Shell Script - How to get stuff between two separators

I've got something like that:

xinput --list
⎡ Virtual core pointer                      id=2    [master pointer  (3)]
⎜   ↳ Virtual core XTEST pointer                id=4    [slave  pointer  (2)]
⎜   ↳ Microsoft Basic Optical Mouse             id=8    [slave  pointer  (2)]
⎣ Virtual core keyboard                     id=3    [master keyboard (2)]
    ↳ Virtual core XTEST keyboard               id=5    [slave  keyboard (3)]
    ↳ Power Button                              id=6    [slave  keyboard (3)]
    ↳ Power Button    开发者_如何学Go                          id=7    [slave  keyboard (3)]
    ↳ CHICONY HP Basic USB Keyboard             id=9    [slave  keyboard (3)]

And I want to get the id number. I've try something like awk -F 'id=' '[print $2} but text after the id number is still here! How i can only get the id number!

Thanks in advance.

---

This may not be the shortest, but should be easy to understand for anyone with a little bit of awk knowledge:

awk '{ for (i = 1; i <= NF; i++) if ($i ~ /id=/) { print substr($i, 4); break; } }'

---

Try:

xinput --list | grep -Eo 'id=[0-9]+' | grep -Eo '[0-9]+'

or

xinput --list | grep -Eo 'id=[0-9]+' | awk -F= '{print $2}'

---

one way

xinput --list | awk -F"id=" 'NF{split($2,a," ");print a[1]}' file

with the shell(bash)

#!/bin/bash
xinput --list | while read -r line
do
  case "$line" in
   *id=* )
     line=${line##*id=}
     echo ${line%% *}
  esac
done

---

try something like this:

xinput --list | while read line; do
    ID=`echo ${line} | cut -d= -f2 | cut -d\[ -f1 | xargs`
    echo ${ID}
done

---

Bash only:

while read line ; do
    if [[ $line =~ id=([[:digit:]]) ]] ; then
        echo "${BASH_REMATCH[1]}"
    fi
done < <( xinput --list )

---

I find these things simplest with sed, using substitution of a group to capture the data you want and ensuring the rest of the line is matched in the regex.

sed 's/.*id=\([0-9]*\).*/\1/'

---

Try:

cat x | awk -Fid= '{print $2}' | awk  '{print $1}'