[SOLVED] Getting selected file name(s) in bash script

ThomasHunter · 01-04-2013, 07:04 PM

Hello, I want to point out that I am very new to the world of bash script programming. Okay, so I made a couple very simple scripts to compile C++/Allegro/Java programs for me like this one:

Code:

#!/bin/bash
c++ *_src.cpp -o Out.cpp

I put this in my ~/.gnome2/nautilus-scripts folder so that I could right-click a _src file and compile it. HOWEVER this (obviously) only works if I have a single _src file in a directory. I need to be able to select a file, right-click, and compile it; having an output of the same file name and that one only. I do not know how to get a selected file name through script though, so I just used an asterisk. Thanks for any help in advance!

ThomasHunter · 01-05-2013, 08:35 PM

So for anyone who's looking at this in the future, I found that you can use the $* variable to get the selected file name. I edited my above code to this:

Code:

#!/bin/bash
c++ $* -o ConsoleOut.cpp
if [ $? -eq 0 ] #$? is the variable that says whether the last command is true or false
then
	zenity --info --title="Success - Compilation"  --text="Compilation Successful."
else
	zenity --error --title="Error - Ccompilation"  --text="Something went wrong - compile in terminal."
fi

So I hope this helps someone.