How to compare two dates in a shell?

Question

How can two dates be compared in a shell?

Here is an example of how I would like to use this, though it does not work as-is:

todate=2013-07-18
cond=2013-07-15

if [ $todate -ge $cond ];
then
    break
fi

How can I achieve the desired result?

What do you want to loop for? Do you mean conditional rather than loop? — Mat, Jul 25 '13 at 07:46
Why you tagged [tag:files]? Are those dates actually file times? — manatwork, Jul 25 '13 at 07:47
Check out this on stackoverflow: http://stackoverflow.com/questions/8116503/how-to-compare-two-datetime-strings-and-return-difference-in-hours-bash-shell — slm, Jul 25 '13 at 14:04

score 205 · Answer 1 · edited May 23 '19 at 16:49

205

The right answer is still missing:

todate=$(date -d 2013-07-18 +%s)
cond=$(date -d 2014-08-19 +%s)

if [ $todate -ge $cond ];
then
    break
fi

Note that this requires GNU date. The equivalent date syntax for BSD date (like found in OSX by default) is date -j -f "%F" 2014-08-19 +"%s"

edited May 23 '19 at 16:49

Community

1

answered Dec 02 '14 at 07:42

user93501

2,067

14

Donno why someone downvoted this, it's pretty accurate and doesn't deal with concatenating ugly strings. Convert your date to unix timestamp (in seconds), compare unix timestamps. – Nicholi May 02 '15 at 23:16
I think, me main advantage of this solution is, that you can do additions or subtractions easily. For example, I wanted to have a timeout of x seconds, but my first idea (timeout=$(\date +%Y%m%d%H%M%S` + 500)`) is obviously wrong. – iGEL Jul 20 '18 at 09:59
2

You can include nanoseconds precision with date -d 2013-07-18 +%s%N – daparic May 31 '19 at 09:43
5

It's probably better to use (( ... > ... )) than [ ... -gt ... ] or similar. [[ "$x" -lt "$y" ]] returns true when $x is empty, in fact [[ "" -eq 0 ]] is TRUE. If something goes wrong populating $x and it winds up empty, I'd want "" < 1 to return false. (( "$x" < "$y" )) properly errors when $x is empty without needing extra checks. – user1169420 Mar 11 '21 at 00:39
@Nicholi it's downvoted because it doesn't work, I get an error saying command not found. ¯\(ツ)/¯ – Nathan McKaskle Feb 14 '23 at 15:26
@NathanMcKaskle Do you mean command date? Which platform you are using? – jarno Feb 16 '23 at 20:55
Nevermind it was the fact that it needs a space between the variables and the brackets on either side, so rigid. – Nathan McKaskle Feb 17 '23 at 19:58

score 61 · Answer 2 · edited Nov 03 '22 at 06:28

61

Can use a ksh-style string comparison to compare the chronological ordering [of strings in a year, month, day format].

date_a=2013-07-18
date_b=2013-07-15
if [[ "$date_a" > "$date_b" ]] ;
then
    echo "break"
fi

Thankfully, when [strings that use the YYYY-MM-DD format] are sorted* in alphabetical order, they are also sorted* in chronological order.

(* - sorted or compared)

Nothing fancy needed in this case. yay!

edited Nov 03 '22 at 06:28

Stéphane Chazelas

544,893

answered Jul 21 '14 at 23:16

dajon

801
6
6

Where is the else branch here? – Vladimir Despotovic Dec 15 '16 at 14:12
This is the only solution that worked for me on Sierra MacOS – Vladimir Despotovic Dec 15 '16 at 14:22
This is simpler that my solution of if [ $(sed -e s/-//g <<< $todate) -ge $(sed -e s/-//g <<< $cond) ]; … This date format was designed to be sorted using standard alpha numeric sort, or for - to be stripped and sorted numerically. – ctrl-alt-delor Apr 08 '17 at 17:59
2

This is far and away the smartest answer on here – Ed Randall Feb 23 '19 at 11:34
2

