Searching a CSV File Using Grep

Question

Lets say I have a csv file like this:

a,b1,12,
a,b1,42,
d,e1,12,
r,12,33,

I want to use grep to return only only the rows where the third column = 12. So it would return:

a,b1,12,
d,e1,12,

but not:

r,12,33,

Any ideas for a regular expression that will allow me to do this?

Answer 1

grep "^[^,]\+,[^,]\+,12," file.csv

Answer 2

Here's a variation:

egrep "^([^,]+,){2}12," file.csv

The advantage is that you can select the field simply by changing the number enclosed in curly braces without having to add or subtract literal copies of the pattern manually.

Answer 3

I'd jump straight to awk to test the value exactly

awk -F, '$3 == 12' file.csv

This, and any regexp-based solution, assumes that the values of the first two fields do not contain commas

Answer 4

when you have csv files, where you have distinct delimiters such as commas, use the splitting on field/delimiters approach, not regular expression. Tools to break strings up like awk, Perl/Python does the job easily for you (Perl/Python has support for csv modules for more complex csv parsing)

Perl,

$ perl -F/,/ -alne  'print if $F[2]==12;' file
a,b1,12,
d,e1,12,

$ awk -F"," '$3==12' file
a,b1,12,
d,e1,12,

or with just the shell

while IFS="," read a b c d
do
    case "$c" in
        12) echo "$a,$b,$c,$d"
    esac
done <"file"