Extract lines that have two or more dots

Question

I need to extract (or count) the lines (in a file) that have two or more dots. The lines should not start with dot (it’s OK if they end with a dot), and there must not be two dots in a row (i.e., the dots are all separated with non-dot characters).

Output Example:

a.b.
a.b.com
a.b.c.
a.b.c.com

But not:

a.com
a..b
a.b.c..d

I did this command:

grep -P '^[^.]+\.([^.]+\.)+[.]+' file.txt | wc -l

but it didn't find any matching lines.  How should I do this?

Answer 1

1. \. and [.] are equivalent — they both match a literal dot, and not any other character.  As a matter of style, pick one and use it consistently.
2. Your problem is that your regular expression (i.e., pattern) has ([^.]+\.)+ followed by [.]+.  That’s really (sort of) equivalent to [^.]+\. followed by [.], with the result that your grep is looking for lines that contain text.text.., i.e., two dots in a row.  If you check, you’ll see that your command matches a.b...
3. OK, I believe that the fix is fairly simple:

   grep -P '^[^.]+\.([^.]+\.)+[^.]*$'

   I.e., change the [.] to [^.] (perhaps that’s what you meant originally?), change the following + to an *, and add a $.  After some number of text. groups, require/allow any number (zero or more) characters other than dot, up to the end of the line.
4. An even simpler approach (easier to understand) would be

   grep -P '^[^.]+\..*\.' file.txt | grep -v '\.\.'

   The first grep finds lines that begin with a non-dot character and include at least two dots.  The second grep removes lines that have two consecutive dots.
5. Rather than do grep … | wc -l, just do grep -c ….

Answer 2

Using awk:

$ cat file
.com
.c.c.c.c
a.b.
a.b.com
a.b.c.
a.b.c.com
a.com
a..b
a.b.c..d

$ awk -F . 'NF > 2 && !/^\./ && !/\.\./' file
a.b.
a.b.com
a.b.c.
a.b.c.com

The awk program here uses the dot as a field separator. A line having two or more dots is the same as a line having more than two fields. This is what the NF > 2 test tests. The first regular expression discards lines that starts with a dot, and the second regular expression discards lines that contain two dots or more in a row. The rest of the lines are printed.

The same thing with grep:

grep '\..*\.' file | grep -v -e '^\.' -e '\.\.'

The first expression extracts lines that contain at least two dots, and the two others delete line that start with a dot or contains two consecutive dots.

Or with sed,

sed -n '/^\./d; /\.\./d; /\..*\./p' file

Answer 3

You can do this with lookarounds, like as shown:

$ grep -Pc '^(?!\.)(?!.*\.\..*)(?=.*\..*\.)' file.txt

To be read as:

• I am standing at the beginning of the line and looking to my right ^
• I see the line not begin with a literal dot (?!\.)
• On looking further ahead nowhere do I see two consecutive literal dots (?!.*\.\..*)
• But I do see two dots, but since consecutive dots were ruled out in my previous lookahead, that means these two dots MUST be separated by at least one non-dot character (?=.*\..*\.)
• Q.E.D.