bash: local: `0': not a valid identifier - bash

Question

When I call for example encrypt HI the result is:
bash: local: `0': not a valid identifier
bash: local: `1': not a valid identifier
Here is the code:
#!/bin/bash
encrypt(){
local s="$1";
local lenght=${#s};
local i=0
while [ $i -lt $lenght ]
do
local j=1
local letter=expr substr $s $i $j;
letterToNumber $letter;
echo $number;
i=$[$i+$j];
done
}

Answer 1

> local letter=expr substr $s $i $j;
Looks like this is your problem. This is a newbie mistake where apparently you meant to write
local letter=$(expr substr "$s" "$i" "$j")
Newcomers sometimes seem to have a hard time with shell syntax. The shell simply tokenizes your command from the left; the first token which does not contain = is a keyword or a command, and (in the general case) the rest of the tokens are just text to pass as arguments to that command. So letter=expr substr will assign the value expr to the variable letter and (try to) run the command substr; and local foo=bar baz quux will declare the variables foo, baz, and quux as local, and assign the value bar to foo while we are at it.
Using expr is very rarely what you want to do in a modern shell script. Bash has a built-in substring operator; ${s:$i-1:$j}
encrypt () {
for((i=0; i<${#1}; ++i)); do
letterToNumber "${1:i:1}"
done
}
(I am guessing you also meant to say number=$(letterToNumber "${1:i:1}"); echo "$number" but that's just a useless use of echo.)

Answer 2

tripleee is right.
And though I do not have your letterToNumber func/program, I think there's 2 other problems:
#!/bin/bash
#set -v # always handy
#set -e # handy too
encrypt(){
local s="$1";
local lenght=${#s};
local i=1 # <-------- start at 1, not 0
while [ $i -le $lenght ] # <-------- le, not lt
do
local j=1
local letter=$(expr substr $s $i $j);
echo letter $letter
#letterToNumber $letter;
echo $number;
i=$[$i+$j];
done
}
encrypt HELLO
Output:
$ ./foobar.sh
letter H
letter E
letter L
letter L
letter O

Answer 3

Instead of local letter=expr substr $s $i $j;, which is incorrect (to know why, see answer by tripleee), you could do:
# Insert this line before `local s="$1";` or anywhere else before while loop
local letter
# Inside the "while loop"...
letter=${s:$i:$j}
To calculate i=i+j, its better to use shorter (than yourth) BASH internal math syntax:
# Replace `local i=0` with this line:
local -i i=0 j=1
# i=i+j . Note: you dont need "$" here!
i+=j
Finally overall encrypt function code will be:
encrypt () {
local s="$1" letter length=${#s}
local -i i=0 j=1
for ((; i<length; i+=j )); do
letter=${s:$i:$j}
letterToNumber "$letter"
echo $number
done
return 0
}
This code may be replaced by the simplest one-liner, but... you dont answer (yet) about how to do that :)