Shell – How to print lines containing 3 words

shelltext processing

I would like to print lines in my file which containing 3 words seperated my 2 spaces.

for example:

AAA BBB CCC
BB AA
CCCCCCCC

only AAA BBB CCC sshould be printed.

I already use grep for it but I cant handle the spaces.

try

 awk 'NF==3' file.txt

this will grep line with 3 field (NF).

With GNU grep:

grep -o '\b\w*\.swf\b' file

Output:

example.swf
example2.swf
example3.swf

\b: a zero-width word boundary

\w: word character

\.: match one dot

grep -Exv '[A-Za-z]*([A-Z][a-z]|[a-z][A-Z])[A-Za-z]*'

Explanation

The idea is to match the opposite of what you want first, i.e. those lines that contain only upper- and lower-case letters. This uses grep -Ex, i.e. grep with extended regex, match the whole line. The -v flag then negates the regex, i.e. return those lines that do not match the following regex.
The central part ([A-Z][a-z]|[a-z][A-Z]) matches a single upper-case letter followed by a lower-case letter, or vice versa.
The outer part [A-Za-z]*...[A-Za-z]* means that the rest of the line must comprise upper- or lower-case letters only.