Ohai, got another Bash problem.
for i in ${#input}; do echo ${input:$i:1}; done
works in the interactive bash, but not in a shell script. Input is a variable, based on an argument. It's set properly. Any ideas?
2 Answers
Not sure how it is working in interactive shell for you, this expression:
${#input}
- will return number of characters in the first element if
inputis array variable
otherwise - will return number of characters in the
inputvariable
EDIT: based on your comments
I think you are trying to do this code:
for ((i=0; i<${#input}; i++))
do
echo ${input:$i:1}
done
This code will iterate for each character in input string and then echo each character.
3 Comments
anubhava
But what is
$i in this case? As per your code for loop will execute only once.anubhava
@thkala: lol :) I wish I had such power. But
for i in ${#input}; made no sense and as per his code it just appeared to me perhaps he/she's trying to print input string character by character (of course I may be completely wrong).
braindump
Yep, that's what I try to do. $i is the counter of a for loop.
I cannot see the purpose of this piece of code - it's equivalent to a single echo.
Explanation:
${#input} is the length of the contents of the input variable. It is always a single number, hence the loop will always be executed once. $i is then modified to contain that variable length.
Then ${input:$i:1} means "1 character from the input variable, starting at offset $i, which is equal to the length of the variable, and, therefore, past its end. I.e. ${input:$i:1} is always an empty string. So your code can be simplified to:
echo
As an example, if $input is foobar, then ${#input} is 6, and your loop executes once this command: echo ${input:6:1}. Offsets here are zero-based, therefore 6 is just past the last character of $input. If you are trying to get the last character of a variable, try this:
$ input=foobar; echo ${input:${#input}-1:1}
r
What exactly are you trying to do?
#!/bin/shor#!/bin/bash?