'select subdomains using print command

cat a.txt

a.b.c.d.e.google.com
x.y.z.google.com

rev a.txt | awk -F. '{print $2,$3}' | rev

This is showing:

e google
x google

But I want this output

a.b.c.d.e.google
b.c.d.e.google
c.d.e.google
e.google
x.y.z.google
y.z.google
z.google

Thank You in Advance

linuxawksedprintingcommand


Solution 1:[1]

With your shown samples, please try following awk code. Written and tested in GNU awk should work in any awk.

awk '
BEGIN{
  FS=OFS="."
}
{
  nf=NF
  for(i=1;i<(nf-1);i++){
    print
    $1=""
    sub(/^[[:space:]]*\./,"")
  }
}
' Input_file

Solution 2:[2]

Here is one more awk solution:

awk -F. '{while (!/^[^.]+\.[^.]+$/) {print; sub(/^[^.]+\./, "")}}' file

a.b.c.d.e.google.com
b.c.d.e.google.com
c.d.e.google.com
d.e.google.com
e.google.com
x.y.z.google.com
y.z.google.com
z.google.com

Solution 3:[3]

Using sed

$ sed -En 'p;:a;s/[^.]+\.(.*([^.]+\.){2}[[:alpha:]]+$)/\1/p;ta' input_file
a.b.c.d.e.google.com
b.c.d.e.google.com
c.d.e.google.com
d.e.google.com
e.google.com
x.y.z.google.com
y.z.google.com
z.google.com

Solution 4:[4]

Using bash:

IFS=.
while read -ra a; do
    for ((i=${#a[@]}; i>2; i--)); do
        echo "${a[*]: -i}"
    done
done < a.txt

Gives:

a.b.c.d.e.google.com
b.c.d.e.google.com
c.d.e.google.com
d.e.google.com
e.google.com
x.y.z.google.com
y.z.google.com
z.google.com

(I assume the lack of d.e.google.com in your expected output is typo?)

Solution 5:[5]

For a shorter and arguably simpler solution, you could use Perl.

To auto-split the line on the dot character into the @F array, and then print the range you want:

perl -F'\.' -le 'print join(".", @F[0..$#F-1])' a.txt

-F'\.' will auto-split each input line into the @F array. It will split on the given regular expression, so the dot needs to be escaped to be taken literally.

$#F is the number of elements in the array. So @F[0..$#F-1] is the range of elements from the first one ($F[0]) to the penultimate one. If you wanted to leave out both "google" and "com", you would use @F[0..$#F-2] etc.

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 RavinderSingh13
Solution 2 anubhava
Solution 3
Solution 4 dan
Solution 5 mivk

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