Help on sed/awk

08-07-2012

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neutronscott's proposal

Quote:

Originally Posted by neutronscott

Is the date MMDDYY? Seems that way. I understand that you want to sum the bytes transfered during those times.

Code:

awk -F: '
    /DEST_FN/ { m = 0+substr($3,1,2) }
    /bytes sent/ { bytes[m] += 0+$0 }
    END {
        for (i in bytes)
            printf("%d\t%.2f KB\n", i, bytes[i]/1024)
    }' input

output is like this:

Code:

5       1.71 KB
6       0.86 KB

does exactly what you want, except maybe for the headers and months' names. Here's an enhanced version of his command:

Code:

awk -F: 'BEGIN    {
        print "Month\tTotal kB\n=====\t========";
        split("Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec", mn, " ")
        }

    /DEST_FN/ { m = 0+substr($3,1,2) }
    /bytes sent/ { bytes[m] += 0+$0 }

    END {
        for (i in bytes)
        printf("%s\t%.2f kB\n", mn[i], bytes[i]/1024)
    }' inputfile

---------- Post updated at 01:46 PM ---------- Previous update was at 01:38 PM ----------

Quote:

Originally Posted by raj_saini20

Code:

awk -F":" 'NR==1{mm=substr($3,1,2);s=0}
($0 ~ /Kbytes/){split($0,a," ");s=s+a[1];getline;if(mm+0!=substr($3,1,2)+0)
{printf "%s %.2f KB\n",mm,s/1024;mm=substr($3,1,2);s=0}}' logfile

This code works fine for the example given, but it's got problems if
- months are not in ascending order
- the month changes in the DEST_FN: line (which was required by the OP) and not in the paragraph's first line.

Last edited by RudiC; 08-08-2012 at 05:40 AM.. Reason: improved wording

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08-08-2012

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Try this for irregular date occurrence

Code:

awk -F":" '
NR==1{
	i=1;mm=substr($3,1,2);a[mm]=0;b[i]=mm;i++;s=0
	}
($0 ~ /Kbytes/){
		split($0,c," ");
		a[mm]=a[mm]+c[1];
		getline;
		n_mm=substr($3,1,2);
		if((mm+0!=n_mm+0)&&length(n_mm))
			{
				mm=substr($3,1,2);
				if(!a[mm])
					{
						a[mm]=0;b[i]=mm;i++
					}
			}
		}
END{
	for(k=1;k<i;k++)
		{
			printf "%s %.2f KB\n",b[k],a[b[k]]/1024
		}
}' logfile

raj_saini20

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