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Home/ Questions/Q 8965793
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T16:53:13+00:00 2026-06-15T16:53:13+00:00

Basically I get line from ls -la command: -rw-r–r– 13 ondrejodchazel staff 442 Dec

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Basically I get line from ls -la command:

-rw-r--r--  13 ondrejodchazel  staff  442 Dec 10 16:23 some_file

and want to get size of file (442). I have tried cut and sed commands, but was unsuccesfull. Using just basic UNIX tools (cut, sed, awk…), how can i get specific column from stdin, where delimiter is / +/ regexp?

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  1. Editorial Team
    Editorial Team
    2026-06-15T16:53:14+00:00Added an answer on June 15, 2026 at 4:53 pm

    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 '[^ ]+$'
    
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