带有正则表达式的字符串之间的 Grep 值

Question

$ acpi

Battery 0: Charging, 18%, 01:37:09 until charged

How to grep the battery level value without percentage character (18)?

This should do it but I'm getting an empty result:

acpi | grep -e '(?<=, )(.*)(?=%)'

Answer 1

Your regex is correct but will work with experimental -P or perl mode regex option in gnu grep . You will also need -o to show only matching text.

Correct command would be:

grep -oP '(?<=, )\d+(?=%)'

However, if you don't have gnu grep then you can also use sed like this:

sed -nE 's/.*, ([0-9]+)%.*/\1/p' file

18

Answer 2

Using awk :

 awk -F"," '{print $2+0}'

Using GNU sed :

sed -rn 's/.*\, *([0-9]+)\%\,.*/\1/p'

Answer 3

Could you please try following, written and tested in link https://ideone.com/nzSGKs

your_command | awk 'match($0,/Charging, [0-9]+%/){print substr($0,RSTART+10,RLENGTH-11)}'

Explanation: Adding detailed explanation for above only for explanation purposes.

your_command |                              ##Running OP command and passing its output to awk as standrd input here.
awk '                                       ##Starting awk program from here.
match($0,/Charging, [0-9]+%/){              ##Using match function to match regex Charging, [0-9]+% in line here.
  print substr($0,RSTART+10,RLENGTH-11)     ##Printing sub string and printing from 11th character from starting and leaving last 11 chars here in matched regex of current line.
}'

Answer 4

You can use sed :

$ acpi | sed -nE 's/.*Charging, ([[:digit:]]*)%.*/\1/p'
18

Or, if Charging is not always in the string, you can look for the , :

$ acpi | sed -nE 's/[^,]*, ([[:digit:]]*)%.*/\1/p'

Answer 5

Using bash :

s='Battery 0: Charging, 18%, 01:37:09 until charged'
res="${s#*, }"
res="${res%%%*}"
echo "$res"

Result: 18 .

res="${s#*, }" removes text from the beginning to the first comma+space and "${res%%%*}" removes all text from end till (and including) the last occurrence of % .