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Comparing 2 dates in bash – Debian

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.

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2

Answers


  1. 0

    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
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  2. 0

    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
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