AWK to take last previous column

You can use the below command to print last previous column.

awk '{print $(NF-1)}' access.log | sort | uniq | wc -l

AWK IF ELSE block

nawk -F"," ' BEGIN{OFS=","} {if (NR==1) {
print "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXx"
} else {
print $1,$2,$3,$4,"S",1
print $1,$2,$3,$5,"S",2
print $1,$2,$3,$6,"D",1
print $1,$2,$3,$7,"D",2
}}' in.txt

AWK to calculate sum, min, max of particular field in txt

Recently, I came across situation to calculate the sum, min, max of particular field using awk.
Below is the snippet I've used to calculate the field values.

awk 'BEGIN{
FS=OFS=",";
max = -999999999; min = 9999999999;
print "----------------------------"
}
{
sub(" Total requests: ","",$2);
val = int($2);
lines+=1;sum+=val;
if( val > max ) max = val;
if ( val < min ) min = val;
}
END{
print "| Total No of Lines: "lines " Sum: " sum " Average: " sum/lines " Max: " max " Min: " min
}' with_out.txt

Text File Contents for parsing:

09/Sep/2008:03:57:09, Total requests: 36, New requests: 34
09/Sep/2008:03:57:10, Total requests: 61, New requests: 58
09/Sep/2008:03:57:11, Total requests: 40, New requests: 38
09/Sep/2008:03:57:12, Total requests: 101, New requests: 97
09/Sep/2008:03:57:13, Total requests: 33, New requests: 35

AWK Samples

Here are some of the useful awk program collections!

awk '{print $2,$1}' filename
awk '{print $2,$1}' filename
awk '$1 > $2 {print $1,$2,$1-$2}' filename

awk '$1 > $2'           filename
awk '$1 > $2{print}'    filename
awk '$1 > $2{print $0}' filename

awk '$1=="foo"{print $2}' filename
awk '/foo.*bar/{print $1,$3}' filename
awk '$2~/foo/{print $3,$1}' filename
awk '$2!~/foo/{print $3,$1}' filename

awk '/foo/,/bar/' filename

awk 'BEGIN{print"fee"} $1=="foo"{print"fi"}
  END{print"fo fum"}' filename

awk '{print $1,$NF }' filename
awk '{print NR,$0 }' filename
awk -F: '{print $1,$3 }' /etc/passwd
awk '{$10=""; print }' filename
awk '{for(i=1;i<=NF;i++) print $i }' filename awk '{printf("%s %03d %02d %.15g\n",$1,$2,$3,$3/$2); }' filename
awk '{for(i=NF;i > 0;i--) printf("%s",$i); printf("\n"); }' filename

Power of Shell Script

Scripts makes the job more easy !

Below is the sample script which I wrote to retrieve the count of files modified for all the modified dates in home directory. This script also gives the sum of bytes for files modified on the corresponding dates.

prakash@prakash-laptop:~$ for filedate in `ls -l | awk '{print $6}' | sort | uniq -c | awk '{print $2}'`;do ls -l | grep $filedate | awk 'BEGIN{print "----------------------------"} {count+=1;sum+=$5} END{print $6" | " count " | " sum }' ; done


Sample Output
----------------------------
Date | Count | Sum of bytes
----------------------------
2007-10-30 | 1 | 26
----------------------------
2007-11-21 | 1 | 4096
----------------------------
2008-02-06 | 1 | 332341
----------------------------
2008-02-25 | 2 | 8192
----------------------------
2008-03-03 | 1 | 999
----------------------------
2008-03-04 | 1 | 219
----------------------------
2008-03-05 | 1 | 3402
----------------------------
2008-03-06 | 2 | 739209
----------------------------
2008-03-11 | 4 | 6648
----------------------------
2008-03-14 | 10 | 50918
----------------------------
2008-03-15 | 4 | 106573
----------------------------
2008-03-16 | 1 | 10635


Really the scripting makes your job more easy.