Linux Terminal: Finding number of lines longer than x

Question

I come to you with a problem that has me stumped. I'm attempting to find the number of lines in a file (in this case, the html of a certain site) longer than x (which, in this case, is 80).

For example: google.com has (by checking with wc -l) has 7 lines, two of which are longer than 80 (checking with awk '{print NF}'). I'm trying to find a way to check how many lines are longer than 80, and then outputting that number.

My command so far looks like this: wget -qO - google.com | awk '{print NF}' | sort -g

I was thinking of just counting which lines have values larger than 80, but I can't figure out the syntax for that. Perhaps 'awk'? Maybe I'm going about this the clumsiest way possible and have hit a wall for a reason.

Thanks for the help!

Edit: The unit of measurement are characters. The command should be able to find the number of lines with more than 80 characters in them.

Answer 1

If you want the number of lines that are longer than 80 characters (your question is missing the units), grep is a good candidate:

grep -c '.\{80\}'

So:

wget -qO - google.com | grep -c '.\{80\}'

outputs 6.

Answer 2

Blue Moon's answer (in its original version) will print the number of fields, not the length of the line. Since the default field separator in awk is ' ' (space) you will get a word count, not the length of the line.

Try this:

wget -q0 - google.com | awk '{ if (length($0) > 80) count++; } END{print count}'

Answer 3

Using awk:

wget -qO - google.com | awk 'NF>80{count++} END{print count}'

This gives 2 as output as there are two lines with more than 80 fields.

If you mean number of characters (I presumed fields based on what you have in the question) then:

wget -qO - google.com | awk 'length($0)>80{c++} END{print c}'

which gives 6.