从bash脚本中的变量添加两个字母

Question

im working to make my code take an input (word) and output the sum of all the letters in the input, the letters would be equal to there numeric value- a=1 b=2 c=3 etc,here is my unfinished code so far:-

echo enter word
read word
for let in $word
do
echo $let

*here is where it should take the input and calculate the output (w+o+r+d = ??)

Answer 1

Here's a solution that uses an associative array to map (English) letters to their ordinal values. Note that associative arrays require bash 4.0 or higher .

#!/usr/bin/env bash

# Declare variables.
declare -i i sum  # -i declares integer variables
declare -A letterValues # -A declares associative arrays, which requires bash 4+
declare letter # a regular (string) variable

# Create a mapping between letters and their values
# using an associative array.
# The sequence brace expression {a..z} expands to 'a b c ... z'
# $((++i)) increments variable $i and returns the new value.
i=0
for letter in {a..z}; do
    letterValues[$letter]=$((++i))
done

# Prompt for a word.
read -p 'Enter word: ' word

# Loop over all chars. in the word
# and sum up the individual letter values.
sum=0
for (( i = 0; i < ${#word}; i++ )); do
  # Extract the substring of length 1 (the letter) at position $i.
  # Substring indices are 0-based.
  letter=${word:i:1}
  # Note that due to having declared $sum with -i,
  # surrounding the following statement with (( ... ))
  # is optional.
  sum+=letterValues[$letter]
done

# Output the result.
echo "Sum of letter values: $sum"

Answer 2

To iterate over the characters of a string, do this:

string="hello world"
for ((i=0; i < ${#string}; i++)); do
    char=${string:i:1}       # substring starting at $i, of length 1
    echo "$i -> '$char'"
done

0 -> 'h'
1 -> 'e'
2 -> 'l'
3 -> 'l'
4 -> 'o'
5 -> ' '
6 -> 'w'
7 -> 'o'
8 -> 'r'
9 -> 'l'
10 -> 'd'