How to get nth column with regexp delimiter [duplicate]

bash unix text text-processing

If you want to do it with cut you need to squeeze the space first (tr -s ' ') because cut doesn't support +. This should work:

ls -la | tr -s ' ' | cut -d' ' -f 5

It's a bit more work when doing it with sed (GNU sed):

ls -la | sed -r 's/([^ ]+ +){4}([^ ]+).*/\2/'

Slightly more finger punching if you use the grep alternative (GNU grep):

ls -la | grep -Eo '[^ ]+( +[^ ]+){4}' | grep -Eo '[^ ]+$'

bash unix text text-processing

Parsing ls output is harder than you think. Use a dedicated tool such as stat instead.

size=$(stat -c '%s' some_file)

One way ls -la some_file | awk '{print $5}' could break is if numbers use space as a thousands separator (this is common in some European locales).

bash unix text text-processing

Pipe your output with:

awk '{print $5}'

Or even better us to use stat command like this (On Mac):

stat -f "%z" yourFile

Or (on Linux)

stat -c "%s" yourFile

that will output size of file in bytes.

CodeHunter