使用 shell 脚本在星期日创建的 Listout 文件

Question

**I have files in my directory. Files name is as below.

1. Backup-2018-07-29.tar.gz
2. Backup-2018-07-30.tar.gz
3. Backup-2018-08-05.tar.gz
4. Backup-2018-08-06.tar.gz
5. Connectly-2018-08-05-05.tar.gz
6. meraBoxDump-2018-08-05.tar.gz

Now I need to listout files created on sunday. Expected out put :

1. Backup-2018-08-05.tar.gz

   2.Connectly-2018-08-05-05.tar.gz

** How can I do this task in Shell Script

Answer 1

May be done using bash regex and GNU date

for file in *; do
    if [[ $file =~ .*([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]).* ]] &&
      [[ Sunday = `LANG=C /usr/bin/date -d${BASH_REMATCH[1]} +%A` ]]; then
        echo "$file"
    fi
done