Date extraction from file name in bash

I am having a directory which contains multiple files with name in pattern abc12gddmmyyyy.zip.
I am working in bash script and I want to extract the pattern ddmmyyyy from the file name. Could anyone please help. I am new to linux enviornment. Thanks

Sample code

for filename in data/*; do
    date=<This part needed>
    echo $date
done

There are, of course, many solutions. One way could be to use sed:

date="$(echo "$filename" | sed -r 's/^.*12g([0-9]{8})\.zip//')"

With bash and a regex:

[[ "$filename" =~ (.{8})\.zip$ ]] && date="${BASH_REMATCH[1]}"

files:

ls ./files/
abc12g01022020.zip  abc12g02022020.zip

script:

for file in ./files/*
do
  echo $file | awk -F'.zip' '{print $1}' | tail -c 9
done

output:

01022020
02022020

Though to be honest it's a bit of a one-use-case scenario you've painted for us.

Here is another pure bash option:

for filename in data/*.zip; do
    date="${filename: -12:8}"
    echo "$date"
done

This method assumes that all your files have exactly the pattern as you describe: <random_string>DDMMYYYY.zip. Since the date is starting at the 12th character from the back, we can extract it as a substring.

> ${parameter:offset:length} Substring Expansion. Expands to up to length characters of
parameter starting at the character specified by offset. <snip> If offset evaluates to a number less than zero, the value is used
as an offset from the end of the value of parameter. <snip>
A negative offset is taken relative to one greater than the max‐
imum index of the specified array. <snip> Note that a
negative offset must be separated from the colon by at least one
space to avoid being confused with the :- expansion.

> _{source: man bash}

note: be aware that a date-format of the type DDMMYYYY is not sortable and generally will lead one day to a lot of annoyance. It is always useful to use sortable formats like YYYYMMDD