When do we need to set UseShellExecute to True?

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the process object must have the useshellexecute property set to false in order to use
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//
// Summary:
//     Gets or sets a value indicating whether to use the operating system shell
//     to start the process.
//
// Returns:
//     true to use the shell when starting the process; otherwise, the process is
//     created directly from the executable file. The default is true.
[DefaultValue(true)]
[MonitoringDescription("ProcessUseShellExecute")]
[NotifyParentProperty(true)]
public bool UseShellExecute { get; set; }

If we spawn a new process, when do we need to set UseShellExecute to True?

c# - When do we need to set UseShellExecute to True?, In order to do this I know UseShellExecute = true is the only option. I get the "​The process object must have the UseShellExecute property set  The UseShellExecute boolean property is related to the use of the windows ShellExecute function vs the CreateProcess function – the short answer is that if UseShellExecute is true then the Process class will use the ShellExecute function, otherwise it will use CreateProcess. The longer

I think mostly for non-executables. For instance if are trying to open a .html file, if you'll have to set UseShellExecute to true and that will open the .html in a browser that's set as default by the user.

ProcessStartInfo.UseShellExecute Property (System.Diagnostics , Without this set, you can only execute an EXE file directly. By setting this, you UseShellExecute must be true for ProcessStartInfo.ErrorDialog  I need to set the useshellexecute to true for my windows form application project before the application process start, but I can not find where in the project I have to make the change? It's a windows form application project in visual studio 2015.

From MSDN:

Setting this property to false enables you to redirect input, output, and error streams.

UseShellExecute must be false if the UserName property is not null or an empty string, or an InvalidOperationException will be thrown when the Process.Start(ProcessStartInfo) method is called.

When you use the operating system shell to start processes, you can start any document (which is any registered file type associated with an executable that has a default open action) and perform operations on the file, such as printing, with the Process component. When UseShellExecute is false, you can start only executables with the Process component.

UseShellExecute must be true if you set the ErrorDialog property to true.

C# UseShellExecute = TRUE. Solutions, So, to make this work, we need to set UseShellExecute to false and then, UseShellExecute is true, you can start any document if its file type is  By setting UseShellExecute to true, process.Start() will run the string represented by startInfo.FileName via the shell (either cmd.exe or PowerShell, I'm not sure which). This allows us as end-users to specify .cmd, .bat, or an executable that includes commandline arguments (I think).

If we want to hide the current Application executable window , Then UseShellExecute should be set to true

compatibility consideration for UseShellExecute with Process.Start , RedirectStandardError = true, UseShellExecute = false, }); var errors = new StringBuilder With UseShellExecute true, you can do the following: Specify a path to a file or document rather than Should you need to do so—while launching a file or document—a workaround is to set UseShellExecute to false and invoke the  In order to do this I know UseShellExecute = true is the only option. I am totally fine with that but then I get the "The process object must have the UseShellExecute property set to false in order to start a process as a user."

When the path contains a space or some other special (i.e. accented) characters, CreateProcess (UseShellExecute=false) seems to be using short file names ("DOS" 8.3 notation), ShellExecute (UseShellExecute=true) uses long file names. So when you use UseShellExecute=false, make sure to convert your directory and file names to 8.3 names (google ".net how to get 8.3 filename"). (Not exectly sure what Windows versions and/or file systems do it this way, tested on Windows 7, NTFS.)

Process.Start() Quips – Living .NET…, NET code behind this example is very simple, so we'll run through it very quickly. and RedirectStandardOutput, which we need to set to true, so that any output  Another simple tip is that, when UseShellExecute is true, you can start any document if its file type is associated with an executable that has the default action as open. For e.g. The following code launches test.txt in default text editor (notepad on my m/c)

C# 8.0 in a Nutshell: The Definitive Reference, Initializer Syntax You will often want to set certain properties when you create an Start(new ProcessStartInfo { FileName = "cmd.exe", UseShellExecute = true,  Turning ShellExecute on tells the ProcessClass to use the ShellExecute API function to execute the command instead of using the CreateProcess function. If ShellExecute is used to start the process you lose access to the input and output streams. There is no way to get at the streams if launched by ShellExecute.

Pro SQL Server 2005 Assemblies, So why set the UseShellExecute as false ? Now, to capture the input/output/​error, we need to create the process directly RedirectStandardOutput = true;. UseShellExecute is false because we're specifying // an executable directly and in this case depending on it being in a PATH folder. By setting // RedirectStandardOutput to true, the output of cl.exe is directed to the Process.StandardOutput stream // which is then displayed in this console window directly.

Programming C# 8.0: Build Cloud, Web, and Desktop Applications, Here we run a program called dcm2jpg.exe—it converts a certain image format. Part 1: This example first creates a ProcessStartInfo. We use CreateNoWindow and UseShellExecute to control some command-line options. Part 2: We set some arguments to indicate to the executable what directories were are using.

Comments
  • Great Stuff, but you write that (with ShellExecute), "It[you claim] isn't possible to redirect the standard input / output / error handles" <-- Surely that is incorrect or inaccurate. Even with useShellExecute set to true, while indeed you can't do processStartInfo.RedirectStandardOutput=true, it seems to me you can still redirect standard output by doing process.Arguments= "cmd /c dir >c:\\crp\\a.a". Likewise from a run dialog box you can do cmd /c dir>c:\crp\a.a
  • also, you say that when UseShellExecute=false i.e. CreateProcess, won't check the path, but I see that even when I do "UseShellExecute=false" i.e. supposedly not checking the path, then process.FileName="cmd.exe" works so it is checking c:\windows\system32. And if I copy cmd.exe to c:\windows and name it cmmmd.exe then I do process1.FileName="cmmmd.exe" that works too so it's checking c:\windows so it seems like it's checking the path, or some bunch of directories.
  • MSDN docs agree with @barlop: "When UseShellExecute is false, the FileName property can be either a fully qualified path to the executable, or a simple executable name that the system will attempt to find within folders specified by the PATH environment variable."
  • By setting UseShellExecute to true I was able to share an environment variable (that was only created in the calling process). Very handy
  • Could it be that it is just cutting off the path at the space? Putting quotes around the "path/program name" solves this.