I am trying to compare a string converted to date and a file creation time in bash.
#!/bin/bash
test='2020-05-13 08:00'
testConverted=$(date -d "$test" +'%Y %m %d %H:%M')
[ "~/fileToCompare" -nt "$testConverted" ] && echo "yes"
This always returns false no matter what date I put for test. Is the conversion of the date wrong? Is this possible to do?
The way to do this is to convert the date string into Unix Epoch time (seconds since Jan 1, 1970) with the date command and then similarly get the Epoch time modification date of the test file using the stat command and compare using arithmetic evaluation
#!bin/bash
testDate='2020-05-13 08:00'
testFile="$HOME/fileToCompare"
if (( $(date -d "$test" +%s) > $(stat "$testFile" -c %Z) )); then
echo "testDate ($testDate) is newer than $testFile"
fi
Related
I'm trying to make a script that is working from 12/24/2020 through 06/01/2021. For now I olny have it working only for 2 dates with this:
if [ "$(date +'%m%d')" != "1224" ] && [ "$(date +'%m%d')" != "1226" ];
So this script should be working from 12/24 - 01/06.
Any suggestions? Thanks
You can use the +%s option in date to convert the start/finish dates to epoch time. You supply the dates using the -d option.
Then, you retrieve the current date using +%s again and compare using standard integer comparison, like this:
start=$(date +%s -d '12/24/2020')
finish=$(date +%s -d '06/02/2021')
now=$(date +%s)
if [ $now -ge $start ] && [ $now -lt $finish ]
then
echo "Do something"
else
echo "Skip"
fi
Edit: For the finish date, its necessary to use the day after your intended finishing date and compare using -lt. This is because date will return the timestamp for the start of the specified date. So doing this, you end up having the comparison succeed until midnight on 06/01/2021. Thanks to #gordon-davisson for pointing this out.
beg='20201224'
end='20210601'
now=$(date +'%Y%m%d')
if (( beg <= now )) && (( now <= end)); then
Using BASH I want to loop from a start to end date at ten-minute intervals.
I tried
begin_date="2015-01-01 00:00:00"
end_date="2015-02-20 00:00:00"
d=$begin_date
while [ "$d" != "$end_date" ]; do
echo $d
d=$(date -d "${d} + 10 min" +"%Y-%m-%d %H:%M")
done
But it didn't work. Looking at Bash:Looping thru dates
#This works
d=$(date -I -d "${d} + 1 day")
#This doesn't work
d=$(date -d "${d} + 1 day" +"%Y-%m-%d")
What am I missing in the format string?
The expression date -d "${d} + 10 min" seems not to produce a date with an offset of 10 minutes. In fact, when I run your code, I see a date counter going backwards. (Posting this diagnostic as part of your question would help others see where the problem is; you should not require others to run your code just to see what it does.)
Anyway, the sane way to do this is to convert the dates to Unix epoch, then take it from there.
for ((d=$(date -d "$begin_date" +%s); d <= $(date -d "$end_date" +%s); d += 600))
do
date -d #$d +"%F %H:%M"
done
Doing date arithmetic in the shell is probably going to be rather inefficient; converting this to e.g. Awk or Perl might be worth your time if you find it's too sluggish, or need to run it lots of times.
The example you linked to just needs to be adjusted slightly:
#!/bin/bash
## User-specified dates.
# See [GNU Coreutils: Date] for more info
# [GNU Coreutils: Date]: https://www.gnu.org/software/coreutils/manual/html_node/Combined-date-and-time-of-day-items.html#Combined-date-and-time-of-day-items
begin_date="2015-01-01T00:00"
end_date="2015-01-01T00:40"
# Run through `date` to ensure iso-8601 consistency
startdate=$(date --iso-8601='minutes' --date="${begin_date}")
enddate=$(date --iso-8601='minutes' --date="${end_date}")
# Do loop
d="$startdate"
while [ "$d" != "$enddate" ]; do
echo $d
d=$(date --iso-8601='minutes' --date="$d + 10 minutes")
done
Note that the options -I and -d are equivalent to --iso-8601 and --date respectively.
How to validate the date in Unix.
Where the date format is DD-MM-YYYY.
For example if i run one script ./ValidateDate.sh DD-MM-YYYY then it should check that the date is in this format and also that the date exists.
If date is valid and in above format then proceed for next else print message 'Invalid date'
Thanks.
Well... This is a fine can o' worms.
Any shell script that you create may not work on all of the various flavors of Unix/Linux.
The BSD date command (found on OS X and Solaris) does a great job at taking in the date and verifying the format. It requires you to specify your date format, but once you do, it has no problems:
if date -j -f "%d-%m-%Y" "$DATE" 2>&1 > /dev/null # No output
then
echo "ERROR: Date '$DATE' is not a valid format."
exit 2
else
echo "Date is '$DATE'."
fi
Linux and other systems that use GNU's date can also validate the date, but use a different syntax:
date -d "$DATE" 2>&1 /dev/null # With a bit of luck this will work...
if $? -ne 0
then
echo "ERROR: Date '$DATE' is not a valid format."
else
echo "Date is '$DATE'."
fi
I say With a bit of luck because it's up to the date command to figure out your date's format. This normally works, and will work with your YYYY-MM-DD format, but it can be confusing:
Imagine this:
$ DATE="15/6/2014" # It's June 15, 2014
$ date -d "$DATE"
date: invalid date `15/6/2014' # Invalid?
