Count times an expression is true in awk.

Question

In unix/awk code

Essentially, I need to iterate through the file (data1.txt) and count the number of times the substr($0,29,2) of each line of the file == "04".

data1.txt:

6597243042 20160305001100003140152852153019000127000200
6597243042 20160305001100003140170306190306020000000200
6597243042 20160305001100003140170552190552020000000200
6597243042 20160305001100003140201430201543000113000400
6592311319 20160305041100003460072719072839000120001200
6592311319 20160305041100003460072927072952000025001200

In this case only 2 lines of the file (data1.txt) fulfill the condition (substr($0,29,2)=="04")

I am stuck at finding a way to do so

Below is my unix/awk code

Filename="def"
file="data1.txt"
#awk '{count1=0}'
while IFS= read line
do

   awk '{ if (substr($0,29,2)=="04") {print substr($0,29,4)}}' 

done <"$file"

How can I count this?

Answer 1

Another approach (thanks to Archemar + kos for avoiding use of deprecated egrep and using -c)

grep -cE '^.{28}04' data1.txt

Answer 2

simply try

awk 'BEGIN { count = 0 ;} substr($0,29,2) == "04" { count++ ; } END { print count ;}' files

where

• substr($0,29,2) == "04" search fo condition
• { count++ ; } if found increase count
• END at the end
• { print count ;} print it.

edit:

• count=0 thanks to @steeldriver

awk command line can be entered as below for readability.

awk 'BEGIN { count = 0 ;} 
     substr($0,29,2) == "04" { count++ ; } 
     END { print count ;}' files