Trying to take a string and break each letter into a separate variable

Shell Programming and Scripting
1

I am trying to make a script that takes a word and each letter up and turns it into a separate variable. My code currently does not work but I feel I just need to tweak one thing that I am unsure of.

(ex: if forum was typed in letter1=f; letter2=o; letter3=r;...)

Thank you

count=1;

for i in `echo $1 | sed 's/./& /g'`
do
        echo $i
        letter$count=$i
        count=`expr $count + 1`
done


echo $var1
2

Hi!
I hope this may help,

for x in $(seq 1 ${#mystr}) do                
eval "echo letter$x=${mystr:$((x-1)):1}"
done

It uses parameter substitution which is very handy when working with strings, and also some goodies like seq and arithmetics.
Best regards,
Lakris

1 Like
3

The following is kind of what you were going for, though Lakris' example above is why I come to this site, now I can spend Saturday examining the ${#var} (length of var ?)and ${var:$offset:$length} (substring ?) functions.

#!/bin/bash

for i in $(echo $1 | sed 's/./& /g'); do
        letters[$count]=$i
        count=$(($count+1))
done
for i in $(seq 0 $((${#letters[@]}-1))) ; do
        echo $i letter is ${letters[$i]}
done
4

:wink:
But all credit should go to where i found it, 10 Steps to Beautiful Shell Scripts .
A wonderful site that can keep You going through sunday as well.

/Lakris

5

When I try to do this, it doesn't save anything into the variables. it leaves them all blank

6

Looks good to me. What does your output look like?

$
$
$ cat -n var.sh
     1  #!/bin/bash
     2
     3  for i in $(echo $1 | sed 's/./& /g'); do
     4          letters[$count]=$i
     5          count=$(($count+1))
     6  done
     7  for i in $(seq 0 $((${#letters[@]}-1))) ; do
     8          echo $i letter is ${letters[$i]}
     9  done
    10
$
$
$ ./var.sh forum
0 letter is f
1 letter is o
2 letter is r
3 letter is u
4 letter is m
$
$
$ echo $SHELL
/bin/bash
$
$

tyler_durden

7

I am trying to echo out

echo letter1
echo letter2
echo letter3

and it returns nothing

I am unsure about what this does:

for i in $(seq 0 $((${#letters[@]}-1))) ; do
echo $i letter is ${letters[$i]}
done

but it appears that it does not save the letters into seperate variables? Sorry for the confusion

8

Either that eval serves no purpose or that echo is unintended, but both of them together result in a pointlessly convoluted printing statement that is vulnerable to shell metacharacters.

echo "letter$x=${mystr:$((x-1)):1}" accomplishes the same thing in a safe, simple manner.

Anyone not using linux may find that their system does not have seq, a non-standard gnu tool. Also, the use of echo can be problematic. Aside from the lack of portability (not all echoes support the same options), if the value of the parameter expansion begins with a dash, it can cause problems. printf is a superior tool when dealing with unknown data.

To simply print out the letters one per line:

w=$word
while [ "$w" ]; do
    printf %c\\n "$w"
    w=${w#?}
done

To assign each letter to a variable whose name ends with an incrementing number (starting with 1):

w=$word i=0
while [ "$w" ]; do
    eval "letter$((i+=1))=\${w%\$w{#?}}"
    w=${w#?}
done

Regards,
Alister