AWK: how to select lines by the number of words in one field

awktext processing

Here's a text file I have:

1|this|1000
2|that|2000
3|hello|3000
4|hello world|4000
5|lucky you|5000
6|awk is awesome|6000
.
.
.

How do I only print the lines that have two and only two words (line 4 and 5) in the $2?

This is what I have tried but it counts the number of letters instead of words:

awk -F"|" '{if(length($2==2) print $0}'

You can use the return value of the awk split function:

$ awk -F'|' 'split($2,a,"[ \t]+") == 2' file
4|hello world|4000
5|lucky you|5000

awk solution:

awk 'v && NR==n{ print $6,v > "result.txt" }/^!/{ v=$5; n=NR+1 }' file

<condition1> { <statement> ... }<condition2>{ <statement> ... } - conditions with respective statements will be evaluated consecutively
/^!/{ v=$5; n=NR+1 } - on encountering line starting with ! - capture the 5th field value $5 and plan the next line number NR+1 (assigning to variable n)
v && NR==n - if we have the 1st crucial number v and the current record number NR is the needed "next line number" n - print the values into file result.txt

The result.txt file contents:

188 -9744.24963670
140 -9744.30001681
155 -9744.33953891
164 -9744.36584201
154 -9744.37925372
153 -9744.39185493
160 -9744.39836617