计算字符串中出现的次数。 的Linux

Question

Okay so what I am trying to figure out is how do I count the number of periods in a string and then cut everything up to that point but minus 2. Meaning like this:

string="aaa.bbb.ccc.ddd.google.com"

number_of_periods="5"

number_of_periods=`expr $number_of_periods-2`

string=`echo $string | cut -d"." -f$number_of_periods`

echo $string

result: "aaa.bbb.ccc.ddd"

The way that I was thinking of doing it was sending the string to a text file and then just greping for the number of times like this:

 grep -c "." infile

The reason I don't want to do that is because I want to avoid creating another text file for I do not have permission to do so. It would also be simpler for the code I am trying to build right now.

EDIT

I don't think I made it clear but I want to make finding the number of periods more dynamic because the address I will be looking at will change as the script moves forward.

Answer 1

If you don't need to count the dots, but just remove the penultimate dot and everything afterwards, you can use Bash's built-in string manuipulation .

${string%substring} Deletes shortest match of $substring from back of $string .

Example:

$ string="aaa.bbb.ccc.ddd.google.com"
$ echo ${string%.*.*} 
aaa.bbb.ccc.ddd

Nice and simple and no need for sed , awk or cut !

Answer 2

What about this:

echo "aaa.bbb.ccc.ddd.google.com"|awk 'BEGIN{FS=OFS="."}{NF=NF-2}1'

(further shortened by helpful comment from @steve)

gives:

aaa.bbb.ccc.ddd

The awk command:

    awk 'BEGIN{FS=OFS="."}{NF=NF-2}1'

works by separating the input line into fields ( FS ) by . , then joining them as output ( OFS ) with . , but the number of fields ( NF ) has been reduced by 2. The final 1 in the command is responsible for the print.

This will reduce a given input line by eliminating the last two period separated items.

This approach is " shell-agnostic " :)

Answer 3

Perhaps this will help:

#!/bin/sh

input="aaa.bbb.ccc.ddd.google.com"

number_of_fields=$(echo $input | tr "." "\n" | wc -l)
interesting_fields=$(($number_of_fields-2))

echo $input | cut -d. -f-${interesting_fields}

Answer 4

grep -o "\." <<<"aaa.bbb.ccc.ddd.google.com" | wc -l
5