如何通过单击从 Windows 中的 Go 可执行文件打开特定文件

Question

I have a Go program from which I have built a Windows OS executable. This program just takes in command line arguments and prints them.

What I want to do is: in Windows, I want to click on a specific file (let's say a text file) and I want my Go executable to print the name of the file. I am not sure how Windows passes the name of the file to an executable.

So far, I have right-clicked on my text file > Open with > choose another program > Look for another program in my pc > selected my Go executable. But I am unsure about the next steps.

I want my Go executable to find the location of the text file and then manipulate it.

Answer 1

Windows passes the file name to the executable in the arguments so in your case if you are opening the file named process.txt using

process.txt > Open with > Choose another program > Look for another program in my pc > Selected my Go executable

I assume that your Golang code would be following following.

package main

import (
    "log"
    "os"
    "time"
)

func main() {
    log.Println(os.Args[1:])
    time.Sleep(10 * time.Second)
}

Output:

[.../.../process.txt]

This sample program will open up a command prompt and display the arguments on the console which will contain the file name which is passed via Open With . Now you will have the file path which you can process in your program.