awk shell script cannot run properly in crontab

please post the crontab line
try to use full path to all program in your crontab

I can't omit the "8 2006" after formatit.awk in the script because it will generate the output with missnig the month and year.
Please help me.

The crontab line is as follow:

22 16 05 09 * cal 8 2006 | grep -v "[A-Z]" | cut -c4-17 | awk -f /aaa/home/scripts/aaa_oli_valo_all/formatit.awk - 8 2006 > /aaa/home/scripts/aaa_oli_valo_all/1.txt 2>&1

>I can't omit the "8 2006" after formatit.awk in the script because it will generate the output with missnig the month and year.
Looks like 8 and 2006 are parameters for formatit.awk and not input file. In this case you need to modify your call to awk in something like the following:
awk -f /aaa/home/scripts/aaa_oli_valo_all/formatit.awk -v month=8 -v year=2006 > /aaa/home/scripts/aaa_oli_valo_all/1.txt 2>&1

Replace "month" and "year" by the name of the variables you use in your script for the month and year.

> The crontab line is as follow:
please use full path to all programs in crontab and try again

My formatit.awk is as follow:

BEGIN {
     MON=ARGV[2]
     if ( length(MON) == 1 ){
        MON= "0" MON
     }
     YEAR=ARGV[3]
}
{
     for (i=1; i<=NF; i++)
         if (length($i)==1){
            print "0" $i "/" MON "/" YEAR;
         }else{
            print $i "/" MON "/" YEAR;
         }
}


Would you advise me how to se the parameters ? Thanks.

>  MON=ARGV[2]
what should this do?
awk has no ARGV, gawk has.
But if you want to pass parameters to the awk program itself (beside the text it should process), you better use gawk with its -v option.

As I suggested several times now: please use full path to awk. I guess that your cron uses tradtional AT&T awk while in your shell you use gawk (probably aliased or found in PATH first). Please correct me if I'm wrong.

Try the following:
cal 9 2006 | grep -v "[A-Z]" | cut -c4-17 | awk -v mon=9 -v year=2006 '{for (i=1;i<=NF;i++) printf "%02d/%02d/%d\n", $i, mon, year;}'

As ahoffmann as already suggested, you probably have different 'flavor' of awk (e.g. awk, nawk, gawk, ...) on your system and that's the reason why he suggest you to set the full path to awk. If you don't know which version you use, when you are connected in the shell simply issue a "which awk" command. This command will display the full path to awk.

For example:
$ which awk
/usr/bin/awk
$

Now that you know the full path to awk, just edit your crontab like this:
22 16 05 09 * cal 8 2006 | grep -v "[A-Z]" | cut -c4-17 | /usr/bin/awk -f /aaa/home/scripts/aaa_oli_valo_all/formatit.awk - 8 2006 > /aaa/home/scripts/aaa_oli_valo_all/1.txt 2>&1

or

22 16 05 09 * cal 9 2006 | grep -v "[A-Z]" | cut -c4-17 | /usr/bin/awk -v mon=9 -v year=2006 '{for (i=1;i<=NF;i++) printf "%02d/%02d/%d\n", $i, mon, year;}' > /aaa/home/scripts/aaa_oli_valo_all/1.txt 2>&1

I still got error......

here is the script ....

the filename is called "gen_weekday.sh"

#!/bin/ksh
MON=`date '+%m'`
YEAR=`date '+%Y'`
if [ $MON -eq 1 ]
then
   MON=12
   YEAR=`expr $YEAR - 1`
else
   MON=`expr $MON - 1`
fi
cal $MON $YEAR | grep -v "[A-Z]" | cut -c4-17 | awk -v mon=$MON -v year=$YEAR '{for (i=1;i<=NF;i++) printf "%02d/%02d/%d\n", $i, mon, year;}'



in crontab

11 17 08 09 * /aaa/home/scripts/gen_weekday.sh > /aaa/home/scripts/date.txt 2>&1

the date.txt contains the error:
awk: syntax error near line 1
awk: bailing out near line 1


Please suggest me how to solve the problem. Thanks.

mikekwok, why do you not follow the suggestions?