Print Rows if End of Field Matches a String in AWK - csv

Question

I have a csv file and I am trying to print rows using awk if a certail field ends with a specific string. So for example, I have the below CSV file:
col1,col2,col3
1,abcd,.abcd_efg
2,efgh,.abcd
3,ijkl,.abcd_mno
4,mnop,.abcd
5,qrst,.abcd_uvw
This is the result I am seeking after:
2,efgh,.abcd
4,mnop,.abcd
But I am getting a different result. This is the awk command I am using:
cat file.csv | awk -F"," '{if ($3 ~ ".abcd" ) print $0}'
and This is the result I am getting:
1,abcd,.abcd_efg
2,efgh,.abcd
3,ijkl,.abcd_mno
4,mnop,.abcd
5,qrst,.abcd_uvw
I event tried the below, but no matched is returned so it didn't work:
cat file.csv | awk -F"," '{if ($3 ~ ".abcd$" ) print $0}'
Any clue what the issue might be? Am I using the wrong expression to get this result?
EDIT: This is another command I tried where I tried Kent's solution, but it didn't work:
cat file.csv | awk -F"," '$3 ~ "[.]abcd"'

Answer 1

First of all the cat in cat file|awk ... is useless, just awk ... file
Your input text has no single comma, how come you set FS=","?
If you want to do exact String compare, use $3 == "whatever" instead of $3 ~ /regex/
So your codes could be changed into:
awk '$3 == ".abcd"' file
If you really love regex, and want to do it in regex match way:
awk '$3 ~ "[.]abcd$"' file
or
awk '$3 ~ /^[.]abcd$/' file
depends on what you required.

Answer 2

You may modify your awk command as followed,
$ cat file.csv | awk '$3 ~ /\.abcd$/ {print $0}'
2 efgh .abcd
4 mnop .abcd
Brief explanation,
$3 ~ /.abcd$/: if $3 matches the regex .abcd$, print $0
According to your modified question, you may change the awk command to:
cat file.csv | awk -F, '$3 ~ /\.abcd$/ {print $0}'