Comparing 2 dates in bash

Question

So I am trying to compare 2 dates in bash , but my dates contains characters like Jan or Monday . Can I compare them directly or do i need to format them like this "20201607"(and how do i do this plz) and them numerically compare them? Thank you For example:

today=$(date)
day='21 Jan 2021'

if [ $today < $day ]
   echo "$day"

basically my function will just return the dates who does not happened atm.

Is the input date complete free-form? Is it something you get from the user with no control over the date formatting? — glenn jackman, Jan 26 '21 at 14:31
What date format is `20201607` -- when I look at that I think "there is no 16th month" — glenn jackman, Jan 26 '21 at 14:33
Actually i get it from --calendar option on zenity . it is then send to a text file . i suppose it is a string at this time. — Justin Mayer, Jan 26 '21 at 15:10

score 3 · Answer 1 · edited Jan 26 '21 at 14:31

3

Use the date to seconds since the epoch (1970-01-01 UTC) through specifying %s with date and so:

day='21 Jan 2021'
if [[ "$(date +%s)" -lt "$(date -d "$day" +%s)" ]];
then 
    echo "$day";
fi

edited Jan 26 '21 at 14:31

glenn jackman

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answered Jan 26 '21 at 13:39

Raman Sailopal

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score 2 · Accepted Answer · answered Jan 26 '21 at 15:23

2

Zenity can help you: Tell zenity what output date format you want:

while ! date=$(zenity --calendar --date-format '%Y%m%d'); do
    # ...........................^^^^^^^^^^^^^^^^^^^^^^
    echo "Don't cancel the zenity window, select a day"
done

# bash v4.2+ can to date formatting
today=$(printf '%(%Y%m%d)T' -1)

if [ "$date" -gt "$today" ]; then
    echo "you entered a future date"
fi

answered Jan 26 '21 at 15:23

glenn jackman

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Thank you , to fix it , i transformed the date in a integer format with year month and day and then compare it numerically. – Justin Mayer Mar 06 '21 at 06:00

Comparing 2 dates in bash

2 Answers2