bash - Use sed to replace regex with a variable

Question

I'm trying to create a bash script where I can replace a date in a filename with the current date, however, I'm not being able to do so.

Here's what I have so far:

#!/bin/bash

my_file="FILENAME_20170410235908_GTT_DEV_20170410235400_20170410235408_XX_YY.nc"

my_date=`date "+%Y%m%d"`

echo "$my_file" | sed  's/([0-9]{12})/"${my_date}"/g'

I'm currently getting this:

FILENAME_"${my_date}"08_GTT_DEV_"${my_date}"00_"${my_date}"08_XX_YY.nc

Howerver, this is what I'd like to have:

FILENAME_2019070135908_GTT_DEV_20190701235400_20190701235408_XX_YY.nc

How can I achieve this?

Answer 1

You can try

sed "s/([0-9]+)/${my_date}/g"

single quote will not replace variable data. my_date will have only date. If you want the timestamp also, add it from the date command.