How to count occurrence of a given character in a unix line ( crontab)

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Hello ,

I have data that look like the floowing exepmles :

a ;kkkhkhk ;hgvhgvhgvh ;123 ;t

a ;bbbg ;yuj ;1237

x ;25 ;z ;t

I want to count the occurrences of ";"(semicolon) in every line. and, when the number of the given character is above 4, it should store the corresponding line into a file .

I have tried the code below :

awk -F ';' 'NF==31' $nom'_temp2.csv' > $nom'_temp3.csv'

it did work on command line but not in crontab.

Waiting for a solution.
Thanks you in advance

What I have tried:

awk -F ';' 'NF==31' $nom'_temp2.csv' > $nom'_temp3.csv'

Posted 29-Mar-17 21:33pm

Member 13094135

Updated 30-Mar-17 0:49am

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Tomas Takac · Answer 1 · 2017-03-30T00:49:00

Solution 1

This should work:

Python

awk -F ';' 'NF > 4 {print}' myinputfile > myoutputfile

This sets the field separator to semicolon and then outputs all the lines only where are more than 4 fields i.e. at least 4 semicolons.

Posted 30-Mar-17 0:49am

Tomas Takac