Get only numbers in output

Question

I need to get only numbers from this: release/M_0.1.0 thus, need to extract with bash to have in output this: 0.1.0 . I have tried this but cannot finish it: echo "release/M_0.1.0" | awk -F'/' '{print $2}' echo "release/M_0.1.0" | awk -F'/' '{print $2}'

And what about if given such string? relea234se/sdf23_4Mm0.1.0.8 . How to get only 0.1.0.8 ? Please pay attention that this can be random digits such as 0.2 or 1.9.1 .

Answer 1

You could also use general parameter expansion parsing to literally remove characters up through the last that isn't digits or dots.

$: ver() { echo "${1//*[^.0-9]/}"; }
$: ver release/M_0.1.0
0.1.0
$: ver relea234se/sdf23_4Mm0.1.0.8
0.1.0.8

Answer 2

请检查此grep命令是否有效

echo "release/M_0.1.0" | egrep -o '[0-9.]+'

Answer 3

With sed you can do:

echo "release/M_0.1.0" | sed 's@.*_@@'

Output :

0.1.0

Answer 4

Considering that your Input_file will be same as shown samples.

echo "$var" | awk -F'_' '{print $2}'

OR could use sub :

echo "$var" | awk '{sub(/.*_/,"")} 1'

---

With simple bash you could use:

echo "${var#*_}"

Answer 5

echo release/M_0.1.0 | awk -F\_ '{print $2}'
0.1.0

Answer 6

Take your pick:

$ var='relea234se/sdf23_4Mm0.1.0.8'

$ [[ $var =~ .*[^0-9.](.*) ]] && echo "${BASH_REMATCH[1]}"
0.1.0.8

$ echo "$var" | sed 's/.*[^0-9.]//'
0.1.0.8

$ echo "$var" | awk -F'[^0-9.]' '{print $NF}'
0.1.0.8

Answer 7

如果数据在d文件中，请尝试使用 gnu sed：

sed -E 's/relea.*/.*([0-9][0-9.]*)$/\1/' d