So I'm trying to write a simple script in bash that asks user for input date in following format (YYYY-dd-mm). Unfortunately I got stuck on first step, which is verifying that input is in correct format. I tried using 'date' with no luck (as it returns actual current date). I'm trying to make this as simple as possible. Thank you for your help!
2 Answers
Using regex:
if [[ $date =~ ^[0-9]{4}-[0-3][0-9]-[0-1][0-9]$ ]]; then or with bash globs:
if [[ $date == [0-9][0-9][0-9][0-9]-[0-3][0-9]-[0-1][0-9] ]]; then Please note that this regex will accept a date like 9999-00-19 which is not a correct date. So after you check its possible correctness with this regex you should verify that the numbers are correct.
IFS='-' read -r year day month <<< "$date" This will put the numbers into $year $day and $month variables.
answered Sep 11, 2013 at 18:46
Aleks-Daniel Jakimenko-A.
10.8k3 gold badges43 silver badges41 bronze badges
1 Comment
lukabix22
Perfect, it does exactly what it is supposed to! Thank you!
date -d "$date" +%Y-%m-%d The latter is the format, the -d allows an input date. If it's wrong it will return an error that can be piped to the bit bucket, if it's correct it will return the date.
The format modifiers can be found in the manpage of date man 1 date. Here an example with an array of 3 dates:
dates=(2012-01-34 2014-01-01 2015-12-24) for Date in ${dates[@]} ; do if [ -z "$(date -d $Date 2>/dev/null)" ; then echo "Date $Date is invalid" else echo "Date $Date is valid" fi done Just a note of caution: typing man date into Google while at work can produce some NSFW results ;)