That's because in my timezone, the dd/mm/yyyy format isn't a valid format. In my timezone, the date should be mm/dd/yyyy. To get around this you can use Wintermute's suggestion and format the date into ISO format before using GNU's date format.
A more universal possibility is to use Perl or Python to figure out if the date is correct:
if perl -mTime::Piece -e "Time::Piece->strptime(\"$DATE\", \"%Y-%m-%d\")" 2> /dev/null
then
echo "ERROR: Date '$DATE' is not a valid format."
else
echo "Date is '$DATE'."
fi
This Perl oneliner will return a non-zero error status if $DATE isn't a valid date in %Y-%m-%d format.
You can validate the date with the date utility, but you first have to transform the input into something it can understand. I suggest ISO representation (always). It could, for example, look like this:
#!/bin/sh
PATTERN='\([0-9]\{1,2\}\)-\([0-9]\{1,2\}\)-\([0-9]\+\)'
# if the pattern doesn't match, this will give date an empty string, so it will fail as expected.
if date -d $(echo "$1" | sed -n "/$PATTERN/ { s/$PATTERN/\3-\2-\1/; p }") > /dev/null 2>&1 ; then
echo valid
else
echo invalid
fi
awk way
awk 'split($0,a,"-"){print strftime("%d-%m-%Y",mktime(a[3]" "a[2]" "a[1]" 00 00 00"))==$0?"valid":"not valid"}' <<< "31-12-1992"
It Converts the string to epoch,converts epoch back then checks against the original.
Edit:
Thought i would add after testing this works for dates
FROM 01-01-0
TO 31-12-2147483647
Although a drawback is after you go below the year 1000 you have to remove leading zeros from the year.
You can do this fairly easily if you are willing to break the validation into two steps. First, check that the string is in the right format:
date=$1
[[ $date =~ ([0-9][0-9])-([0-9][0-9])-([0-9]+) ]] || { printf "Invalid date format\n"; exit 1; }
If that test passes, you can extract the day, month, and year fields, then verify that each falls in the correct range.
day=${BASH_REMATCH[1]}
month=${BASH_REMATCH[2]}
year=${BASH_REMATCH[3]}
thirty_one_days='0[1-9]|[12][0--9]|3[01]'
thirty_days='0[1-9]|[12][0--9]|30'
twenty_eight_days='0[1-9]|1[0-9]|2[0-8]'
case $month in
01|03|05|07|08|10|12)
[[ $day =~ $thirty_one_days ]] ;;
04|06|09|11)
[[ $day =~ $thirty_days ]] ;;
02)
# 1-28, but 29 OK in a leap year.
[[ $day =~ $twenty_eight_days ]] ||
(( year % 4 == 0 && $year % 400 == 0 && day == 29 ))
*)
false
esac || { print "Invalid date\n"; exit 1; }
I trying to print date between 2 dates using while loop in a bash script.
But when i execute i am getting below error:
test.sh: line 8: [: 02-12-14: integer expression expected
Below is my code, can anyone help me out
#!/bin/bash
sdate=02-12-14
edate=02-25-14
while [ "$sdate" -le "$edate" ]
do
echo $sdate
sdate=$(date +%m-%d-%y -d "$sdate + 1 day")
done
You should store them as timestamps:
#!/bin/bash
sdate=$(date -d '2014-02-12' +%s)
edate=$(date -d '2014-02-25' +%s)
while [[ sdate -le edate ]]; do
date -d "#$sdate" '+%m-%d-%y'
sdate=$(date -d "$(date -d "#${sdate}" ) + 1 day" +%s)
done
Output:
02-12-14
02-13-14
02-14-14
02-15-14
02-16-14
02-17-14
02-18-14
02-19-14
02-20-14
02-21-14
02-22-14
02-23-14
02-24-14
02-25-14
Always prefer [[ ]] over [ ] when it comes to conditional expressions in Bash. (( )) may also be a preference.
It requires GNU date. e.g. date --version = date (GNU coreutils) 8.21 ...
mm-dd-yy is not a format acceptable by date for input so I used yyyy-mm-dd which is acceptable.
I am trying to test how old ago a file was created (in seconds) with bash in an if statement. I need creation date, not modification.
Do you have any idea how to do this, without using a command like find with grep?
I'm afraid I cann't answer the question for creation time, but for last modification time you can use the following to get the epoch date, in seconds, since filename was last modified:
date --utc --reference=filename +%s
So you could then so something like:
modsecs=$(date --utc --reference=filename +%s)
nowsecs=$(date +%s)
delta=$(($nowsecs-$modsecs))
echo "File $filename was modified $delta secs ago"
if [ $delta -lt 120 ]; then
# do something
fi
etc..
Update
A more elgant way of doing this (again, modified time only): how do I check in bash whether a file was created more than x time ago?
Here is the best answer I found at the time being, but it's only for the modification time :
expr `date +%s` - `stat -c %Y /home/user/my_file`
If your system has stat:
modsecs=$(stat --format '%Y' filename)
And you can do the math as in Joel's answer.
you can use ls with --full-time
file1="file1"
file2="file2"
declare -a array
i=0
ls -go --full-time "$file1" "$file2" | { while read -r a b c d time f
do
time=${time%.*}
IFS=":"
set -- $time
hr=$1;min=$2;sec=$3
hr=$(( hr * 3600 ))
min=$(( min * 60 ))
totalsecs=$(( hr+min+sec ))
array[$i]=$totalsecs
i=$((i+1))
unset IFS
done
echo $(( ${array[0]}-${array[1]} ))
}