Counting the records based on condition

Hi Unix team,

I have a file with 30 columns with tab delimited. I need to count the records based on column 18 & 21 data.
So I cut the data from the file using

 
awk -F"\t" '{print $18,$21}' foo.txt

Following is the output:

 
USED SEDAN
USED SUV
NEW  SUV
USED Truck
USED Truck
USED SEDAN
USED SUV
NEW  Small

Now I need to count the records that are
A. Non SEDAN or Non SUV in $2
B. Records which are USED in $1, but only SEDAN or SUV.

 
Expected result:
3
4

I tried as:

 
awk -F"\t" '{print $18,$21}' foo.txt | \
while read line
do
if [ "$2" != "SEDAN" ] && [ "$2" != "SUV" ] 
then
((count+=1))
fi
done

Thanks.

Is this homework?

No it's not a homework.

awk ' BEGIN { uss=0; nss=0; } { 
 if($1=="USED" && ($2=="SEDAN" || $2=="SUV")) uss++; if($2!="SEDAN" && $2!="SUV") nss++; 
 } END { 
 printf "%d\n%d\n", nss, uss; 
} ' infile
3
4

Note: change positions $1 $2 according to your file.

try also:

awk '{if ($2!~/SUV/ && $2!~/SEDAN/) {a++} else {if ($1~/USED/) b++}} END {print a; print b}' input

Thanks.

Why don't you do it in one go:

$ awk '{if ($21!~/SEDAN|SUV/) s++; else if ($18=="USED") u++} END{print s;print u}' file
3
4

Note that the matching operator ~ can match a composed regular expression and thus may be faster than running it twice on two strings. If only one single string needs to be found, the immediate string compare == will be more efficient.