Develop Reference

How to take window ID of a created process in c#?

I am working on a C# .net core project.I created a process to run "xdotool windowactivate $windowpid".I should store the windowID which process run on it.The solution could be any property of xdotool which i couldn't find,or Is there any way to take windowId of a process when it is created?
Another Try is that:
I created my pages with this method. I tried to take mainwindowtitle of process;because of single process,i couldn't take the titles.
static List<string> chromeTitles = new List<string>();
public static Process StartChrome(string filePath)
{
string dataDirectory = Path.Combine(AppDomain.CurrentDomain.BaseDirectory, "Chrome-UserData");
filePath += " --user-data-dir=" + dataDirectory;
var Chrome = new Process
{
StartInfo =
{
FileName = "C:/Program/chrome.exe",
Arguments = filePath,
UseShellExecute = false,
CreateNoWindow=true,
WindowStyle = ProcessWindowStyle.Maximized,
}
};
Chrome.Start();
string title = Chrome.MainWindowTitle;
chromeTitles.Add(title);
}
Then I call it :
StartChrome("https://tr.wikipedia.org/wiki/Anasayfa");
Thread.Sleep(2000);
StartChrome("https://tureng.com/");

You can use the Process class for accessing more capabilities.
System.Diagnostics.Process process = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.FileName = "xdotool.exe";
startInfo.UseShellExecute = false;
startInfo.RedirectStandardOutput = true;
startInfo.Arguments = $"windowactivate $windowpid";
process.StartInfo = startInfo;
process.Start();
To get the PID of the process that got run by the code, you can use Process.ID property:
process.Id;
if you want to read the output, you can add this code:
string output = process.StandardOutput.ReadToEnd();
To get Output, startInfo.RedirectStandardOutput should be True.

How to give 2nd input to a '.exe' file after the Process.start method is called?

I have to execute a '.exe' file which produces a output and asks for another input again. I am able to run the first part but I am not able to pass the second parameter to complete the process.
Here is my code:
ProcessStartInfo startInfo = new ProcessStartInfo();
startInfo.FileName = #"C:\Context.exe";
startInfo.Arguments = "xyz";
startInfo.UseShellExecute = false;
startInfo.CreateNoWindow = true;
Process.Start(startInfo);

I would use StandardInput in conjunction with RedirectStandardInput. You can pass any data that a user would enter on the command line using this StandardInput StreamWriter object. If this application has a user interface, you may need to do something else entirely.
ProcessStartInfo startInfo = new ProcessStartInfo();
startInfo.FileName = #"C:\Context.exe";
startInfo.Arguments = "xyz";
startInfo.UseShellExecute = false;
startInfo.CreateNoWindow = true;
startInfo.RedirectStandardInput = true;
var p = Process.Start(startInfo);
// Write whatever data you need to send to the application here.
p.StandardInput.Write("y");

You can use startInfo.RedirectStandardInput. See here for more details

run command promt application from code C#

This is my code:
var startupPath = Directory.GetParent(Directory.GetCurrentDirectory()).Parent.Parent.FullName; // +\\Common
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 " + startupPath + #"\Common\sound 1.wav result.wav -tempo=35";
process.StartInfo = startInfo;
process.Start();
Pathes are correct. A wanna to start application and set this parameters sound 1.wav result.wav -tempo=35. What am I doing wrong?
content of startInfo.Arguments:
"/c C:\\SOUNDS\\New folder\\TrunscribeHelper\\Common\\sound 1.wav result.wav -tempo=35"
I've tryed to start it directly:
ProcessStartInfo startInfo = new ProcessStartInfo(string.Concat(startupPath + #"\Common", "\\", "sound.exe"));
startInfo.Arguments = "1.wav result.wav -tempo=35";
startInfo.UseShellExecute = false;
System.Diagnostics.Process.Start(startInfo);
But Am I doing wrong now?

