Crontab schedule every two week

Question

There is a question happened this week on my crontab job.
It's be set as below and works normal every two weeks until now.

10 06 * * 1 test $(($(date +\%W)\%2)) -eq 0 && echo 'test' > /tmp/test.log

The problem is

$(($(date +\%W)\%2)) would be 08, over 7 in February.

And it will show error message if you run in bash: value too great for base (error token is "08").

There is also not working when I try to revise it for forcing decimal base purpose :

10 06 * * 1 test $((10#$(date +\%W)\%2)) -eq 0 && echo 'test' > /tmp/test.log

Does someone know how to solve this issue?
Many thanks.

Answer 1

Make condition in your script

The use of % sign is discouraged in crontab!

To ensure having jod started on monday, every two weeks you've better to create a small script.

#!/bin/bash

if [ "$1" == "-f" ] ;then # use "-f" switch to force execution on odd weeks
    shift
else
    printf -v d '%(%W)T' -1
    ((10#$d%2)) || exit 0  # stop here if odd week
fi

printf "Real job begin at %(%c)T, there...\n" -1

This could be shortened to one single line:

#!/bin/bash
[ "$1" == "-f" ]&&shift||{ printf -v d '%(%W)T' -1;((10#$d%2))||exit;}

So you could add standard lines in your crontab:

10 06 * * 1 path/to/myScript > /tmp/test.log

Sample: year 2016

date -d 2016-1-4\ 10:06  +%s
1451898360
for i in {0..52};do
    printf -v d '%(%W)T' $((c=i*7*86400+1451898360))
    ((10#$d%2)) && printf "%(%a %d %b, week: %W)T\n" $c
  done | cat -n
     1  Mon 04 Jan, week: 01
     2  Mon 18 Jan, week: 03
     3  Mon 01 Feb, week: 05
     4  Mon 15 Feb, week: 07
     5  Mon 29 Feb, week: 09
     6  Mon 14 Mar, week: 11
     7  Mon 28 Mar, week: 13
     8  Mon 11 Apr, week: 15
     9  Mon 25 Apr, week: 17
    10  Mon 09 May, week: 19
    11  Mon 23 May, week: 21
    12  Mon 06 Jun, week: 23
    13  Mon 20 Jun, week: 25
    14  Mon 04 Jul, week: 27
    15  Mon 18 Jul, week: 29
    16  Mon 01 Aug, week: 31
    17  Mon 15 Aug, week: 33
    18  Mon 29 Aug, week: 35
    19  Mon 12 Sep, week: 37
    20  Mon 26 Sep, week: 39
    21  Mon 10 Oct, week: 41
    22  Mon 24 Oct, week: 43
    23  Mon 07 Nov, week: 45
    24  Mon 21 Nov, week: 47
    25  Mon 05 Dec, week: 49
    26  Mon 19 Dec, week: 51
    27  Mon 02 Jan, week: 01

For comparission with idea of using 15th and 30th each months:

for i in {1..12}/{15,30};do
    date -d '2016/'$i +'%a %d %b, week: %W' 2>/dev/null
  done | cat -n
 1  Fri 15 Jan, week: 02
 2  Sat 30 Jan, week: 04
 3  Mon 15 Feb, week: 07
 4  Tue 15 Mar, week: 11
 5  Wed 30 Mar, week: 13
 6  Fri 15 Apr, week: 15
 7  Sat 30 Apr, week: 17
 8  Sun 15 May, week: 19
 9  Mon 30 May, week: 22
10  Wed 15 Jun, week: 24
11  Thu 30 Jun, week: 26
12  Fri 15 Jul, week: 28
13  Sat 30 Jul, week: 30
14  Mon 15 Aug, week: 33
15  Tue 30 Aug, week: 35
16  Thu 15 Sep, week: 37
17  Fri 30 Sep, week: 39
18  Sat 15 Oct, week: 41
19  Sun 30 Oct, week: 43
20  Tue 15 Nov, week: 46
21  Wed 30 Nov, week: 48
22  Thu 15 Dec, week: 50
23  Fri 30 Dec, week: 52

As you could see, there miss one operation on weeks: 6, 9, 10, 21, 32 and 45. At least there is 3 operation less in one year.

Answer 2

Every two week at midnight

   0 0 */15 * * echo 'test'  > tmp.txt


Your cron job will be run at: (5 times displayed)

2016-03-15 00:00:00 UTC
2016-03-30 00:00:00 UTC
2016-04-15 00:00:00 UTC
2016-04-30 00:00:00 UTC
2016-05-15 00:00:00 UTC