Bash – Extracting the second word from a string variable

bashstringvariable

I have a string "rtcpOnNbActive true" stored in a variable x. I want to extract "true" as substring and store in a variable. How can I do this?

Best Answer

Try this way:

y=$(echo $x | awk '{print $2}')
echo $y

echo $x display the value of x.
awk '{print $2}' prints the second field of the previously displayed x.
$(...) hold the output and let assign it to y.

Related Solutions

Bash – Remove specific word in variable

Try:

$ printf '%s\n' "${FOO//$WORDTOREMOVE/}"
CATS DOGS FISH

This also work in ksh93, mksh, zsh.

POSIXLY:

FOO="CATS DOGS FISH MICE"
WORDTOREMOVE="MICE"

remove_word() (
  set -f
  IFS=' '

  s=$1
  w=$2

  set -- $1
  for arg do
    shift
    [ "$arg" = "$w" ] && continue
    set -- "$@" "$arg"
  done

  printf '%s\n' "$*"
)

remove_word "$FOO" "$WORDTOREMOVE"

It assumes your words are space delimited and has side effect that remove spaces before and after "$WORDTOREMOVE".

Shell – Extracting substring from environment variable

You can use parameter expansion, which is available in any POSIX compliant shell.

$ export FOO=http://unix.stackexchange.com/questions/ask
$ tmp="${FOO#*//}" # remove http://
$ echo "${tmp%%/*}" # remove everything after the first /
unix.stackexchange.com

A more reliable, but uglier method would be to use an actual URL parser. Here is an example for python:

$ echo "$FOO" | python -c 'import urlparse; import sys;  print urlparse.urlparse(sys.stdin.read()).netloc' 
unix.stackexchange.com

Best Answer

Related Solutions

Bash – Remove specific word in variable

Shell – Extracting substring from environment variable

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