Sorting dates in array with Bash

Question

I'm writing a bash script for some automation, and I can't seem to find the best way of sorting an array that contains dates.

An example array:

dates=('180212', '110112', '040312')

The output should be as follows:

110112
180212
040312

I've tried to use sort but it just sorts it as if it were a integer.

Could anyone advise on the best method to use? It doesn't necessarily have to sort either, I just need the oldest date from the array.

Thanks!

user570500 · Accepted Answer · 2013-01-30 12:08:44Z

1

You could swap the year and the day digits, and use sort to sort the resulting values. Then take the oldest value and undo the swap. You can use sed like this to do the swapping:

echo 112233 | sed 's/\(..\)\(..\)\(..\)/\3\2\1/'

This swaps the first two with the last two digits.

answered Jan 30, 2013 at 12:08

user570500

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Patrick B. · Accepted Answer · 2013-01-30 12:15:32Z

1

Example bash-script:

dates="180212 110112 040312"
min="999999"
for d in $dates
do
    rdate="${d:4}${d:2:2}${d:0:2}"
    if [ "$min" -gt "$rdate" ]
    then
        min=$rdate
    fi
done

min="${min:4}${min:2:2}${min:0:2}"
echo $min

Works only with dates which are ddmmyy. The idea is to reverse the date in the format yymmdd and then find the minimum. Does the job in bash. Not beautiful though.

answered Jan 30, 2013 at 12:15

Patrick B.

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