Develop Reference

Making directories with command line on C#

System.Diagnostics.Process process = new System.Diagnostics.Process ();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo ();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = "md " + Environment.GetFolderPath(Environment.SpecialFolder.DesktopDirectory);
process.StartInfo = startInfo;
process.Start ();
I'm attempting to make a directory on the desktop with this command, it doesn't make one however. Can anyone tell me why?

Just do this:
Directory.CreateDirectory(Path.Combine(
Environment.GetFolderPath(Environment.SpecialFolder.DesktopDirectory),
"my new folder name"));
Always prefer using the .NET class library instead of invoking external processes to do your work, unless you have a very specific reason not to do so.
One of the reasons your code is not working is because you are using the wrong syntax for cmd.exe. In order to pass a command as an argument, you have to use the following with the /K switch (use cmd /? for more information):
cmd.exe /K MD "c:\test\blah"
Another reason your code won't work is that the path you're providing to the MD command is just the path to the desktop itself:
Environment.GetFolderPath(Environment.SpecialFolder.DesktopDirectory)
You have forgotten to append the name of the folder you want to create on the desktop.

Run multiple net use commands for a process using c#

System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new
System.Diagnostics.ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Maximized;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = "/C net use W: \\\\abcdpc25\\views\\Projects ";
process.StartInfo = startInfo;
process.Start();
I want to run multiple net use commands and i am not sure how to go about it
I am not sure if something like below would work .
startInfo.Arguments = "/C net use W: \\\\abcdpc25\\views\\Projects
net use H: \\\\abcdpc25\\views\\Project456 net use E:
\\\\abcdpc25\\views\\Project123";

Use & as the separator between commands and it should work.
You can also put all the net use statements in a batch file and put the batch file name on your command line.
You can also launch multiple processes, one for each net use.

Unable to execute cmd commands from C#

I want to execute cmd commands from Visual Studio (using c#). There are two commands which I want to run.
I referred this post, but not working in my case.
My code is:
private void ExecuteCmdCommands()
{
string strCmdText;
strCmdText = #"cd C:\Test + makecab /f wsp.ddf";
Process process = new Process();
ProcessStartInfo startInfo = new ProcessStartInfo();
startInfo.WindowStyle = ProcessWindowStyle.Hidden;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = strCmdText;
process.StartInfo = startInfo;
process.Start();
}
When I run this code, only command prompt is open no commands are executed. What am I missing?

Don change directory, simply shell open the file:
strCmdText = #"C:\Test + makecab /f wsp.ddf";
Edit: Set the working directory:
startInfo.WorkingDirectory = #"C:\Test";

In order to call 2 command from one line, you need to use the & sign.
In your case:
#"/c cd C:\Test & makecab /f wsp.ddf";
Also dont forget the /c flag, telling the cmd to execute the command.

try to change to strCmdText = #"C:\Test + makecab /f wsp.ddf";

Just a guess, but it looks like you are attempting to execute 2 different command on the same line??
First changing the directory is not necessary, and you don't need to execute cmd.exe. Just create a process directly for the makecab program.
startInfo.Filename = #"makecab.exe";
startInfo.Argumanets = #"/f c:\test\wsp.ddf";

The issue here is that the commands you're passing in are arguments and not commands (which would have to be piped in through the StandardInput pipe). Fortunately, you can use the "/c" argument as mentioned here. I'm not sure if it will work with the "+" operator.
Note, as someone else mentioned, you should also set the working directory using the available property or your program will fail if not run with a "C:" working directory.

you have to do run shell commands from C#
string strCmdText;
strCmdText= "/C copy /b Image1.jpg + xyz.rar Image2.jpg"; System.Diagnostics.Process.Start("CMD.exe",strCmdText);
**EDIT:**
This is to hide the cmd window.
System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = "/C copy /b Image1.jpg + xyz.rar Image2.jpg";
process.StartInfo = startInfo;
process.Start();

