extract lines according to a particular column

Question

I have a CSV file like

CK,ck
XYZ,xyz,xyzs
ABC,abc,abcs
PQR,pqr,pqrs
LMN,lmn,
IJK,ijk,

I have to check if something is written in column 3 except whitespace, then display the whole line.the output will be:

XYZ,xyz,xyzs
ABC,abc,abcs
PQR,pqr,pqrs

Answer 1

A sed approach:

sed -n '/[^,]*,[^,]*,[^, ]\+/p' file

Or a grep solution:

grep -oE '^[^,]*,[^,]*,[^, ]+' file

And awk:

awk -F, '$3 ~ /[^, ]+/' file

Answer 2

Try:

$ awk -F, '$3 && $3 != " "' file
XYZ,xyz,xyzs
ABC,abc,abcs
PQR,pqr,pqrs

or:

$ awk -F, '$3 ~ /[^[:blank:]]/' file

Answer 3

You could do something like this in awk :

awk -F',' '{gsub(/[ \t]+$/,"",$3)}$3' foo.csv

This will remove trailing whitespaces and only print if there's something left in the third field.

Answer 4

Using GNU awk:

awk -F, '$3' input_file