Crontab schedule every two week - bash

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

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

Answer 2

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.