CentOS - Jan 2011 - variable in loop

$ ASDF=hello; a=0; a=$(( 70 - $(echo $ASDF | awk '{print length}') ));
echo "$a $ASDF"$(for i in {1..$a}; do printf "."; done)
65 hello.
$ 


Why doesn't it print:
65 hello.................................................................



What am i missing?

On 1/2/11 4:27 PM, S Mathias wrote:
> $ ASDF=hello; a=0; a=$(( 70 - $(echo $ASDF | awk '{print length}')
)); echo "$a $ASDF"$(for i in {1..$a}; do printf "."; done)
> 65 hello.
> $
>
>
> Why doesn't it print:
> 65 hello.................................................................
>
>
>
> What am i missing?
>
Order of operations.  Brace expansion happens before variable substitution 
(echo $i to see the actual value you are getting).

-- 
   Les Mikesell
    lesmikesell at gmail.com

On Sun, Jan 2, 2011 at 2:27 PM, S Mathias <smathias1972 at yahoo.com>
wrote:
> $ ASDF=hello; a=0; a=$(( 70 - $(echo $ASDF | awk '{print length}')
)); echo "$a $ASDF"$(for i in {1..$a}; do printf "."; done)
> 65 hello.
> $
>
> Why doesn't it print:
> 65 hello.................................................................
>
> What am i missing?
>
This is not a group for basic shell programming questions, nor is the
Ubuntu list where you also posted this exact same question.

Please learn to do your own homework and ask list-appropriate
questions in the right lists.

I'm guessing that a little basic netiquette is what you are missing.

From: S Mathias <smathias1972 at yahoo.com>
> $ ASDF=hello; a=0; a=$(( 70 - $(echo $ASDF | awk '{print length}')
)); echo "$a
>$ASDF"$(for i in {1..$a}; do printf "."; done)
> 65 hello.
> Why  doesn't it print:
> 65  hello.................................................................
> What  am i missing?
$ for i in {1..3}; do echo $i; done
1
2
3
$ a=3; for i in {1..$a}; do echo $i; done
{1..3}

Try with:
for ((i=1;i<=$a;i++))

JD