Use this answer, it is far more robust as it does not rely on an external command. – BitByteDog Jun 13 '20 at 22:20
1

That will not work in POSIX shell: – sena May 18 '22 at 12:47
1

@sena You're right it won't. So just use [ ] instead of [[ ]]. – Tripp Kinetics Jul 13 '22 at 17:10
@TrippKinetics No you can't, bash treats as a redirection. [ "0" > "1" ] && echo ok gives you ok, and creates the file 1. – elboulangero Feb 10 '23 at 07:41
@elboulangero The correct syntax for [ ] would be -gt. Of course, this requires the dates to be numbers, so no dashes. – Tripp Kinetics Feb 13 '23 at 16:32
1

@TrippKinetics I just realized that you can also quote the >, so [ 2023-02-15 \> 2023-01-01 ] && echo ok works. – elboulangero Feb 15 '23 at 04:20

score 40 · Answer 3 · edited Apr 24 '15 at 09:49

40

You are missing the date format for the comparison:

#!/bin/bash

todate=$(date -d 2013-07-18 +"%Y%m%d")  # = 20130718
cond=$(date -d 2013-07-15 +"%Y%m%d")    # = 20130715

if [ $todate -ge $cond ]; #put the loop where you need it
then
 echo 'yes';
fi

You are missing looping structures too, how are you planning to get more dates?

edited Apr 24 '15 at 09:49

Niels Basjes

113

answered Jul 25 '13 at 08:35

LPoblet

509

This doesn't work with the date shipped with macOS. – Flimm Dec 08 '16 at 09:49
1

date -d is non standard and does not work on a typical UNIX. On BSD it even attempts to set the kenel value DST. – schily Dec 07 '18 at 10:13
1

The above works if you use gdate on macos, it's available in brew. – slm Feb 11 '20 at 07:37

sergut · Answer 4 · 2018-09-27T07:47:23.743

This is not a problem of looping structures but of data types.

Those dates (todate and cond) are strings, not numbers, so you cannot use the "-ge" operator of test. (Remember that square bracket notation is equivalent to the command test.)

What you can do is use a different notation for your dates so that they are integers. For example:

date +%Y%m%d

will produce an integer like 20130715 for July 15th, 2013. Then you can compare your dates with "-ge" and equivalent operators.

Update: if your dates are given (e.g. you are reading them from a file in 2013-07-13 format) then you can preprocess them easily with tr.

$ echo "2013-07-15" | tr -d "-"
20130715

score 8 · Answer 5 · edited Apr 24 '15 at 10:01

The operator -ge only works with integers, which your dates aren't.

If your script is a bash or ksh or zsh script, you can use the < operator instead. This operator is not available in dash or other shells that don't go much beyond the POSIX standard.

if [[ $cond < $todate ]]; then break; fi

In any shell, you can convert the strings to numbers while respecting the order of dates simply by removing the dashes.

if [ "$(echo "$todate" | tr -d -)" -ge "$(echo "$cond" | tr -d -)" ]; then break; fi

Alternatively, you can go traditional and use the expr utility.

if expr "$todate" ">=" "$cond" > /dev/null; then break; fi

As invoking subprocesses in a loop can be slow, you may prefer to do the transformation using shell string processing constructs.

todate_num=${todate%%-*}${todate#*-}; todate_num=${todate_num%%-*}${todate_num#*-}
cond_num=${cond%%-*}${cond#*-}; cond_num=${cond_num%%-*}${cond_num#*-}
if [ "$todate_num" -ge "$cond_num" ]; then break; fi

Of course, if you can retrieve the dates without the hyphens in the first place, you'll be able to compare them with -ge.

Where is the else branch in this bash? – Vladimir Despotovic Dec 15 '16 at 14:12 — Vladimir Despotovic, Dec 15 '16 at 14:12
This seems to be the most correct answer. Very helpful! – Mojtaba Rezaeian Oct 21 '18 at 00:50 — Mojtaba Rezaeian, Oct 21 '18 at 00:50

slm · Answer 6 · 2018-10-08T23:51:38.470

2

There is also this method from the article titled: Simple date and time calulation in BASH from unix.com.

