[SOLVED] how do I extract a date stamp from a file name stored in a bash variable

Issue

I have a bash variable $filename that looks like this

[rmerritt@server]$ echo $filename
/opt/project/xxxx/async_transaction__2023-05-28.csv

what I am hoping to do is extract that date stamp 2023-05-28 and store it in another bash variable $dateis that allows me to do this

 cp /opt/project/xxxx/*`echo $dateis`*.csv|/opt/project/jdev/data/landing/landing_xxxx/

I.E. copy all csv files with that datestamp to another directory any idea how I could do this in a oneliner ? I have tried using awk

cp `ls  -Art   /opt/project/xxxx/*__*.csv` /opt/project/jdev/data/landing/landing_xxxx/

Solution

With GNU bash and its builtin parameter expansion:

filename="/opt/project/xxxx/async_transaction__2023-05-28.csv"
date="${filename##*_}"
date="${date%.*}"
echo "$date"

Output:

2023-05-28

Answered By - Cyrus

Answer Checked By - Marilyn (WPSolving Volunteer)