check if argument is a valid date in bash shell - linux

I am writing a bash shell script in Linux, this program will accept a date (mm-dd-yyyy) as a parameter. I am wondering if there is a simply way to check if the date is valid? is there an operator and I can just use test to check?

You can check with date -d "datestring"
So date -d "12/31/2012" is valid, but using hyphens, e.g. date -d "12-31-2012", is not valid for date.
You can also use words: date -d 'yesterday' or date -d '1 week ago' are both valid.

You can extract the day, month, and year values from the input date value MM-DD-YYYY and validate it as the unambiguous (ISO) format YYYY-MM-DD instead (you can validate a DD-MM-YYY formatted date as "correct" using date, e.g. 25-12-2010, but it is not a valid MM-DD-YYY date, hence the need to change the date format first)
A valid date in the correct format is OK
30th November 2005 is valid:
$ DATE=11-30-2005; d=${DATE:3:2}; m=${DATE:0:2}; Y=${DATE:6:4}; echo "year=$Y, month=$m, day=$d"; if date -d "$Y-$m-$d" &> /dev/null; then echo VALID; else echo INVALID; fi
year=2005, month=11, day=30
VALID
$ DATE=11-30-2005; if date -d "${DATE:6:4}-${DATE:0:2}-${DATE:3:2}" &> /dev/null; then echo VALID; else echo INVALID; fi
VALID
An invalid date in the correct format is NOT OK
31st November 2005 does not validate:
$ DATE=11-31-2005; d=${DATE:3:2}; m=${DATE:0:2}; Y=${DATE:6:4}; echo "year=$Y, month=$m, day=$d"; if date -d "$Y-$m-$d" &> /dev/null; then echo VALID; else echo INVALID; fi
year=2005, month=11, day=31
INVALID
$ DATE=11-31-2005; if date -d "${DATE:6:4}-${DATE:0:2}-${DATE:3:2}" &> /dev/null; then echo VALID; else echo INVALID; fi
INVALID
A valid date in the incorrect format is NOT OK
20th April 1979 in DD-MM-YYYY format does not validate as a MM-DD-YYYY date:
$ DATE=20-04-1979; d=${DATE:3:2}; m=${DATE:0:2}; Y=${DATE:6:4}; echo "year=$Y, month=$m, day=$d"; if date -d "$Y-$m-$d" &> /dev/null; then echo VALID; else echo INVALID; fi
year=1979, month=20, day=04
INVALID
$ DATE=20-04-1979; if date -d "${DATE:6:4}-${DATE:0:2}-${DATE:3:2}" &> /dev/null; then echo VALID; else echo INVALID; fi
INVALID
Alternate simpler method: use BASH variable string replace hyphens to slashes
$ DATE="04-30-2005"; [[ $(date -d "${DATE//-/\/}" 2> /dev/null) ]] && echo VALID || echo INVALID
VALID
$ DATE="04-31-2005"; [[ $(date -d "${DATE//-/\/}" 2> /dev/null) ]] && echo VALID || echo INVALID
INVALID

For script use, I kept it as simple as I could. Testing the date value with the date function then checking the exit code of the process.
date -d "02/01/2000" 2>: 1>:; echo $?
This will redirect the standard in and standard error to null : and using echo to return the exit code with $? allows me to check for 0=good date and 1=bad date.

The following worked well for me. Many thanks to my co-worker, Tyler Chamberlain, for the OSX solution.
# Validate a given date/time in Bash on either Linux or Mac (OSX).
# Expected date/time format (in quotes from the command line): YYYY-MM-DD HH:MM:SS
# Example(s): ./this_script "2012-02-29 13:00:00" # IS valid
# ./this_script "2013-02-29 13:00:00" # Is NOT valid
START_DATETIME=$1
function report_error_and_exit
{
local MSG=$1
echo "$MSG" >&2
exit 1
}
# We can use OSTYPE to determine what OS we're running on.
# From http://stackoverflow.com/questions/394230/detect-the-os-from-a-bash-script
# Determine whether the given START_DATETIME is valid.
if [[ "$OSTYPE" == "linux-gnu" ]]
then
# Validate the date on a Linux machine (Redhat or Debian). On Linux, this is
# as easy as adding one minute and checking the return code. If one minute
# cannot be added, then the starting value is not a valid date/time.
date -d "$START_DATETIME UTC + 1 min" +"%F %T" &> /dev/null
test $? -eq 0 || report_error_and_exit "'$START_DATETIME' is not a valid date/time value. $OSTYPE"
elif [[ "$OSTYPE" == "darwin"* ]]
then
# Validate the date on a Mac (OSX). This is done by adding and subtracting
# one minute from the given date/time. If the resulting date/time string is identical
# to the given date/time string, then the given date/time is valid. If not, then the
# given date/time is invalid.
TEST_DATETIME=$(date -v+1M -v-1M -jf "%F %T" "$START_DATETIME" +"%F %T" 2> /dev/null)
if [[ "$TEST_DATETIME" != "$START_DATETIME" ]]
then
report_error_and_exit "'$START_DATETIME' is not a valid date/time value. $OSTYPE"
fi
fi
echo "The date/time is valid."
I tested this script on a Red Hat-based system, a Debian-based system and OSX, and it worked as expected on all three platforms. I did not have time to test on Windows (Cygwin).