These functions are an excerpt from a script in that thread!

date2stamp () {
    date --utc --date "$1" +%s
}

dateDiff (){
    case $1 in
        -s)   sec=1;      shift;;
        -m)   sec=60;     shift;;
        -h)   sec=3600;   shift;;
        -d)   sec=86400;  shift;;
        *)    sec=86400;;
    esac
    dte1=$(date2stamp $1)
    dte2=$(date2stamp $2)
    diffSec=$((dte2-dte1))
    if ((diffSec < 0)); then abs=-1; else abs=1; fi
    echo $((diffSec/sec*abs))
}

Usage

# calculate the number of days between 2 dates
    # -s in sec. | -m in min. | -h in hours  | -d in days (default)
    dateDiff -s "2006-10-01" "2006-10-31"
    dateDiff -m "2006-10-01" "2006-10-31"
    dateDiff -h "2006-10-01" "2006-10-31"
    dateDiff -d "2006-10-01" "2006-10-31"
    dateDiff  "2006-10-01" "2006-10-31"

edited Oct 08 '18 at 23:51

answered Jul 25 '13 at 14:11

slm

369,824

2

This doesn't work with the date that ships with macOS. – Flimm Dec 08 '16 at 09:51
If you install gdate on MacOS via brew this can be easily modified to work by changing date to gdate. – slm Oct 08 '18 at 20:40
1

This answer was very helpful on my case. It should rate up! – Mojtaba Rezaeian Oct 21 '18 at 00:40

score 1 · Answer 7 · answered Feb 03 '15 at 13:12

1

I convert the Strings into unix-timestamps (seconds since 1.1.1970 0:0:0). These can compared easily

unix_todate=$(date -d "${todate}" "+%s")
unix_cond=$(date -d "${cond}" "+%s")
if [ ${unix_todate} -ge ${unix_cond} ]; then
   echo "over condition"
fi

answered Feb 03 '15 at 13:12

InuSasha

11

1

This doesn't work with the date that ships with macOS. – Flimm Dec 08 '16 at 09:50
Seems to be the same as http://unix.stackexchange.com/a/170982/100397 which was posted some time before yours – Chris Davies Apr 08 '17 at 16:02

F. Hauri - Give Up GitHub · Answer 8 · 2019-08-28T09:48:01.437

bash specific answer

As I like to reduce forks and bash do permit a lot of tricks, there is my purpose:

todate=2013-07-18
cond=2013-07-15

Well, now:

{ read todate; read cond ;} < <(date -f - +%s <<<"$todate"$'\n'"$cond")

This will re-populate both variables $todate and $cond, using only one fork, with ouptut of date -f - wich take stdio for reading one date by line.

Finally, you could break your loop with

((todate>=cond))&&break

Or as a function:

myfunc() {
    local todate cond
    { read todate
      read cond
    } < <(
      date -f - +%s <<<"$1"$'\n'"$2"
    )
    ((todate>=cond))&&return
    printf "%(%a %d %b %Y)T older than %(%a %d %b %Y)T...\n" $todate $cond
}

Using bash's builtin printf wich could render date time with seconds from epoch (see man bash;-)

This script only use one fork.

Alternative with limited forks and date reader function

This will create a dedicated subprocess (only one fork):

mkfifo /tmp/fifo
exec 99> >(exec stdbuf -i 0 -o 0 date -f - +%s >/tmp/fifo 2>&1)
exec 98</tmp/fifo
rm /tmp/fifo

As input and output are open, fifo entry could be deleted.

The function:

myDate() {
    local var="${@:$#}"
    shift
    echo >&99 "${@:1:$#-1}"
    read -t .01 -u 98 $var
}

Nota In order to prevent useless forks like todate=$(myDate 2013-07-18), the variable is to be set by the function himself. And to permit free syntax (with or without quotes to datestring), the variable name must be the last argument.

