How to find number of `_`-separated words?

Question

I want to make a Unix command that reads a string and counts how many words it contains. In my case the words are separated by a _ (underscore) character, not spaces.

For example, TABLE_PERSON contains two words, so the command should print "2".

How can I do that?

Answer 1

If your string does not contain spaces you can substitute the _ with a space (with tr (translate character) or sed (stream editor)) and use wc (word count) to count the words:

 $ echo TABLE_PERSON | tr '_' ' ' | wc -w
 2

or

 $ echo TABLE_PERSON | sed 's/_/ /' | wc -w
 2

if your string contains spaces which do not have to be considered as a separator (e.g., if "TABLE_PERSON 2" counts as two words separated by _: TABLE and PERSON 2) you can first remove the spaces with any character (e.g., -) and use the same command:

 $ echo 'TABLE_PERSON 2' | tr ' ' '-' | tr '_' ' ' | wc -w
 2

Answer 2

Many ways, most of which are easy to figure out:

#!/bin/sh
n=$(tr -dc "_" | wc -c)
echo $((n+1))

#!/bin/sh
tr " _" "x " | wc -w

#!/usr/bin/env bash
read -r string
len1=${#string}
string=${string//_/}
len2=${#string}
echo $((len1-len2+1))

#!/usr/bin/env perl
print (scalar split /_/, <STDIN>)."\n";

#!/usr/bin/env python3
import sys
print(len(sys.stdin.readline().split("_")))

#!/usr/bin/env ruby
puts gets.split(/_/).length

#include <stdio.h>
int main(void) {
    int c, n = 1;
    for (;;) {
        c = getchar();
        if (c == EOF || c == '\n')
            break;
        else if (c == '_')
            n++;
    }
    printf("%u\n", n);
    return 0;
}