.sql");
Actually I need a specific file from that directory. What I have tried so far as follows but don't work!
var localizationSqlFile = Directory.GetFiles($"{AppDomain.CurrentDomain.BaseDirectory}Content\\DbScripts\\Localizations.sql").FirstOrDefault();
It throws expection:
The directory name is invalid.\r\n
Is there any method in C# to get a single file from a directory? If not then what will be most efficient way?

If you want to get the bytes of a certain file and you already have the full path, you can use the static method File.ReadAllBytes
var fileBytes = File.ReadAllBytes(myPath);
If you want to get file infos, you can create a new FileInfo object
var fileInfo = new FileInfo(myPath);
If you just want to check, if a file exists, you can also use the method File.Exist
if (File.Exists(myPath))

You can get the file from your sqlFiles returned
var sqlFiles = Directory.GetFiles($"{AppDomain.CurrentDomain.BaseDirectory}Content\\DbScripts\\","*.sql");
var yourFile = sqlFiles.FirstOrDefault(x=> Path.GetFileName(x) == "Localizations.sql");

Download a file with ServiceStack HttpResult: how to specify a file name for downloaded content?

I am using ServiceStack for a simple web application. The main purpose is to let a user download a file. I am using an HttpResult as follows:
public class FileDownloadService : Service
{
public object Any()
{
string fileFullPath = "...";
string mimeType = "application/pdf";
FileInfo fi = new FileInfo(fileFullPath);
byte[] reportBytes = File.ReadAllBytes(fi.FullName);
result = new HttpResult(reportBytes, mimeType);
return result;
}
}
This opens a dialog in the user's browser and the user can specify where to save the file. A default name is specified, which is the name of the restPath of ServiceStack.
My question is: is it possible to specify a custom file name for when the user chooses to save (changing the default one)? I tried to work with HttpResult properties, but no luck.
Thanks in advance!

You should set the 'Content-Disposition' header on the HTTP result. That allows to set the filename:
public object Any()
{
string fileFullPath = "...";
string mimeType = "application/pdf";
FileInfo fi = new FileInfo(fileFullPath);
byte[] reportBytes = File.ReadAllBytes(fi.FullName);
result = new HttpResult(reportBytes, mimeType);
result.Headers.Add("Content-Disposition", "attachment;filename=YOUR_NAME_HERE.pdf;");
return result;
}

I have found the solution. I was deceived by the readonly properties of the HttpResult.
In order to change the default file name, I discovered that I have to add the following line, that treats the content-disposition:
result.Headers[HttpHeaders.ContentDisposition] = "attachment; filename=\"UserSurname_UserName.pdf\"";
Thank you all for your time!

If I understood your question correctly, what you want is the user wants to rename the file. This is nothing to do with the server. The user has to do some changes in the brwoser he is uses.
e.g Firefox - Tools > Options > General
Then check the "Always ask me where to save files"
Chrome - Settings > Downloads
Then check the "Ask where to save each file before "
Then it will open file browser where to save, you may change your file name what ever your wanted

c# .dat file assign to variable

how can i assign a to a variable, which is located at the same project, for example at my project i created a folder named App_Data and for example the file is file.dat , how can i assign the file at a variable,.. for example:
var file = App_Data/file.dat
I need it to be assigned to a variable because i will be using that variable as a parameter to a method,.. it used to be :
var file= HttpContext.Current.Request.MapPath("/App_Data/file.dat");
but now i want the path to be at the same project
if it should be absolute path it should be fine too

The MapPath should give you the absolute location of the file on disk from a relative url to the root of your website:
var absoluteFileLocation = HostingEnvironment.MapPath("~/App_Data/file.dat");
This should return something like:
c:\inetpub\wwwroot\MyWebSite\App_Data\file.dat
UPDATE:
It looks like you are trying to retrieve the contents of the file, not the location. Here's how this could be done:
var absoluteFileLocation = HostingEnvironment.MapPath("~/App_Data/file.dat");
string fileContents = System.IO.File.ReadAllText(absoluteFileLocation);

You need to read the file using one of the available methods (Streams, Readers, etc).
The easiest would be:
string fileContent = File.ReadAllText(fileNameAndPath);
where the variable fileNameAndPath contains the full path and file name to the file as described by Darin Dimitrov.

Your intention isn't exactly clear, anyway:
if you want file stats:
System.IO.File file = new System.IO.File("~/App_Data/file.dat");
if you want the file content use:
public static string readFileContent(String filename)
{
try
{
using (System.IO.StreamReader sr = new System.IO.StreamReader(filename))
return sr.ReadToEnd();
}
catch { return String.Empty; }
}

Get file path from inputstream?

