Output file lines from last to first in Bash

Question

I want to display the last 10 lines of my log file, starting with the last line- like a normal log reader. I thought this would be a variation of the tail command,

Answer 1

You can do that with pure bash:

#!/bin/bash
readarray file
lines=$(( ${#file[@]} - 1 ))
for (( line=$lines, i=${1:-$lines}; (( line >= 0 && i > 0 )); line--, i-- )); do
    echo -ne "${file[$line]}"
done

./tailtac 10 < somefile

./tailtac -10 < somefile

./tailtac 100000 < somefile

./tailtac < somefile

Answer 2

I ended up using tail -r, which worked on my OSX (tac doesn't)

tail -r -n10

Answer 3

tac does what you want. It's the reverse of cat.

tail -10 logfile | tac

Answer 4

This is the perfect methods to print output in reverse order

tail -n 10 <logfile>  | tac

Answer 5

GNU (Linux) uses the following:

tail -n 10 <logfile> | tac

tail -n 10 <logfile> prints out the last 10 lines of the log file and tac (cat spelled backwards) reverses the order.

BSD (OS X) of tail uses the -r option:

tail -r -n 10 <logfile>

For both cases, you can try the following:

if hash tac 2>/dev/null; then tail -n 10 <logfile> | tac; else tail -n 10 -r <logfile>; fi

NOTE: The GNU manual states that the BSD -r option "can only reverse files that are at most as large as its buffer, which is typically 32 KiB" and that tac is more reliable. If buffer size is a problem and you cannot use tac, you may want to consider using @ata's answer which writes the functionality in bash.