Using bash, I need to be able to find a file in a specific position when listed alphabetically. For example, if I had the files a, b, c, d, e in a directory, and I wanted to find the third file, I would need it to return c. If I wanted the 5th file, it would return e.
Thanks for any help, sorry if this is phrased poorly, I'll rephrase it later if I can think of a way
With zsh:
printf '%s\n' *([5]) Gives you the 5th non-hidden file in lexical order. Change to *(D[5]) to include hidden files (note that . and .. are never included).
In any Bourne-like shell, you can do the same with:
set -- * printf '%s\n' "$5" 
This work in any shell: ls | awk "NR==$fileIndex{ print; }"
Explanation:
ls returns all the files in a directory in alphabetical order, and piping ls runs each file on its own newline, and awk "NR==$fileIndex{ print; }" will print the line number defined by $fileIndex.
ls; 2) the bash variable won't expand inside single quotes -- pass the variable with awk's -v option.