I am trying to get the last modified date from a file, but need its path? Could someone please show me how i can get the file path?
[HttpGet]
public string uploadfile(string token, string filenameP, DateTime modDate, HttpPostedFileBase file)
{
MemoryStream target = new MemoryStream();
file.InputStream.CopyTo(target);
byte[] data = target.ToArray();
//ModDate = File.GetLastWriteTimeUtc("Path");
}

You are creating a new file on the server when you upload. The last modified date will be "now" (the time the file is created). There is no way to snoop the user's machine to get this information (which is not part of the file itself). Can't be done with an HTTP form upload.
Now, some file types may contain metadata in the file which may have pertinent information. If you know the file type and it does contain such metadata then you can open the file and have a look.

You just don't. Most (if not all) browsers do not provide this information for security reasons in internet sceanrios.

You can read date by javascript (HTML5) and send it as hidden input field of form.
Something like
<script>
function handleFileSelect(evt) {
var files = evt.target.files; // FileList object
// files is a FileList of File objects. List some properties.
var output = [];
for (var i = 0, f; f = files[i]; i++) {
output.push(f.lastModifiedDate ? f.lastModifiedDate.toLocaleDateString() );
}
document.getElementById('list').innerHTML = '<ul>' + output.join('') + '</ul>';
}
document.getElementById('files').addEventListener('change', handleFileSelect, false);
</script>
http://www.html5rocks.com/en/tutorials/file/dndfiles/

SharpZipLib : Compressing a single file to a single compressed file

I am currently working with SharpZipLib under .NET 2.0 and via this I need to compress a single file to a single compressed archive. In order to do this I am currently using the following:
string tempFilePath = #"C:\Users\Username\AppData\Local\Temp\tmp9AE0.tmp.xml";
string archiveFilePath = #"C:\Archive\Archive_[UTC TIMESTAMP].zip";
FileInfo inFileInfo = new FileInfo(tempFilePath);
ICSharpCode.SharpZipLib.Zip.FastZip fZip = new ICSharpCode.SharpZipLib.Zip.FastZip();
fZip.CreateZip(archiveFilePath, inFileInfo.Directory.FullName, false, inFileInfo.Name);
This works exactly (ish) as it should, however while testing I have encountered a minor gotcha. Lets say that my temp directory (i.e. the directory that contains the uncompressed input file) contains the following files:
tmp9AE0.tmp.xml //The input file I want to compress
xxx_tmp9AE0.tmp.xml // Some other file
yyy_tmp9AE0.tmp.xml // Some other file
wibble.dat // Some other file
When I run the compression all the .xml files are included in the compressed archive. The reason for this is because of the final fileFilter parameter passed to the CreateZip method. Under the hood SharpZipLib is performing a pattern match and this also picks up the files prefixed with xxx_ and yyy_. I assume it would also pick up anything postfixed as well.
So the question is, how can I compress a single file with SharpZipLib? Then again maybe the question is how can I format that fileFilter so that the match can only ever pick up the file I want to compress and nothing else.
As an aside, is there any reason as to why System.IO.Compression not include a ZipStream class? (It only supports GZipStream)
EDIT : Solution (Derived from accepted answer from Hans Passant)
This is the compression method I implemented:
private static void CompressFile(string inputPath, string outputPath)
{
FileInfo outFileInfo = new FileInfo(outputPath);
FileInfo inFileInfo = new FileInfo(inputPath);
// Create the output directory if it does not exist
if (!Directory.Exists(outFileInfo.Directory.FullName))
{
Directory.CreateDirectory(outFileInfo.Directory.FullName);
}
// Compress
using (FileStream fsOut = File.Create(outputPath))
{
using (ICSharpCode.SharpZipLib.Zip.ZipOutputStream zipStream = new ICSharpCode.SharpZipLib.Zip.ZipOutputStream(fsOut))
{
zipStream.SetLevel(3);
ICSharpCode.SharpZipLib.Zip.ZipEntry newEntry = new ICSharpCode.SharpZipLib.Zip.ZipEntry(inFileInfo.Name);
newEntry.DateTime = DateTime.UtcNow;
zipStream.PutNextEntry(newEntry);
byte[] buffer = new byte[4096];
using (FileStream streamReader = File.OpenRead(inputPath))
{
ICSharpCode.SharpZipLib.Core.StreamUtils.Copy(streamReader, zipStream, buffer);
}
zipStream.CloseEntry();
zipStream.IsStreamOwner = true;
zipStream.Close();
}
}
}

This is an XY problem, just don't use FastZip. Follow the first example on this web page to avoid accidents.