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AWK matching problem

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# 1  
Old 10-22-2009
AWK matching problem


i have a list as follow:
Taba 4
aba 1
Taba 2
aba 1
aba 0
where "aba" can be zero or one. I want match all "aba" and count how many are set to one. I mean in the previous case i should get just "2":

I tried as follow:

awk '/^aba/ BEGIN{i=0}{if($2==1) i=i+1}END{print i}' file

but i get some errors.
If i do:

awk '/^aba/ {print $0}' do

I get the list with all "aba " i think that the problem refers to the BEGIN. Can anyone suggest a solution?


# 2  
Old 10-22-2009
awk '$1 ~ /^aba/  && $2 ~ /1/ {i = i+1} END {print i}' a.txt


awk '$1 == "aba"  && $2 == "1" {i = i+1} END {print i}' a.txt

# 3  
Old 10-22-2009
to be more precise in case you have things like "abalone 1" , use equality
awk '$1=="aba" && $2==1{c++}END{print c}' file

# 4  
Old 10-22-2009
thanks for both solutions


EDIT: i found out also that in this way works:

awk '/^aba/ {if($2==1) i=i+1}END{print i}' file


Last edited by Dedalus; 10-22-2009 at 12:57 PM..
# 5  
Old 10-22-2009
awk '$0=="aba 1"{i++}END{print i}' file

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