how to use basic linux command to count the number of dot in every line then insert the value in each line in bash

Question

I process a text file with many domain like

abcd.efgh.ijkl.mnop

I hope to add a 3 at the same line no matter how it is at the beginning or at the end.

current I use

sed 's/[^.]//g' thetextfile | awk '{ print length }' > thenumber

then I use librecalc to combine them.

Answer 1

The blindingly obvious answer is

awk -F . '{ print NF-1, $0 }' thetextfile >thenumber

Combining the results to a single file in an external utility seems like a humongous overcomplication; just process all the files in one go.

awk -F . '{ print NF-1, $0 }' * >allnumbers

or if your needs are more complex maybe something like

for file in *; do
    echo "$0: processing $file ..." >&2
    awk -F . '{ print NF-1, $0 }' "$file"
done >allnumbers

The -F. says to use the literal dot as field separator, and NF is the number of fields (which will be one more than the number of separators; look up "fencepost error" if you don't understand why).

In more complex cases still, maybe append the output from each file with >>allnumbers inside the loop if you really need to (but understand that this is less efficient than only writing once after the done ).

Awk's default output field separator is a single space; if you want to produce TSV (tab-separated output) or CSV, you can set OFS="\t" or OFS="," as required.

Answer 2

This should do it:

while read z;do echo $(echo -n "$z"|sed 's/[^\.]//g'|wc -m) $z;done < input.txt

which will produce something like:

3 abcd.efgh.ijkl.mnop
6 abcd.efgh.ijkl.mnop...
4 abcd.efgh.ijkl.mnop.

Answer 3

Using a perl one-liner, taking advantage of tr returning the number of changes:

$ perl -ne 'print tr/././, " ", $_' input.txt
3 abcd.efgh.ijkl.mnop
6 abcd.efgh.ijkl.mnop...
4 abcd.efgh.ijkl.mnop.

Answer 4

In plain bash , without using any external utility:

#!/bin/bash

while read -r line; do
    dots=${line//[^.]}
    printf '%d\t%s\n' ${#dots} "$line" 
done < file