Opening .exe file using OpenFileDialog( ) in C# .net

[dvp10]

Hi, i want open an .exe file using OpenFileDialog( ) in C#
The code i ve used is -


Stream strm;
StreamReader reader;
try
{
strm = OpenFileDialog1 .OpenFile();
reader = new StreamReader(st rm);
}
catch (Exception err)
{
string Message = err.Message + "\n\nCannot open "
+ OpenFileDialog1 .FileName;
MessageBox.Show (Message, "Open error",
MessageBoxButto ns.OK,
MessageBoxIcon. Error);
return;
}
reader.ReadToEn d();
reader.Close();
strm.Close();


But, I m getting an error..
Can anybody tell me the cause....
Thanx in advance

[r035198x]

Hi, i want open an .exe file using OpenFileDialog( ) in C#
The code i ve used is -


Stream strm;
StreamReader reader;
try
{
strm = OpenFileDialog1 .OpenFile();
reader = new StreamReader(st rm);
}
catch (Exception err)
{
string Message = err.Message + "\n\nCannot open "
+ OpenFileDialog1 .FileName;
MessageBox.Show (Message, "Open error",
MessageBoxButto ns.OK,
MessageBoxIcon. Error);
return;
}
reader.ReadToEn d();
reader.Close();
strm.Close();


But, I m getting an error..
Can anybody tell me the cause....
Thanx in advance

What is the error you are getting?
Do you want to read the bytes making up the file or do you want to run the .exe program?

[dvp10]

What is the error you are getting?
Do you want to read the bytes making up the file or do you want to run the .exe program?

I want to open the .exe file and also ro run it.
The error message is- 'Index was outside the bounds of the array'

[r035198x]

I want to open the .exe file and also ro run it.
The error message is- 'Index was outside the bounds of the array'

To run an executable program, you need to start a process for it.
Here's an example that I use to run the javac.exe command

using System;
 
using System.Diagnostics;
 
class Commands {
 
    public static String compileJava (String file) {
 
        Process p = new Process();
 
            p.StartInfo.FileName = "javac";
 
               p.StartInfo.Arguments = file;
 
            p.StartInfo.UseShellExecute = false;
                            p.StartInfo.RedirectStandardOutput = true;
 
            p.Start();
 
                  return p.StandardOutput.ReadToEnd();
 
 
 
 
    }
 
 
        public static void Main(String[] args) {
                                    String output = Commands.compileJava("/home/Java/HelloCSharp.java");
 
        Console.WriteLine(output);
        }
 
} 
 

[dvp10]

Hi r035198x,
Thanx for reply

Actually i want to run the .exe file..so it must be opened using OpenFileDialog( )..
Also even if i try to open .txt file, its not opening...

I ve tried ur code in C# bt its not working...Can u plz send me C# code for the process..' If possible'....

Thnx in advance

[r035198x]

Hi r035198x,
Thanx for reply

Actually i want to run the .exe file..so it must be opened using OpenFileDialog( )..
Also even if i try to open .txt file, its not opening...

I ve tried ur code in C# bt its not working...Can u plz send me C# code for the process..' If possible'....

Thnx in advance

It doesn't matter where you get the path from. set the full path of you .exe to be the process's StartInfo.FileN ame. If you get any errors then post the code you used and the error message that you got.

[Plater]

It should be pointed out that if the EXE is running, you probably cannot open it to read in the bytes at the same time.

I am assuming you are showing the OpenFileDialog and handling the OK/Cancel button presses correctly before jumping to that line of code.

[dvp10]

Thanx,

My problem is solved..
i used System.Diagnost ic.Process.Star t("Path")

Thanx very much

[r035198x]

Thanx,

My problem is solved..
i used System.Diagnost ic.Process.Star t("Path")

Thanx very much

Glad you got it to work.