Print only lines with more than $1 words

Question

I'd like to print only the lines with $1 number of words or more. Please help.

while read line ; do
    echo $line | wc -w 
done t1.txt

Answer 1

Assuming you're defining a word as characters delimited by spaces, then awk would do this easily:

awk -v COUNT=$1 'NF>COUNT' t1.txt

It passes the first arg in as an awk variable named count, and prints rows where the number of space delimited fields is above the count provided.

e.g.

$ echo $COUNT
3
$ cat t1.txt
hey
hey hey hey hey hey
hey hey hey
hey hey hey
hey hey hey hey hey
hey hey hey hey hey
hey hey hey

$ awk -v COUNT=$COUNT 'NF>COUNT' t1.txt
hey hey hey hey hey
hey hey hey hey hey
hey hey hey hey hey