How to count words in a line

Question

I have a text file called "shoplist.txt" which one have:

drinks water cola fanta
fruit banana orange

And I want to get how many items per line I have. I'm able to extract drinks and fruit with function "cut" but how can I count how many words I have in each line?

My actually code is:

fileLine=`cat file.txt`
#Here I get each line saving it to fileLine
for line in $fileLine; do
 echo
((aux++))
done

But this code dosen't work because it save to %fileLine each work (drinks, then water,then cola,...)

How can I get the first line and then count the words on that line?

Answer 1

If you can use awk, NF is the number of fields in the current line (by default, a field is a word delimited by any amount of whitespace).

Use

awk '{ print NF, $0 }' inputfile

With your sample input, this will print

4 drinks water cola fanta
3 fruit banana orange

Answer 2

In Bash and wc:

IFS=$'\n'
while read line; do
    wc -w <<< "$line"
done < file.txt

wc counts lines, words, bytes in files. With a shell loop you can make it count words in a line.

Answer 3

Maybe slightly out of context (though similar), as it concerns numbers aka (single) digits, but this supersimple awk-rule helped me out a lot to keep only those lines of a file with less than four times the high digits on each line:

awk -F [5,6,7,8,9] 'NF-1<4' IN > OUT 

'NF-1' to obtain an exact count of 0, 1, 2 or 3 high digits (not more).