在bash脚本中基于名称作为日期删除文件夹（以mm / dd / yy为单位）

Question

I want to delete all directories apart from the current date's directory. directory names are taken as date in this MM-DD-YY format so directory name is

10-12-17

10-11-17

10-10-17 ..etc

    #!/bin/bash
    echo Hello World!

    one_day=$(date -d "1 days ago" +%m%d%y)
    for f in [0-9][0-9]-[0-9][0-9]-[0-9][0-9]; do
    [ -d "$f" ] || continue
    (( $f < $one_day )) &&  sudo rm -rf "$f"
    done

While running my script I am getting the following error:

./script.sh: line 9: ((: 10-08: value too great for base (error token is "08")

./script.sh: line 9: ((: 10-09: value too great for base (error token is "09")

Answer 1

You seem to be trying to check an inequality between "11-10-17" and "111017" (one is an int and the other is a string). Bash will let you check for a string inequality using !=

#!/bin/bash

one_day=$(date -d "1 days ago" +%m-%d-%y)
for f in [0-9][0-9]-[0-9][0-9]-[0-9][0-9]; do
  [ -d "$f" ] || continue
  [ "$f" != "$one_day" ] && sudo rm -rf "$f" && echo "$f" && continue
  [ "$f" == "$one_day" ] && echo "Leaving $f"
done