Ubuntu – “command not found” when running “lessc” from bash

Better use:

if [[ "$(uname -m)" == "x86_64" ]]; then

Notice the space between [[ and first parameter, two = signs , and the space between "x86_64" and ]]

Also, it is not a good idea to include ! inside echo :)

I think that that's the best place to refer to when doing such operations: http://mywiki.wooledge.org/BashPitfalls

The -A sudo option allows you to specify a helper program (in the SUDO_ASKPASS variable) that will ask for the password.

Create a script to ask the password (myaskpass.sh):

#!/bin/bash
zenity --password --title=Authentication

Then insert this line at the beginning of your script:

export SUDO_ASKPASS="/path/to/myaskpass.sh"

and replace all occurences of sudo <command> with:

sudo -A <command>

You can use whatever password asking program you want instead of zenity. I had to encapsulate it within a script because SUDO_ASKPASS must point to a file, so it won't work with the --password option required by zenity.

The above works like a charm if it runs from command line or if you choose Run in terminal after double click the script file in the file manager, but if you choose Run or try to launch it from a .desktop file every sudo will ask for the for password again.

If you don't want a terminal window at all, you can store the password in a variable and pipe it to sudo -S. Maybe there's some security concerns, but I think it's pretty safe (read the comments on this answer).

Insert this line at the beginning of your script:

PASSWD="$(zenity --password --title=Authentication)\n"

and replace all occurences of sudo <command> with:

echo -e $PASSWD | sudo -S <command>