Commandline Arguments in C#

Hello again Stackoverflow community,
Today I am trying to execute an application with commandline parameters in C#, that not realy difficult like I tried
Process.Start(foldervar + "cocacola.exe", "pepsi.txt");
Cocacola.exe writes and Log in its current folder. In my commandline I write it manually like this
C:\myfolder>cocacola.exe pepsi.txt
Works wonderful but if I try it in C# a total fail.
I read that C# parses the command as C:\myfolder>cocacola pepsi.txt, without the ".EXE" ending. And I tested it manually without the ending, and this does not work.
Now, my question is what is the correct way to get C# executing it C:\myfolder>cocacola.exe pepsi.txt with the ".EXE"

use ProcessStartInfo
http://www.dotnetperls.com/process-start
example:
System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.WorkingDirectory=#"c:\someplace";
proc.StartInfo.FileName="cocacola.exe";
proc.StartInfo.Arguments="pepsi.txt";
proc.Start();
proc.WaitForExit();
here is docs on the StartInfo properties:
http://msdn.microsoft.com/en-us/library/system.diagnostics.processstartinfo.aspx

Try setting the StartInfo properties.
Process process = new Process();
process.StartInfo.FileName = #"C:\myfolder\cocacola.exe";
process.StartInfo.Arguments = #"C:\myfolder\pepsi.txt";
process.Start();

ProcessStartInfo has the WorkingDirectory property you should set to C:\myfolder
see: http://msdn.microsoft.com/en-us/library/system.diagnostics.processstartinfo.workingdirectory.aspx

You need to set the working directory first
string foldervar = #"C:\myfolder";
Process process = new Process();
process.StartInfo.WorkingDirectory = foldervar;
process.StartInfo.FileName = #"cocacola.exe";
process.StartInfo.Arguments = #"pepsi.txt";
process.Start();
Setting the WorkingDirectory is equivilent to cding into the proper directory before running programs. It's what relative paths are relative to.

Open txt file from C# application

The following code is suppose to open CMD from my C# application and open the file text.txt.
I tried to set the file path as an environment variable but when notepad opens it looks for %file%.txt instead of text.txt
Any idea why?
System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.EnableRaisingEvents=false;
proc.StartInfo.EnvironmentVariables.Add("file", "c:\\text.txt");
proc.StartInfo.UseShellExecute = false;
proc.StartInfo.FileName = "notepad";
proc.StartInfo.Arguments="%file%";
proc.Start();
proc.WaitForExit();
Console.WriteLine(proc.ExitCode);

If your purpose is to start the editor with a .txt file (like the title of the question says) just use:
Process.Start("C:\\text.txt")

The short version is that I suspect you are going to have to pass the arg more directly, i.e.
proc.StartInfo.Arguments = #"""c:\text.txt""";
Although you can set environment variables (for use within the process), I don't think you can use them during the process start.

What are you trying to accomplish with %file%? The command line argument for notepad.exe is the file you want to open. You need to do something like this:
proc.StartInfo.Arguments = "c:\\text.txt";

set UseShellExecute = true
that way it should use the cmd.exe processor to expand the %file% variable

One obvious problem is that you have UseShellExecute set false. This means you are executing notepad directly without passing via the command shell cmd.exe. Therefore environment variables aren't being expanded.
I'm not sure what you're trying to achieve (why do you need to add an environment variable?) but the following would work:
System.Diagnostics.Process proc =
new System.Diagnostics.Process();
proc.EnableRaisingEvents = false;
proc.StartInfo.EnvironmentVariables.Add("file", "c:\\text.txt");
proc.StartInfo.UseShellExecute = false;
proc.StartInfo.FileName = "cmd.exe";
proc.StartInfo.Arguments = "/c notepad %file%";
proc.Start();
proc.WaitForExit();

Try this:
proc.StartInfo.Arguments = System.Environment.GetEnvironmentVariable("file");

Perhaps it has to do with how the StartInfo.Arguments work. If you can't find anything better, this worked for me:
proc.StartInfo.FileName = "cmd";
proc.StartInfo.Arguments="/c notepad %my_file%";

I am willing to bet you need to set WorkingDirectory to get this to work. NOTEPAD.exe is usually located in %SYSTEMROOT% (C:\windows) however the default is %SYSTEMROOT%\system32. Try out the below.
System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.EnableRaisingEvents=false;
proc.StartInfo.WorkingDirectory = "%SYSTEMROOT%";
proc.StartInfo.EnvironmentVariables.Add("file", "c:\\text.txt");
proc.StartInfo.UseShellExecute = false;
proc.StartInfo.FileName = "notepad";
proc.StartInfo.Arguments="%file%";
proc.Start();
proc.WaitForExit();
Console.WriteLine(proc.ExitCode);