bashshell

Shell script to open a URL

How do I write a simple shell script (say script.sh), so that I can pass a URL as an argument while executing?

I want a browser to start with the page opened on that URL. I want to write the command in the script to open a browser and open the URL given in argument.

Solution

  • Method 1

    Suppose your browser is Firefox and your script urlopener is

    #!/bin/bash
firefox "$1"

    Run it like

    ./urlopener "https://google.com"

    Sidenote

    Replace firefox with your browser's executable file name.

    Method 2

    As [ @sato-katsura ] mentioned in the comment, in *nixes you can use an application called xdg-open. For example,

    xdg-open https://google.com

    The manual for xdg-open says

    xdg-open - opens a file or URL in the user's preferred application xdg-open opens a file or URL in the user's preferred application. If a URL is provided the URL will be opened in the user's preferred web browser.
    If a file is provided the file will be opened in the preferred application for files of that type. xdg-open supports file, ftp, http and https URLs.

    As [ this ] answer points out you could change your preferred browser using say:

    xdg-settings set default-web-browser firefox.desktop

    or

    xdg-settings set default-web-browser chromium-browser.desktop