"The given path's format is not supported."

C#.NetWebformsUploadStream

C# Problem Overview


I have the following code in my web service:

string str_uploadpath = Server.MapPath("/UploadBucket/Raw/");
FileStream objfilestream = new FileStream(str_uploadpath +
                fileName, FileMode.Create, FileAccess.ReadWrite);

Can someone help me resolve the issue with this error message from line 2 of the code.

> The given path's format is not supported.

Permission on the folder is set to full access to everyone and it is the actual path to the folder.

The breakpoint gave me the value of str_uploadpath as C:\\webprojects\\webservices\\UploadBucket\\Raw\\.

What is wrong with this string?

C# Solutions


Solution 1 - C#

Rather than using str_uploadpath + fileName, try using System.IO.Path.Combine instead:

Path.Combine(str_uploadpath, fileName);

which returns a string.

Solution 2 - C#

I see that the originator found out that the error occurred when trying to save the filename with an entire path. Actually it's enough to have a ":" in the file name to get this error. If there might be ":" in your file name (for instance if you have a date stamp in your file name) make sure you replace these with something else. I.e:

string fullFileName = fileName.Split('.')[0] + "(" + DateTime.Now.ToString().Replace(':', '-') + ")." + fileName.Split('.')[1];

Solution 3 - C#

For me the problem was an invisible to human eye "‪" [Left-To-Right Embedding][1] character.
It stuck at the beginning of the string (just before the 'D'), after I copy-pasted the path, from the windows file properties security tab.

var yourJson = System.IO.File.ReadAllText(@"D:\test\json.txt"); // Works
var yourJson = System.IO.File.ReadAllText(@"‪D:\test\json.txt"); // Error

So those, identical at first glance, two lines are actually different. [1]: https://unicode-table.com/en/202A/

Solution 4 - C#

If you are trying to save a file to the file system. Path.Combine is not bullet proof as it won't help you if the file name contains invalid characters. Here is an extension method that strips out invalid characters from file names:

public static string ToSafeFileName(this string s)
{
	    return s
		    .Replace("\\", "")
		    .Replace("/", "")
		    .Replace("\"", "")
		    .Replace("*", "")
		    .Replace(":", "")
		    .Replace("?", "")
		    .Replace("<", "")
		    .Replace(">", "")
		    .Replace("|", "");
    }

And the usage can be:

Path.Combine(str_uploadpath, fileName.ToSafeFileName());

Solution 5 - C#

Among other things that can cause this error:

You cannot have certain characters in the full PathFile string.

For example, these characters will crash the StreamWriter function:

"/"  
":"

there may be other special characters that crash it too. I found this happens when you try, for example, to put a DateTime stamp into a filename:

AppPath = Path.GetDirectoryName(giFileNames(0))  
' AppPath is a valid path from system. (This was easy in VB6, just AppPath = App.Path & "\")
' AppPath must have "\" char at the end...

DateTime = DateAndTime.Now.ToString ' fails StreamWriter... has ":" characters
FileOut = "Data_Summary_" & DateTime & ".dat"
NewFileOutS = Path.Combine(AppPath, FileOut)
Using sw As StreamWriter = New StreamWriter(NewFileOutS  , True) ' true to append
        sw.WriteLine(NewFileOutS)
        sw.Dispose()
    End Using

One way to prevent this trouble is to replace problem characters in NewFileOutS with benign ones:

' clean the File output file string NewFileOutS so StreamWriter will work
 NewFileOutS = NewFileOutS.Replace("/","-") ' replace / with -
 NewFileOutS = NewFileOutS.Replace(":","-") ' replace : with - 

' after cleaning the FileNamePath string NewFileOutS, StreamWriter will not throw an (Unhandled) exception.

Hope this saves someone some headaches...!

Solution 6 - C#

If you get this error in PowerShell, it's most likely because you're using Resolve-Path to resolve a remote path, e.g.

 Resolve-Path \\server\share\path

In this case, Resolve-Path returns an object that, when converted to a string, doesn't return a valid path. It returns PowerShell's internal path:

> [string](Resolve-Path \\server\share\path)
Microsoft.PowerShell.Core\FileSystem::\\server\share\path

The solution is to use the ProviderPath property on the object returned by Resolve-Path:

> Resolve-Path \\server\share\path | Select-Object -ExpandProperty PRoviderPath
\\server\share\path
> (Resolve-Path \\server\share\path).ProviderPath
\\server\share\path

Solution 7 - C#

Try changing:

Server.MapPath("/UploadBucket/Raw/")

to

Server.MapPath(@"\UploadBucket\Raw\")

Solution 8 - C#

This was my problem, which may help someone else -- although it wasn't the OP's issue:

DirectoryInfo diTemp = new DirectoryInfo(strSomePath);
FileStream fsTemp = new FileStream(diTemp.ToString());

I determined the problem by outputting my path to a log file, and finding it not formatting correctly. Correct for me was quite simply:

DirectoryInfo diTemp = new DirectoryInfo(strSomePath);
FileStream fsTemp = new FileStream(diTemp.FullName.ToString());

Solution 9 - C#

Does using the Path.Combine method help? It's a safer way for joining file paths together. It could be that it's having problems joining the paths together

Solution 10 - C#

I had the same issue today. The file I was trying to load into my code was open for editing in Excel. After closing Excel, the code began to work!

Solution 11 - C#

I am using the (limited) Expression builder for a Variable for use in a simple File System Task to make an archive of a file in SSIS.

This is my quick and dirty hack to remove the colons to stop the error: @[User::LocalFile] + "-" + REPLACE((DT_STR, 30, 1252) GETDATE(), ":", "-") + ".xml"

Solution 12 - C#

Image img = Image.FromFile(System.IO.Path.GetFullPath("C:\\ File Address"));

you need getfullpath by pointed class. I had same error and fixed...

Solution 13 - C#

If the value is a file url like file://C:/whatever, use the Uri class to translate to a regular filename:

var localPath = (new Uri(urlStylePath)).AbsolutePath

In general, using the provided API is best practice.

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Solution 1 - C#user586399View Answer on Stackoverflow
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