How launch any file using default program using Rust (like Python's `os.startfile()`)

user_11400858 0
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Is there a rust method equivalent to os.startfile() in python. For example, I need to launch an "mp3 file" using rust. In python it is os.startfile('audio.mp3'). This will open default media player and start playing that file. I need to do the same with Rust Language.

rust operating-system python-3.x python
2 Answers
user_2796270

There are two ways that will work over multiple OS platforms (Mac, Windows, and Linux) found so far. I have personally tested it as well.
Method 1: Use opener crate (link) On Windows the ShellExecuteW Windows API function is used. On Mac the system open command is used. On other platforms, the xdg-open script is used. The system xdg-open is not used; instead a version is embedded within this library.

Use the following code in the rs file(src/main.rs):

// open a file
let result = opener::open(std::path::Path::new("Cargo.toml"));
println!("{:?}", result); // for viewing errors if any captured in the variable result

Use the following code in the "Cargo.toml" file in the dependencies section:

opener = "0.4.1"

Method 2: Use open crate (link) Use this library to open a path or URL using the program configured on the system. It is equivalent to running one of the following: open <path-or-url>(OSX), start <path-or-url> (Windows), xdg-open <path-or-url> || gio open <path-or-url> || gnome-open <path-or-url> || kde-open <path-or-url> || wslview <path-or-url> (Linux).

Use the following code in the rs file(src/main.rs):

// to open the file using the default application
open::that("Cargo.toml");
// if you want to open the file with a specific program you should use the following
open::with("Cargo.toml", "notepad");

Use the following code in the "Cargo.toml" file in the dependencies section:

open = "1.7.0"

Hope it works to all.

user_114008580
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