I'm writing a Perl program to check if and make sure that some system backup tasks were executed from crontab.
I need to read the crontab and interpret when it was supposed to run, in order to check if that backup was done. Here is an example.
00 03 * * 6 system_backup.sh
Let's suppose this task will generate a file called system_backup_20180510.iso
Then my idea was to store a "desirable date" into a var then compare with that date in the filename. The biggest issue is how to build that desirable date with crontab's day of month field filled up.
What you guys can suggest? Thanks
Note the 5th field: this only runs on Saturday.
Here's something you can do with GNU date:
$ date -d "-1 week saturday 03:00" "+%Y%m%d %H:%M:%S"
20180505 03:00:00
$ date -d "-w week saturday 03:00" "+%Y%m%d %H:%M:%S"
20180518 13:00:00
$ date -d "-2 week saturday 03:00" "+%Y%m%d %H:%M:%S"
20180428 03:00:00
the Schedule::Cron::Events suggestion implemented:
crontab -l | perl -MPOSIX=strftime -MSchedule::Cron::Events -lne '
$cron = Schedule::Cron::Events->new($_);
$prev = strftime("%F %T", $cron->previousEvent());
print "$prev => ", $cron->commandLine();
'
Related
Given a Unixtime such as 1551996855 (Thu, 07 Mar 2019 22:14:15 +0000), how can convert and extract just HH:MM, (22:14)?
you could use date for the conversion and specify +%H:%M as the format:
date -d #$"1551996855" +%H:%M
note: the option to use with date will be different than -d if you are not on a Linux system (which uses date from GNU coreutils).. so YMMV
Could someone please suggest a simple and short approach to convert "Thu Sep 22 3:50 2016" to "2016-09-22" in Solaris, through a shell script?
I do not have GNU date available on Solaris as discussed in below post:
Convert date String to number on Solaris shell script gives No such file or directory
I need to query an sql server db, where date is saved in the format, "2016-09-06", hence I need to convert it
Actually, you do have GNU date available but here is anyway one way to achieve this by scripting:
#!/bin/ksh
a="Thu Sep 22 3:50 2016"
echo $a | nawk '
BEGIN {
m=1
m2m["Jan"]=m++;
m2m["Feb"]=m++;
m2m["Mar"]=m++;
m2m["Apr"]=m++;
m2m["May"]=m++;
m2m["Jun"]=m++;
m2m["Jul"]=m++;
m2m["Aug"]=m++;
m2m["Sep"]=m++;
m2m["Oct"]=m++;
m2m["Nov"]=m++;
m2m["Dec"]=m++;
}
{
printf("%s-%02d-%02d\n",$5,m2m[$2],$3)
}'
output:
2016-09-22
Why not use Oracle's sysdate?
select * from your_table where saved_date >= to_char(sysdate,'yyyy-MM-dd')
Since the date in BusyBox is not as powerful as gnu date, I have problems to calculate the date of last saturday.
last_sat=`date +"%Y-%m-%d" -d "last saturday"`
only works fine with gnu date.
I've found something like this to calculate from Epoch
busybox date -D '%s' -d "$(( `busybox date +%s`+3*60 ))"
but my BusyBox (v1.1.0) doesn't recognize the -D argument.
Any suggestions?
For the last Saturday before today, under busybox 1.16:
date -d "UTC 1970-01-01 $(date +"%s - 86400 - %w * 86400"|xargs expr) secs"
How it works: take the current date in seconds, subtract one day, subtract one day times the number of the current weekday, then convert those seconds back to a date.
EDIT: after hacking together a build of 1.1, this works:
date -d "1970.01.01-00:00:$(date +"%s - 86400 - %w * 86400"|xargs expr)"
This working version is based on code-reading:
} else if (t = *tm_time, sscanf(t_string, "%d.%d.%d-%d:%d:%d", &t.tm_year,
&t.tm_mon, &t.tm_mday,
&t.tm_hour, &t.tm_min,
&t.tm_sec) == 6) {
t.tm_year -= 1900; /* Adjust years */
t.tm_mon -= 1; /* Adjust dates from 1-12 to 0-11 */
BusyBox's date command has been the topic of some discussion over the years. Apparently it doesn't always work as documented, and it doesn't always work the same as previous versions.
On a BB system I administer running BusyBox v1.01, I'm able to use the -d option with dates in the format MMDDhhmmYYYY.ss, and in no other format that I've tried. Luckily, output formats work as expected, presumably because date is using a proper strftime() according to comments in the source.
Here's my forward-and-reverse example:
[~] # busybox date '+%m%d%H%M%Y.%S'
090500152016.41
[~] # busybox date -d 090500152016.41
Mon Sep 5 00:15:41 EDT 2016
So .. what can we do with this? It seems that we can't do an arbitrary adjustment of seconds, as it only reads the first two digits:
[~] # busybox date -d 123119001969.65 '+%s'
65
[~] # busybox date -d 123119001969.100 '+%s'
10
Well, it turns out you can load the date fields with "invalid" numbers.
