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?

Possible duplicate of Replace a string in shell script using a variable — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 1, 2019 at 10:11

oguz ismail · Accepted Answer · 2019-07-01 10:08:17Z

4

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.

edited Jul 1, 2019 at 10:08

oguz ismail

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answered Jul 1, 2019 at 10:05

Prakash Krishna

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