Hide shell when spawning a new process

[Alexandre789010]

Hello, I cannot get the shell to not pop-up when my script is executing with the app build (dev does not show the shell for some reason). This app is running on Windows 11.

I have this on top of my main.rs

#![cfg_attr(
    all(not(debug_assertions), target_os = "windows"),
    windows_subsystem = "windows"
)]

and this is where the script gets called:

let main_window = app_handle.get_window("main").unwrap();
let file_path = save_script_to_file(id)?;
let stdout = std::process::Command::new("powershell")
        .arg("-ExecutionPolicy")
        .arg("Bypass")
        .arg("-File")
        .arg(&file_path)
        .stdout(std::process::Stdio::piped())
        .stderr(std::process::Stdio::piped())
        .spawn()
        .map_err(|e| e.to_string())?
        .stdout
        .ok_or_else(|| "Could not capture standard output.".to_string())?;

let reader = std::io::BufReader::new(stdout);

reader
    .lines()
    .filter_map(|line| line.ok())
    .for_each(|line| {
        if let Err(e) = main_window.emit("script-output", &line) {
            eprintln!("Failed to emit line: {}", e);
        }
    });

Ok(())

Am I forgetting something?

[FabianLars]

You're looking for this: https://doc.rust-lang.org/std/os/windows/process/trait.CommandExt.html#tymethod.creation_flags
Add use std::os::windows::process::CommandExt; to your file, then add .creation_flags(0x08000000) to your Command.