[~] # busybox date 090100002016
Thu Sep 1 00:00:00 EDT 2016
[~] # busybox date 093400002016
Wed Oct 4 00:00:00 EDT 2016
[~] # busybox date 09-200002016
Mon Aug 29 00:00:00 EDT 2016
So let's adjust the "day" field using something based on %w.
today=$(busybox date '+%m%d%H%M%Y')
last_sat=$(busybox date -d "${today:0:2}$( printf '%02d' $(( 10#${today:2:2} - 1 - $(busybox date '+%w') )) )${today:4}" '+%F')
This simply subtracts numbers in the second field (the 3rd and 4th characters of the date string). It obviously requires that your shell either be bash or understand bash-style math notation ($((...))). Math-wise, it should work as long as "last saturday" is within the same month, and it MAY work (I haven't tested it) with rollvers to the previous month (per the last test above).
Rather than jumping through these burning hoops, I recommend you just install a GNU date binary, and don't use busybox for this one binary. :-P
Good luck!
20110216_00
20110216_01
...
20110216_23
20110217_00
..
and so on
I have tried with
date +'%Y%m%d_%H'
but it never starts with 00-23 format but from 01-24 like format, hence I get hour part always incorrect.
Can anybody suggest, how can I get above o/p
You can do it by manupulating the hour part. Check the snip below.
#/bin/ksh
s=`date +'%Y%m%d_'`
t=` date +'%H'`
let t=$t+1
echo "Required date is " $s$t
It gives
Required date is 20110316_16
I tried it on SunOS 5.10 This works !
date +%Y%m%d_%H
-> 20130912_02
date +'%Y%m%d_%H'
-> 20130912_02
Can you tell us which Solaris are you using ?
uname -a
Cheers !
What revision of Solaris are you using? It is roughly 22:30 locally and I see:
mph#sol11express:~$ date +'%Y%m%d_%H'
20110216_22
mph#sol11express:~$ uname -a
SunOS sol11express 5.11 snv_151a i86pc i386 i86pc Solaris
mph#sol11express:~$ echo $SHELL
/bin/bash
which looks to me like it is using 0-23 for hours.
you can use date -u +'%Y-%m-%d-%H'
How to check if today is a weekend using bash or even perl?
I want to prevent certain programs to run on a weekend.
You can use something like:
if [[ $(date +%u) -gt 5 ]]; then echo weekend; fi
date +%u gives you the day of the week from Monday (1) through to Sunday (7). If it's greater than 5 (Saturday is 6 and Sunday is 7), then it's the weekend.
So you could put something like this at the top of your script:
if [[ $(date +%u) -gt 5 ]]; then
echo 'Sorry, you cannot run this program on the weekend.'
exit
fi
Or the more succinct:
[[ $(date +%u) -gt 5 ]] && { echo "Weekend, not running"; exit; }
To check if it's a weekday, use the opposite sense (< 6 rather than > 5):
$(date +%u) -lt 6
case "$(date +%a)" in
Sat|Sun) echo "weekend";;
esac
This is actually a surprisingly difficult problem, because who is to say that "weekend" means Saturday and Sunday... what constitutes "the weekend" can actually vary across cultures (e.g. in Israel, people work on Sunday and have Friday off). While you can get the date with the date command, you will need to store some additional data indicating what constitutes the weekend for each locale if you are to implement this in a way that works for all users. If you target only one country, then the solution posed in the other answers will work... but it is always good to keep in mind the assumptions being made here.
Use Perl's localtime operator.
localtime
Converts a time as returned by the time function to a 9-element list with the time analyzed for the local time zone. Typically used as follows:
# 0 1 2 3 4 5 6 7 8
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time);
…
$wday is the day of the week, with 0 indicating Sunday and 3 indicating Wednesday.
For example:
$ date
Sun Aug 15 20:27:29 CDT 2010
$ perl -le 'my $wday = (localtime)[6];
print $wday >= 1 && $wday <= 5 ? "weekday" : "weekend"'
weekend
printf also has date
printf -v day '%(%a)T'
case $day in
Sat|Sun) echo "Hooray!";;
esac
https://ideone.com/wU7C0c - demo
I am not really sure if this is suited to this question but I wanted to share this it to help others out and this is the closest Stack question to what I was looking for.
This is a shell script that I use for starting an application when i boot my computer, the application cannot connect to its server over the weekend (its down for maintenance as the service doesn't run at weekends (forex trading application).
Information about this script, d is 1 to 7 (mon to sun) h is 24 time. You can adjust this to your own settings, maybe you have a midweek maintenance or any other cause. This is my first ever shell script, so I'm sure there is a better way and if anybody wants make edits, feel free I will review for acceptance (or somebody else may do it)
#!/bin/bash
d=$(date +%u)
h=$(date +%H)
case $d in
'5')
# Friday after 10pm
if (($h >= 22)) ; then
exit
fi
;;
'6')
# all day saturday
exit
;;
'7')
#sunday before 10pm
if (($h < 22)) ; then
exit
fi
;;
*)
# any other time run the program
;;
esac
I like this way of running programs on weekdays:
[[ $(date +%u) -lt 6 ]] && weekday-program
It is a bit tricky construction, but compact and easy to add.