Number of fields returned by awk

Question

Is there a way to get awk to return the number of fields that meet a field-separator criteria? Say, for instance, the file contains:

a b c d

So, awk -F=' ' | <something> should return 4.

Answer 1

The NF variable is set to the total number of fields in the input record. So:

echo "a b c d" | awk --field-separator=" " "{ print NF }"

will display

4

Note, however, that:

echo -e "a b c d\na b" | awk --field-separator=" " "{ print NF }"

will display:

4
2

Hope this helps, and happy awking

Answer 2

NF gives the number of fields for a given record:

[]$ echo "a b c d" | gawk '{print NF}'
4

Answer 3

If you would like to know the set of all the numbers of fields in a multiline content you can run:

X | awk '{print NF}' | sort -n | uniq

being X a command that outputs content in the standard output: cat, echo, etc. Example:

With file.txt:

a b
b c
c d
e t a
e u

The command cat file.txt | awk '{print NF}' | sort -n | uniq will print:

2
3

And with file2.txt:

a b
b c
c d
e u

The command cat file2.txt | awk '{print NF}' | sort -n | uniq will print:

2

Answer 4

awk(1) on FreeBSD does not recognize --field-separator. Use -v instead:

echo "a b c d" | awk -v FS=" " "{ print NF }"

It is a portable, POSIX way to define the field separator.