variable in loop

[S Mathias]

$ 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?

[Sam Sharpe]

On 2 January 2011 22:27, 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.................................................................

Because {1..$a} doesn't output the sequence 1..65 like you think it
should. Replacing it with `seq 1 $a` does.
> What am i missing?

To be honest... the point of this list - elementary Bash programming
isn't a matter for the Fedora Users list.

[Chris Tyler]

On Sun, 2011-01-02 at 14:27 -0800, 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?

The expression {1..$a} is being quoted and treated as a string, instead
of being evaluated. If you change the printf from:

printf "."

...to:

printf "[$i]"

...you'll see what's happening.

-Chris

[NoSpaze]

On Sun, 2011-01-02 at 14:27 -0800, 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.................................................................

You wrote:  for i in {1..$a}
I think you mean: for ((i=1;i<$a;i++))

Cheers.
----------------------------------------------
Rodolfo Alcazar Portillo - [e-mail]
otbits.blogspot.com / counter.li.org: #367962
----------------------------------------------
"Früher oder später emulieren wir euch."
- linux.de

[Roberto Ragusa]

Rodolfo Alcazar Portillo wrote:
> On Sun, 2011-01-02 at 14:27 -0800, 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.................................................................
>
> You wrote:  for i in {1..$a}
> I think you mean: for ((i=1;i<$a;i++))

Right, but with

i=0

or

i<=$a

to have an accurate count ;-)