How to retrieve digits including the separator "." - bash

Question

I am using grep to get a string like this: ANS_LENGTH=266.50 then I use sed to only get the digits: 266.50
This is my full command: grep --text 'ANS_LENGTH=' log.txt | sed -e 's/[^[[:digit:]]]*//g'
The result is : 26650
How can this line be changed so the result still shows the separator: 266.50

Answer 1

You don't need grep if you are going to use sed. Just use sed' // to match the lines you need to print.
sed -n '/ANS_LENGTH/s/[^=]*=\(.*\)/\1/p' log.txt
-n will suppress printing of lines that do not match /ANS_LENGTH/
Using captured group we print the value next to = sign.
p flag at the end allows to print the lines that matches our //.
If your grep happens to support -P option then you can do:
grep -oP '(?<=ANS_LENGTH=).*' log.txt
(?<=...) is a look-behind construct that allows us to match the lines you need. This requires the -P option
-o allows us to print only the value part.

Answer 2

You need to match a literal dot as well as the digits.
Try sed -e 's/[^[[:digit:]\.]]*//g'
The dot will match any single character. Escaping it with the backslash will match only a literal dot.

Answer 3

Here is some awk example:
cat file:
some data ANS_LENGTH=266.50 other=22
not mye data=43
gnu awk (due to RS)
awk '/ANS_LENGTH/ {f=NR} f&&NR-1==f' RS="[ =]" file
266.50
awk '/ANS_LENGTH/ {getline;print}' RS="[ =]" file
266.50
Plain awk
awk -F"[ =]" '{for(i=1;i<=NF;i++) if ($i=="ANS_LENGTH") print $(i+1)}' file
266.50
awk '{for(i=1;i<=NF;i++) if ($i~"ANS_LENGTH") {split($i,a,"=");print a[2]}}' file
266.50