CODEWITHSUNDEEP
regexgrepcut
user10143594
user10143594

Reputation: 79

How to write a regex for this?

Requirements: only grep/cut/join/regex.

I have data like this:

  798 /usr/bin/dbus-daemon --system --address=systemd: --nofork --nopidfile --systemd-activation --syslog-only
15386 /usr/bin/nautilus --gapplication-service
16051 /usr/bin/zeitgeist-daemon

I want to extract rows data from the number to second ending space, like

798 /usr/bin/dbus-daemon

using only grep/cut/join with or without regex.

I have tried

grep -oe "[^ ][^ ]*  *[a-zA-Z\]*$"

but the result isn't as expected.

Upvotes: 1

Views: 51

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626896

You may use

# With GNU grep:
grep -oP '^\s*\K\S+\s+\S+' <<< "$s"
# With a POSIX ERE pattern:
grep -oE '[0-9][^ ]* +[^ ]+' <<< "$s"

See the online demo

  • o - match output mode (not line)
  • P - PCRE regex engine is used to parse the pattern

The PCRE pattern details:

  • ^ - start of line
  • \s* - 0+ whitespaces
  • \K - match reset operator discarding the whole text matched so far
  • \S+ - 1+ non-whitespace chars
  • \s+\S+ - 1+ whitespaces and 1+ non-whitespace chars.

The POSIX ERE pattern matches

  • [0-9] - a digit
  • [^ ]* - 0+ chars other than space
  • + - 1 or more spaces
  • [^ ]+ - 1+ chars other than a space.

Upvotes: 1

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