For validation of YYYY-MM-DD (ISO 8601) dates on OSX in the BASH shell, the following approach validates both the format and the date.
isYYYYMMDDdate() {
[[ "$1" =~ ^[0-9]{4}-[0-9]{2}-[0-9]{2}$ ]] && [[ "$1" == $(date -r $(date -j -f "%Y-%m-%d" "$1" "+%s") '+%Y-%m-%d') ]] &> /dev/null; echo "$?"
}
It first uses a regular expression match to check the format.
Then, it converts the date to epoch time and then back to a date.
If the original and twice-converted dates match, then it is valid.
Test a valid date: 2005-11-30
$ isYYYYMMDDdate 2005-11-30
0
Test an invalid date: 2005-11-31
$ isYYYYMMDDdate 2005-11-31
1
Test a valid date formatted incorrectly: 1979-20-04
$ isYYYYMMDDdate 1979-20-04
1

The date command will parse a date given with the -d argument. If the date is invalid, an error message is printed to STDERR and date exits with an error status. If the date is valid, it prints the date on STDOUT and exits with a success status.
Because of this, date -d "$date" can be used directly in a bash if statement.
The first wrinkle is that to prevent printing a message for valid dates, you need to redirect STDOUT to /dev/null using >/dev/null.
The second wrinkle is that date accepts an empty string as a valid date without complaint. In most cases, that should mean that your user didn't enter a date when they should have. You will want to test for an empty date separately using the test [ "z$date" != "z" ]
date also accepts a variety of formats. If you are using actual bash (as opposed to dash or some of ther sh variety, you could use regular expressions against your preferred format in place of a simple check for an empty string. For example to check my preferred ISO format, I would use: [[ "$date" =~ ^[0-9]{4}-[0-9]{2}-[0-9]{2}$ ]]
date=2001-01-01
if [ "z$date" != "z" ] && date -d "$date" >/dev/null
then
echo "VALID DATE"
fi
If you try this with an invalid date (such as 2001-01-53), it doesn't get into the if and it prints out:
date: invalid date ‘2001-01-53’
Alternately, you could check if the date is invalid and exit:
date=2001-01-01
if [ "z$date" == "z" ]
then
echo "No date specified"
exit 1
fi
if ! [[ "$date" =~ ^[0-9]{4}-[0-9]{2}-[0-9]{2}$ ]]
then
echo "Expected date in YYYY-MM-DD format"
exit 1
fi
if ! date -d "$date" >/dev/null
then
exit 1
fi
echo "VALID DATE"

case statements make it easy to support multiple formats and capturing date-parts, i.e.
case ${date} in
[0-3][0-9]-[0-1][0-9]-[0-9][0-9][0-9][0-9] )
yr=...
mn=...
dy=...
;;
[0-1][0-9]-[0-3][0-9]-[0-9][0-9][0-9][0-9] )
yr=...
dy=...
mn=...
;;
.... other formats
;;
* )
echo "ERROR on date format, from value=$date, expected formats ..."
return 1
;;
esac
I hope this helps.

You can use the strptime() function available in Python's time or datetime modules or Perl's Time::Piece module.