Then date comparission:

myDate 2013-07-18        todate
myDate Mon Jul 15 2013   cond
(( todate >= cond )) && {
    printf "To: %(%c)T > Cond: %(%c)T\n" $todate $cond
    break
}

may render:

To: Thu Jul 18 00:00:00 2013 > Cond: Mon Jul 15 00:00:00 2013
bash: break: only meaningful in a `for', `while', or `until' loop

if outside of a loop.

Or use shell-connector bash function:

wget https://github.com/F-Hauri/Connector-bash/raw/master/shell_connector.bash

or

wget https://f-hauri.ch/vrac/shell_connector.sh

(Wich are not exactly same: .sh do contain full test script if not sourced)

source shell_connector.sh
newConnector /bin/date '-f - +%s' @0 0

myDate 2013-07-18        todate
myDate "Mon Jul 15 2013"   cond
(( todate >= cond )) && {
    printf "To: %(%c)T > Cond: %(%c)T\n" $todate $cond
    break
}

Profiling bash contain a good demonstration for how using date -f - could reduce ressources requirement. — F. Hauri - Give Up GitHub, Feb 13 '17 at 00:02

score 0 · Answer 9 · edited Mar 10 '15 at 06:55

0

You can also use mysql's builtin function to compare the dates. It gives the result in 'days'.

from_date="2015-01-02"
date_today="2015-03-10"
diff=$(mysql -u${DBUSER} -p${DBPASSWORD} -N -e"SELECT DATEDIFF('${date_today}','${from_date}');")

Just insert the values of $DBUSER and $DBPASSWORD variables according to yours.

edited Mar 10 '15 at 06:55

chaos

48,171

answered Mar 10 '15 at 06:42

Abid Khan

1

score 0 · Answer 10 · answered Aug 02 '17 at 15:03

FWIW: on Mac, it seems feeding just a date like "2017-05-05" into date will append the current time to the date and you will get different value each time you convert it to epoch time. to get a more consistent epoch time for fixed dates, include dummy values for hours, minutes, and seconds:

date -j -f "%Y-%m-%d %H:%M:%S" "2017-05-05 00:00:00" +"%s"

This may be a oddity limited to the version of date that shipped with macOS.... tested on macOS 10.12.6.

score 0 · Answer 11 · answered Jun 20 '19 at 16:09

Echoing @Gilles answer ("The operator -ge only works with integers"), here is an example.

curr_date=$(date +'%Y-%m-%d')
echo "$curr_date"
2019-06-20

old_date="2019-06-19"
echo "$old_date"
2019-06-19

datetime integer conversions to epoch, per @mike-q's answer at https://stackoverflow.com/questions/10990949/convert-date-time-string-to-epoch-in-bash :

curr_date_int=$(date -d "${curr_date}" +"%s")
echo "$curr_date_int"
1561014000

old_date_int=$(date -d "${old_date}" +"%s")
echo "$old_date_int"
1560927600

"$curr_date" is greater than (more recent than) "$old_date", but this expression incorrectly evaluates as False:

if [[ "$curr_date" -ge "$old_date" ]]; then echo 'foo'; fi

... shown explicitly, here:

if [[ "$curr_date" -ge "$old_date" ]]; then echo 'foo'; else echo 'bar'; fi
bar

Integer comparisons:

if [[ "$curr_date_int" -ge "$old_date_int" ]]; then echo 'foo'; fi
foo

if [[ "$curr_date_int" -gt "$old_date_int" ]]; then echo 'foo'; fi
foo

if [[ "$curr_date_int" -lt "$old_date_int" ]]; then echo 'foo'; fi

if [[ "$curr_date_int" -lt "$old_date_int" ]]; then echo 'foo'; else echo 'bar'; fi
bar

score 0 · Answer 12 · answered Sep 11 '19 at 10:35

The reason why this comparison doesn't work well with a hyphenated date string is the shell assumes a number with a leading 0 is an octal, in this case the month "07". There have been various solutions proposed but the quickest and easiest is to strip out the hyphens. Bash has a string substitution feature that makes this quick and easy then the comparison can be performed as an arithmetic expression:

todate=2013-07-18
cond=2013-07-15

if (( ${todate//-/} > ${cond//-/} ));
then
    echo larger
fi

score 0 · Answer 13 · answered Dec 09 '20 at 20:03

It seems that all or most of the answers so far either assume you get to specify date format (which the question does not explicitly specify one way or the other), or assume that you have a feature rich version of date command or something.

