File Count Based on FileDate using Shell Script

Shell Programming and Scripting
1

I have file listed in my directory in following format

-rwxrwxr-x+ 1 test        test 4.9M Oct  3 16:06 test20141002150108.txt
-rwxrwxr-x+ 1 test        test 4.9M Oct  4 16:06 test20141003150108.txt
-rwxrwxr-x+ 1 test        test 4.9M Oct  5 16:06 test20141005150108.txt
-rwxrwxr-x+ 1 test        test 4.9M Oct  5 16:06 test20141005150108.txt
-rwxrwxr-x+ 1 test        test 4.9M Oct  5 16:06 test20141005150108.txt

I want Output in following format

DateMoved FileDate Count  Size
  Oct 3   20141002   1         4.9
  Oct 4   20141003   1         4.9
  Oct 5   20141005   3         14.7

Already I am using below script to determine DateMoved,Count and Size.

printf '\nDATE\tFILES\tSIZE\n'
cd /Mydir
ls -lrt | grep "^-" | awk '
{	nkey = $6$7
	if(key != nkey && key != "") {
		printf("%s\t%d\t%d\n", key, freq, size)
		freq = 1;
		size = $5
	} else {freq++
		size += $5
	}
	key = nkey
}
END {	printf("%s\t%d\t%d\n", key, freq, size)
}'

Help me to get FileDate as well based on the file name.
Appreciate your help.

Thanks

2

Hello Krish2014,

Could you please try following and let me know if this helps.(Not tested though)

ls -lrht test2014* | grep "^-" | awk '
	BEGIN{printf("%s\t%s\t%s\t%s\n", "DATE MOVED", "FileDate", "Count", "Size")}
{
	nkey = $6$7
	if(key != nkey && key != "") {
		FileDate=substr($NF,5,8);
		printf("%s\t%s\t%d\t%d\n", key, FileDate, freq, size)
		freq = 1;
		size = $5
	} else {freq++
		size += $5
	}
	key = nkey
}
END {	printf("%s\t%s\t%d\t%d\n", key, FileDate, freq, size)
}'

Thanks,
R. Singh

3

Hello RavinderSingh13,

I tried to execute above code. But i did not receive FileDate.
Here is the sample output I got

DATE MOVED      FileDate        Count   Size
Jan15           106     540078423

106=Count
540078423 = Size (Human Readable)

Thanks