How can I get rows with specific columns and specific pattern?

Question

I have a csv file looks like this:(second column is date pattern)

qqq.eee,2015-10-12
rrr-ttt,2015-11-23
aaapppp,2022-01-10
ddd_fff,2017-05-23
zzzbbbb,2013-09-20
nnnmmmm,2015-06-17
wwwwwww,2014-11-20
uuu-uuu,2016-12-10
ppppppp,2015-10-19
kkkkkkk,2016-12-25

I want to get rows that second column matching as follows:

year between two number(for example): 2014<= year =<2016
month between two number(for example): 10<= month =<12
day between two number(for example): 20<= day =<30

and then get output as follows:

rrr-ttt,2015-11-23
wwwwwww,2014-11-20
kkkkkkk,2016-12-25

Is this possible with awk?

Answer 1

awk is best suited for column and multiple condition filtering. However, this particular problem can be also solved with grep

$ grep -E '201[4-6]-1[0-2]-(2[0-9]|30)$' ip.csv 
rrr-ttt,2015-11-23
wwwwwww,2014-11-20
kkkkkkk,2016-12-25

• -E use extended regular expressions
• 201[4-6] range of 2014 to 2016
• 1[0-2] range of 10 to 12
• 2[0-9] range of 20 to 29
  • using grouping and alternation, (2[0-9]|30) gives the range 20 to 30
• $ end of line
• further reading: http://www.regular-expressions.info/numericranges.html

Answer 2

I just compare from the end to the beginning because of the separators

awk -F[-,] '$NF >= 20 && $NF <= 30 && $(NF-1) >= 10 && 
            $(NF-1) <= 12 && $(NF-2) >= 2014 && $(NF-2) <= 2016' yourfile

1. NF is the last field and $NF is the value, and $(NF-1) is the value for the field before the last and so on
2. $NF day, $(NF-1) month, $(NF-2) year