Sometimes though, you are not in control of the date's format (e.g., the expiration date of an SSL certificate where date is given by a month name abbreviation) and you don't have access to a fancier date command. So while this solution is not completely general, it does run on a vanilla bash shell on Linux, Solaris, and FreeBSD, and handles month names (borrows significantly from a few smart answers from https://stackoverflow.com/questions/15252383/unix-convert-month-name-to-number):

#!/usr/bin/bash
function monthnumber {
    month=$(echo ${1:0:3} | tr '[a-z]' '[A-Z]')
    MONTHS="JANFEBMARAPRMAYJUNJULAUGSEPOCTNOVDEC"
    tmp=${MONTHS%%$month*}
    month=${#tmp}
    monthnumber=$((month/3+1))
    printf "%02d\n" $monthnumber
}
Or at the expense of some flexibility and brevity, you get more readability:
function monthnumber2 {
    case $(echo ${1:0:3} | tr '[a-z]' '[A-Z]') in
        JAN) monthnumber="01" ;;
        FEB) monthnumber="02" ;;
        MAR) monthnumber="03" ;;
        APR) monthnumber="04" ;;
        MAY) monthnumber="05" ;;
        JUN) monthnumber="06" ;;
        JUL) monthnumber="07" ;;
        AUG) monthnumber="08" ;;
        SEP) monthnumber="09" ;;
        OCT) monthnumber="10" ;;
        NOV) monthnumber="11" ;;
        DEC) monthnumber="12" ;;
    esac
    printf "%02d\n" $monthnumber
}
TODAY=$( date "+%Y%m%d" )
echo "GET /" | openssl s_client -connect github.com:443 | sed -ne '/-BEGIN CERTIFICATE-/,/-END CERTIFICATE-/p' > tmp.pem
cert_expiry_date=$(openssl x509 -in tmp.pem -noout -enddate | cut -d'=' -f2)
month_name=$(echo $cert_expiry_date | cut -d' ' -f1)
month_number=$( monthnumber $month_name )
cert_expiration_datestamp=$( echo $cert_expiry_date | awk "{printf &quot;%d%02d%02d&quot;,$4,&quot;${month_number}&quot;,$2}" )
echo "compare: [ $cert_expiration_datestamp -gt $TODAY ]"
if [ $cert_expiration_datestamp -gt $TODAY ] ; then
    echo "all ok, the cert expiration date is in the future."
else
    echo "WARNING: cert expiration date is in the past."
fi

score 0 · Answer 14 · answered Nov 03 '22 at 02:18

In case you want to compare the dates of two files, you can use this:

if [ file1 -nt file2 ]
then
  echo "file1 is newer than file2"
fi
# or +ot for "older than"

See https://superuser.com/questions/188240/how-to-verify-that-file2-is-newer-than-file1-in-bash

score 0 · Answer 15 · answered Jul 25 '13 at 07:52

dates are strings, not integers; you can't compare them with standard arithmetic operators.

one approach you can use, if the separator is guaranteed to be -:

IFS=-
read -ra todate <<<"$todate"
read -ra cond <<<"$cond"
for ((idx=0;idx<=numfields;idx++)); do
  (( todate[idx] > cond[idx] )) && break
done
unset IFS

This works as far back as bash 2.05b.0(1)-release.

How to compare two dates in a shell?

15 Answers15

Usage

bash specific answer

Alternative with limited forks and date reader function

Or use shell-connector bash function:

Or at the expense of some flexibility and brevity, you get more readability:

Linked

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