Shell Programming and Scripting

few more Q.. Which would be your file 1 and file2 in this case? also do you have GNU date (Linux)?

Code:

Aggr_New_06_19_2013_1944.txt
Aggr_New_06_20_2013_1956.txt

In above files, file1 is Aggr_New_06_19_2013_1944.txt and file2 is Aggr_New_06_20_2013_1956.txt, previous file should compare with current file(newfile) and difference or new records in file2 should get output to file3.

GNU: Yes, Date command will work in our linux server, but this comparison will do whenever we require like monthly 15 or 20days will run, not daily, requirement is always previous file should compare with new file and get difference output to file3.

in your script define 2 variables (you can directly feed also to awk)

Code:

 
file1=Aggr_New_`date --date='yesterday' +"%m_%d_%Y"`_*.txt
file2=Aggr_New_`date +"%m_%d_%Y"`_*.txt
awk -F"|" 'FNR==NR{a[$2$3]++;next}!a[$2$3]' file1.txt file2.txt > file3.txt

Thanks for your reply Vidyadhar..!!
Its working if we have y'day and today files, the files executes in windows only some times, like continuous 10days then after that again 20th day...so in this case, 10th day file wil be like...

Code:

Aggr_New_06_10_2013_1944.txt
Aggr_New_06_20_2013_1956.txt

In above case, previous file is 10th day file, new file is 20th date file, script is not working in this case...hope you can understand the requirement.

who and what will decide when to run this? If you are the one to decide then use
date --date='10 days ago' +"%m_%d_%y" something like this

its depends on requirement when we get files, we can't hardcode any particular date, again it will be like manual as example below...

Code:

Day1

Aggr_New_06_10_2013_1944.txt
Aggr_New_06_20_2013_1956.txt

Day2

Aggr_New_06_20_2013_1944.txt
Aggr_New_06_21_2013_1956.txt

next run

Aggr_New_06_21_2013_1944.txt
Aggr_New_06_25_2013_1956.txt

It may occur like this...

If your input file names are as described above and you want them sorted by year, month, day, and time; one of the following should work for you.

When there are just two files in the current directory and both files are from the same year, a simple:

Code:

echo *

will give you the two filenames to be processed in the correct order. If the files may be from different years, try:

Code:

ls -f | sort -t_ -k 5,5 -k 3,4 -k6,6

If your output file is also in the same directory (and has a name that doesn't start with Aggr, you can use:

Code:

ls -f Aggr* | sort -t_ -k 5,5 -k 3,4 -k6,6

If there are a bunch of these files in the directory and you just want the last two, you can use:

Code:

ls -f Aggr* | sort -t_ -k5,5 -k3,4 -k6,6 | tail -n2

If you have other files with names starting with Aggr (such as your output files), but they have a slightly different name format, you can use:

Code:

ls -f Aggr_New_[0-9][0-9]_[0-9][0-9]_[0-9][0-9][0-9][0-9]_[0-9][0-9][0-9][0-9].txt |
        sort -t_ -k5,5 -k3,4 -k6,6 | tail -n2

If you keep just giving us bits and pieces we continue to waste time trying to figure out what you're trying to do.

PLEASE, give a complete specification of your input and output requirements!