Error - The Directory name is InValid

Question

User-1499457942 posted

Hi

  I have below lines

string startPath = @"d:\PO.xls";
string zipPath = @"d:\Po.zip";

ZipFile.CreateFromDirectory(startPath, zipPath);

Thanks

Answer 1

User475983607 posted

JagjitSingh

  I have below lines

string startPath = @"d:\PO.xls";
string zipPath = @"d:\Po.zip";

ZipFile.CreateFromDirectory(startPath, zipPath);

Thanks

You're not using the CreateFromDirectory method as documented.  You are supplying a source file name when the method docs clearly state the method is expecting a source folder.

Please try reading the docs rather than making up you own idea how the method functions.

https://msdn.microsoft.com/en-us/library/hh485707(v=vs.110).aspx

Answer 2

User-1499457942 posted

Hi

 I have written this but still getiing same error 

string zipPath = @"d:\Po\Po.zip";

Thanks

Answer 3

User475983607 posted

Hi

 I have written this but still getiing same error 

string zipPath = @"d:\Po\Po.zip";

Thanks

Correct because you have not changed the source to a folder.  Please read the openly published documentation and see the sample code.

using System;
using System.IO;
using System.IO.Compression;

namespace ConsoleApplication
{
    class Program
    {
        static void Main(string[] args)
        {
            string startPath = @"c:\example\start";
            string zipPath = @"c:\example\result.zip";
            string extractPath = @"c:\example\extract";

            ZipFile.CreateFromDirectory(startPath, zipPath);

            ZipFile.ExtractToDirectory(zipPath, extractPath);
        }
    }
}

Answer 4

User-1499457942 posted

Hi

 Still same problem

string startPath = @"d:\PO\PO.xls";
string zipPath = @"d:\PO\PO.zip";

ZipFile.CreateFromDirectory(startPath, zipPath);

Thanks

Answer 5

User475983607 posted

Hi

 Still same problem

string startPath = @"d:\PO\PO.xls";
string zipPath = @"d:\PO\PO.zip";

ZipFile.CreateFromDirectory(startPath, zipPath);

Thanks

I guess you do not know the difference between a file and a folder.  This is a file

@"d:\PO\PO.xls";

This is a folder.

@"d:\PO";

The method expects a source folder not a file.  Change the startPath to...

@"d:\PO";

and the method will zip everything in the PO folder as openly written in the reference documentation.

Answer 6

User-1499457942 posted

Hi

  I gave startpath = "d:\po"  With this it is zipping all the files . i want to Zip single file

Thanks

Answer 7

User475983607 posted

Hi

  I gave startpath = "d:\po"  With this it is zipping all the files . i want to Zip single file

Thanks

Again, It is very helpful to read the documentation rather than making up your own syntax.

One simple method to solve this problem is to create a folder and place inside the folder a single file.  You might need other logic to clean up the folder - not sure.

Another solution is creating a zip archive and adding to the archive.

https://msdn.microsoft.com/en-us/library/hh485720(v=vs.110).aspx