Bash validate date

Question

I'm writing a shell script and am confused as to why my date validation code is not working. I tried the following solutions to similar questions I found, but is_valid is always set to 1:

date "+%m/%d/%Y" -d "$1" 2>1 > /dev/null //or.. date -d "2012-02-29" > /dev/null 2>&1 is_valid=$? #always set to 1, even when given a valid date

How do I correctly validate the date format? The date should only be valid if in the format MM/DD/YYYY

I also tried this solution: Linux Bash - Date Format but it always rejected the date as well.

I'm not experiencing the same result. When I run date -d "2012-02-29 > /dev/null 2>&1" then "echo $?" I get 0`. — lurker
– lurker, Commented Jan 19, 2014 at 19:45
What does date --version print for you? (BTW, I get the same result as mbratch and Aleksey Izmailov.) — ruakh
– ruakh, Commented Jan 19, 2014 at 19:54
It prints: "date: illegal option -- - \n usage: date ....." I'm using Mac OSX btw — Cbas
– Cbas, Commented Jan 19, 2014 at 20:06
As you've now found out, Mac OS X is not Linux, and the GNU extensions don't necessarily work on non-Linux machines. The difficulty, sometimes, is determining when something is an extension. — Jonathan Leffler
– Jonathan Leffler, Commented Jan 19, 2014 at 21:21
BTW, MM/DD/YYYY is a very poorly chosen date format -- it's quite ambiguous with DD/MM/YYYY. Using YYYY-MM-DD sorts properly and is RFC-compliant. — Charles Duffy
– Charles Duffy, Commented Mar 3, 2017 at 16:58

chepner · Accepted Answer · 2015-10-14 17:13:54Z

The BSD date that ships with Mac OS X doesn't support the -d option (or rather, it uses -d for something entirely different). Either install GNU date, or use the following to validate your input string:

date -f "%Y-%m-%d" -j "2012-02-29" >/dev/null 2>&1

The -f provides the input format, and the -j tells date to simply output the date, not attempt to set the system clock.

yǝsʞǝla · Accepted Answer · 2014-01-19 20:24:48Z

3

I came up with this little function:

function isDateInvalid() { date -d "$1" "+%m/%d/%Y" > /dev/null 2>&1 res=$? echo "$res" } isDateInvalid "2012-02-219" 1 isDateInvalid "2012-02-29" 0

edited Jan 19, 2014 at 20:24

answered Jan 19, 2014 at 19:45

yǝsʞǝla

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1 Comment

Cbas Over a year ago

there is nothing in between those lines. I actually an echo $? command right after, and it always prints 1

fedorqui · Accepted Answer · 2014-11-17 12:28:44Z

for y in {2013..2014}; do for m in {01..12}; do for d in {01..31}; do [[ ! "`date -jf %Y%m%d $y$m$d +%Y%m%d`" = "$y$m$d" ]] && continue echo $y.$m.$d done done done

if strings match, loop will proceed ...

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