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

  • should this work recursively, and if so, how does the counting play into that? do you only want files or would directories count? what if there isn't a (say) 5th file? – Jeff Schaller Jun 11 at 15:49
  • If asked for a 5th file, there would be a 5th file. There also won't be any directories to worry about. – Dragon8oy Jun 11 at 15:53

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; }"


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.

  • 2 points: 1) don't parse ls; 2) the bash variable won't expand inside single quotes -- pass the variable with awk's -v option. – glenn jackman Jun 11 at 15:56
  • Hmm, didn't return anything for me – Dragon8oy Jun 11 at 15:58
  • Make sure you replace $fileIndex with the actual index (i.e. 2) of the file you're trying to find. – Dr-Bracket Jun 11 at 16:03
  • I know, replace the single quotes with double quotes and it works fine – Dragon8oy Jun 11 at 16:21
  • Fixed, thanks :P – Dr-Bracket Jun 11 at 16:28

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