Replace everything in the text file after IP Address

Question

I have a text file in which I have something like this-

10.2.57.44      56538154    3028
120.149.20.197  28909678    3166
10.90.158.161   869126135   6025

In that text file, I have around 400,000 rows exactly as above. I have opened the same text file in Notepad++. I needed a way to remove everything from that text file leaving only IP Address (first column in the above text file is IP Address). I think, I can do that using Regular Expression. And notepad++ also has the option of using Regular Expression. But not sure what regular expression I need to use. Can anyone help me out here?

So output should be something like this-

10.2.57.44
120.149.20.197
10.90.158.161

Answer 1

Find:

(\d+\.\d+\.\d+\.\d+).*

And replace with:

\1

The parentheses will capture that part of the regex to a variable, which is named \\1 (since it is the first capture block in the regex). The rest of the line is not captured but simply thrown out by the replace operation.

Answer 2

Just replace \\s.* with nothing! A regex needs only to match what it needs to match, and what you want to match here is a space followed by anything (thus leaving the IP address alone, since it starts the line and there are no spaces in it)

Alternatively, if you have them, this is even better done with a classical Unix command:

sed -i 's/\s.*//' thefile

Example:

$ cat <<EOF | sed 's/\s.*//'
> 10.2.57.44      56538154    3028
> 120.149.20.197  28909678    3166
> 10.90.158.161   869126135   6025
> EOF
10.2.57.44
120.149.20.197
10.90.158.161
$ 

Answer 3

If you have gawk or something:

gawk "{print $1}" filename

In linux(bash) usage of awk will look like following:

suku@ubuntu-vm:~$ cat stack 
10.2.57.44      56538154    3028
120.149.20.197  28909678    3166
10.90.158.161   869126135   6025
suku@ubuntu-vm:~$ cat stack | awk '{ print $1 }'
10.2.57.44
120.149.20.197
10.90.158.161