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up vote -1 down vote favorite

I have a variable into a do loop. I append its content to a file, but I want to append it to itself so I could use it to do other stuff like sending mail.

I tried variable+=$variable but it didn't work.

I want to have access to the variable outside the do...done

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closed as unclear what you're asking by Gilles, jasonwryan, slm♦, rahmu, Chris Down Dec 17 '13 at 3:20

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.
    
The answer is here: stackoverflow.com/questions/4181703/… AND here stackoverflow.com/questions/2250131/… - Cheers :) –  nwildner Dec 16 '13 at 18:03
1  
What do you mean by “didn't work”? Did you have an error message? Was the variable's resulting value not what you expecting? Post your code. –  Gilles Dec 16 '13 at 23:04
    
Why not just variable="$variable$variable"? I'll bet it's more portable than +=. –  Joseph R. Dec 17 '13 at 0:02

2 Answers 2

up vote 2 down vote

I bet your loop is part of a pipeline

seq 5 | while read num; do x+=$num; done; echo $x 
# expect "12345", actually see ""

In bash, when you construct a pipeline, it spawns subshells for all the parts. When the subshell exits, any variables you modify within it are destroyed too.

You have to code more carefully to ensure you use the variable in the same shell where you modify it.

This example echoes the var in the same subshell:

$ seq 5 | { while read num; do x+=$num; done; echo $x; }
12345

This example uses process substitution so the loop runs in the current shell

$ while read num; do x+=$num; done < <(seq 5)
$ echo $x
12345
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up vote 0 down vote

If these are just strings you can append like this:

Example

$ somevar="a string"
$ echo $somevar
a string

$ somevar="$somevar$somevar"

$ echo $somevar
a stringa string

Loops

You can use the same technique in a for loop in Bash.

$ a="0"; for i in $(seq 3); do a="$a$i"; echo "$i | $a"; done
1 | 01
2 | 012
3 | 0123
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