Related

check if argument is a valid date time in shell script

I am writing a bash shell script in Linux, this program will accept a date 08-FEB-18 11.45.18.844 AM as a parameter.
I am wondering if there is a simply way to check if the date time is valid?
You can get a bit creative since you have bash and map the date string into an array which can then be easily parsed with date -d (and the help of another associative array). Once the date/time is mapped to array elements and converted to seconds since epoch with date -d, you simply check the return of the date command to determine if the conversion succeeded or failed. Handle the return appropriately:
#!/bin/bash
[ -n "$1" ] || { ## validate one argument given
printf "error: insufficient input\nusage: %s dd-mmm-yy hh.mm.ss.ms\n" \
"${0##*/}"
exit 1
}
oifs="$IFS" ## save original Internal Field Separator
IFS=$' \t\n-.'; ## set IFS to break on - or .
dt=( $(echo $1) ) ## separate date into indexed array
[ "${#dt[#]}" -lt '7' ] && { ## check all 7 components present
printf "error: date doesn't match dd-mmm-yy hh.mm.ss.ms format\n"
exit 1
}
IFS="$oifs" ## reset original IFS
## create associative array mapping months to numerics
declare -A mo=(
[JAN]=1
[FEB]=2
[MAR]=3
[APR]=4
[MAY]=5
[JUN]=6
[JUL]=7
[AUG]=8
[SEP]=9
[OCT]=10
[NOV]=11
[DEC]=12
)
## any date after 30 considerd 1930, else considered 2000
[ "${dt[2]}" -gt '30' ] && dt[2]=$((${dt[2]} + 1000)) || \
dt[2]=$((${dt[2]} + 2000))
## use date to convert array contents to seconds since epoch
epochsec=$( date -d "${dt[2]}-${mo[${dt[1]}]}-${dt[0]} \
${dt[3]}:${dt[4]}:${dt[5]}.${dt[6]}" +%s )
if [ "$?" -ne '0' ]; then ## check if last return was error
printf "error: invalid date.\n"
else ## output good date
printf "date: %s\n" "$(date -d #$epochsec)"
fi
Example Use/Output
$ bash chkcustomdt.sh "08-FEB-18 11.45.18.844"
date: Thu Feb 8 11:45:18 CST 2018
There are a lot of ways to approach this, this was just the first that came to mind.

Check date whether AM or PM in shell script and take action accordingly

I want to check whether its AM or PM in a shell script. I have tried this:
if ( `date "+%p"` -eq "AM" ); then echo "Yes"; else "NO"; fi
But it shows:
AM: command not found
Also tried = instead of -eq.
Two issues in your current statement:
1) the format specifier %p in GNU date utility may contain lowercase of either AM or PM - depending on your current locale. It's better to use %P specifier as it contains lowercase presentation
2) In bash, -eq - is integer comparison operator, not for string comparison
The right way would be:
if [ `date +%P` = "am" ]; then echo 'Yes'; else echo 'No'; fi
To deal with any locale you may compare the current hour value (given by %H) with midday hour 12:
if [ `date +%H` -lt 12 ]; then echo 'Yes'; else echo 'No'; fi
The locale can specify language- and/or country- specific strings to use in place of am and pm. For example:
$ LC_TIME=hu_HU date +%P
de
$ LC_TIME=fr_FR date +%P
$ LC_TIME=en_EN date +%P
pm
Instead, check the hour to see which half of the day would be used to provide a value for %P:
h=$(date +%H)
if (( h < 12 )); then
echo YES # am
else
echo NO # pm
fi
Or, force a known locale:
ampm=$(LC_TIME=C date +%P)
if [[ $ampm = am ]]; then
...
else
...
fi
here
# String comparision with =
# [ ] instead if ()
if [ `date "+%p"` = "AM" ]; then
echo "Yes";
else
echo "NO"; # Missed echo here
fi

Accept date as user input in bash script

I want to accept date as input in dd-mm-yyyy format as user input (via read) and then convert it as a date object in shell script. How can I achieve this.
I have had success with yyyy-mm-dd but somehow,it is not working for my use case.
This is my script:-
#!/bin/bash
read -p "Enter a date (dd-mm-yyyy): " user_date
if date=$(date -d "$user_date" +'+%d-%m-%Y'); then
# user date was ok
echo $date
This is my input:-
./2.sh
Enter a date (dd-mm-yyyy): 11-1-2011
Error:
date: invalid date `11-1-2011'
Thanks
To compare two strings (there are no date objects, just the string output of the date command), you need to use either [ or preferably a conditional statement [[ ... ]] .
if [[ $user_date = $(date -d "$user_date" +'+%d-%m-%Y') ]]; then
Of course, there's no guarantee that date can parse the string entered by the user as a date.

Bash validate date

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.
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.
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
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 ...

Check a user's input is correctly formatted

I'm trying to validate a user's input when they enter a date into my bash script.
The correct format should be YYYY-MM-DD HH:MM:SS
What I have so far is this, but it's not quite right.
read -p "-" datetime
if [[ $datetime ! "*[^2000-2100]-[^01-12]-[^10-31] [^00-24]:[^00-60]:[^00-60]*" ]]
then
echo "Not a date"
fi
You could say:
if ! date -d "$datetime" >/dev/null 2>&1; then
echo "Not a date"
fi
Using regexp isn't really recommended, but if you insist you could say:
if [[ "$datetime" != 2[0-1][0-9][0-9]-[0-1][0-9]-[0-3][0-9]\ [0-2][0-9]:[0-6][0-9]:[0-6][0-9] ]]; then
then
echo "Not a date"
fi

Resources