Your have to set your path in the args under double quotes:
"/c \"C:\\SOUNDS\\New folder\\TrunscribeHelper\\Common\\sound 1.wav\" result.wav -tempo=35"

Can you try changing this line:
startInfo.Arguments ="/c " + startupPath + #"\Common\sound 1.wav result.wav -tempo=35";
to
startInfo.Arguments ="/c \"" + startupPath + #"\Common\sound"" 1.wav result.wav -tempo=35";

Multiple arguments with cmd.exe via a C# Process

I try to call the esriRegAsm.exe with arguments from a C# program.
The purpose is to register a Dll. Therefore I usually call the esriRegAsm.exe with the Dll as argument plus some additional parameters (/p:Desktop /s). This works fine if I type it into cmd.exe. Somehow I think that the process sends only the first string to the cmd and not the whole argument list, but I need the "" for the space character in the paths.
For debugging I added a message box and the strings seems to be okay.
Backslash or double backslash seems to be unimportant.
string targetDir = this.Context.Parameters["targ"];
string programFilesFolder = this.Context.Parameters["proFiles"];
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 \"" + programFilesFolder + "Common Files\\\\ArcGIS\\\\bin\\\\esriRegAsm.exe\" " + "\"" + targetDir + "RArcGISTest.dll\" /p:Desktop /s";
MessageBox.Show("/C \"" + programFilesFolder + "Common Files\\\\ArcGIS\\\\bin\\\\esriRegAsm.exe\" " + "\"" + targetDir + "RArcGISTest.dll\" /p:Desktop /s");
process.StartInfo = startInfo;
process.Start();
As I cannot attache a picture of the message box... the output is:
/C "C:\Program Files (x86)\Common Files\ArcGIS\bin\esriRegAsm.exe" "C:\install\RArcGISTest.dll" /p:Desktop /s"

Why are you double-escaping things, and why are you routing it through cmd.exe? Just execute the process directly:
string targetDir = this.Context.Parameters["targ"];
string programFilesFolder = this.Context.Parameters["proFiles"];
Process process = new Process();
ProcessStartInfo startInfo = new ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo.FileName = Path.Combine(programFilesFolder, #"Common Files\ArcGIS\bin\esriRegAsm.exe");
startInfo.Arguments = "\"" + Path.Combine(targetDir, "RArcGISTest.dll") + "\" /p:Desktop /s";
process.StartInfo = startInfo;
process.Start();

How to run CL.exe using Process.Start()?

I have following code
using (StreamWriter outfile = new StreamWriter(#"f:\trial.cpp"))
{
outfile.Write(txtCode.InnerText);
}
System.Diagnostics.ProcessStartInfo procStartInfo = new System.Diagnostics.ProcessStartInfo(#"cl.exe", #" 'trial.cpp'");
procStartInfo.RedirectStandardOutput = true;
procStartInfo.UseShellExecute = false;
procStartInfo.UserName = "asdasd";
SecureString secureString = new SecureString();
foreach (char c in "abcded")
{
secureString.AppendChar(c);
}
procStartInfo.Password = secureString;
// Do not create the black window.
procStartInfo.CreateNoWindow = true;
procStartInfo.WorkingDirectory = #"f:\";
// Now we create a process, assign its ProcessStartInfo and start it
System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.StartInfo = procStartInfo;
proc.Start();
// Get the output into a string
string result = proc.StandardOutput.ReadToEnd();
How to pass file name as parameter? Above code doesn't run and I have tried all full path, different path options.
can anyone help?

The argument is set incorrectly. You have:
var procStartInfo = new ProcessStartInfo(#"cl.exe", #" 'trial.cpp'");
Where there are spaces and single quotes in the name. Try:
var procStartInfo = new ProcessStartInfo(#"cl.exe", #"trial.cpp");

EDIT:
startInfo.CreateNoWindow = false;
startInfo.UseShellExecute = false;
startInfo.FileName = "CL.exe";
startInfo.WindowStyle = ProcessWindowStyle.Hidden;
startInfo.Arguments = "trial.cpp";
try
{
using (Process exeProcess = Process.Start(startInfo))
{
exeProcess.WaitForExit();
}
}
catch
{
// error handling
}
The point here is that CL is a command line executable, not a windows GUI application.
http://msdn.microsoft.com/en-us/library/f2ccy3wt.aspx
http://msdn.microsoft.com/en-us/library/kezkeayy.aspx
http://msdn.microsoft.com/en-us/library/9s7c9wdw.aspx

If the cl.exe is not in the system PATH (which by default it is not) then the start process will not find the executable and it will fail to run.
So I suspect you are seeing the fact that the cl.exe is not in the system PATH.