awk + count field separator in csv and print line number

Question

We have csv file with more than 50000 lines This is only example

Dcccrev,wefrwvfr,rfregt,wr4f,rfvrv,ecxwec,ecfrv,rfrf
Grge,gtgr,frfrv,gthtgv,gerg5tgvrt,rvrfvtg,tgt,frfrf,rfrf
Drfrfr,t,tgtg,rf,rgr,grtg,tgt,gtgtg,rg
.
.
.

My task is: in case numbers of separator “,” In each line isn’t equal to 7 then need to print the line number

Is it possible to create one awk line or one perl liner for this task?

Without to use echo or cat that spend time

Answer 1

This is fairly easy with awk. You can set the delimiter to , with -F',' then count the columns with NF. For 7 comma's we would need 8 fields and print the current line number with NR.

awk -F ',' 'NF != 8 {print NR}' test.txt

Contents of test.txt

Dcccrev,wefrwvfr,rfregt,wr4f,rfvrv,ecxwec,ecfrv,rfrf
Grge,gtgr,frfrv,gthtgv,gerg5tgvrt,rvrfvtg,tgt,frfrf,rfrf
Drfrfr,t,tgtg,rf,rgr,grtg,tgt,gtgtg,rg

Output

2
3

Answer 2

Perl's s/// operator(could also have used the tr/// aka y/// as well) returns the number of substitutions it performed which can be used to find the number of ",". Similarly, the m// operator returns the number of matches.

perl -lne 's/,//g == 7 or print $.'      yourfile
perl -lne 'print $. if 7 != (() = /,/g)' yourfile
sed   -ne 's/[^,]//g; /^.\{7\}$/!='      yourfile