Shell Programming and Scripting

Yes zaxxon, we receive 12 file per hour if we receive for exmaple 11 at 05 hour so there is one file missing, i have to list all received files at 05h to find which of them are missing. I hope now you understand me well. Thanks


thank you maskofzorro, Actually i'm new in shell scripting and i need a little time to get more about the SED and GENSUB commands to understand your scripts, but it's working on the example thanks

There is missing the hours 03 in the output. A tad shorter:

Code:

awk '{a=substr($1,12,2); if(a<=10) {_[a]++}} END{printf("%-10s%-10s\n", "Count", "Hour"); for(e in _){printf("%-10s%-10s\n"),  _[e],e}}' infile
Count     Hour
1         10
3         01
3         02
6         03

You can try this

Code:

a=12
b=` date +%Y%m%d`
c=` ls -ltr | grep 'CCD_'$b | tr -s ' ' | cut -d ' ' -f8 | cut -c1-2 | wc -l`
if [$c -eq $a ]
then
echo 'The files are good'
else
echo 'The files are not good'
exit
fi

but you have to run this script in every hour,and this will give you the CCD file counts for that hour,you can schedule it by using crontab.

Thanks,
Pragyan