[SOLVED] bash for loop iteration question.

ncalsmitty1369 · 02-04-2010, 08:12 PM

Hi,

I am having a problem using the following for loop configuration in a bash shell:

array=( 1 2 3 )
num=${#array[*]}

for x in {1..$num}
do
echo $x
done

Instead of returning:
1
2
3
I get:
{1..3}

Can someone tell me what I am doing wrong?

Thanks!

pixellany · 02-04-2010, 08:58 PM

I don't know if your construct is **supposed** to work...if it is valid, it might need some quotes.

This works:

Code:

for x in $(seq 1 $num); do
    echo $x
done

colucix · 02-05-2010, 07:28 AM

Hi ncalsmitty1369 and welcome to LQ!

Answers to your question, here. And for the reasons explained in that thread, I second the suggestion given by pixellany.

ghostdog74 · 02-05-2010, 07:55 AM

Quote:

Originally Posted by ncalsmitty1369

Hi,

I am having a problem using the following for loop configuration in a bash shell:

array=( 1 2 3 )
num=${#array[*]}

for x in {1..$num}
do
echo $x
done

Instead of returning:
1
2
3
I get:
{1..3}

Can someone tell me what I am doing wrong?

Thanks!

$num will not be expanded in brace expansion. you need to do eval

Code:

for i in $(eval echo {1..$num})
do
 echo $i
done

ncalsmitty1369 · 02-05-2010, 05:50 PM

Quote:

Originally Posted by colucix

Hi ncalsmitty1369 and welcome to LQ!

Answers to your question, here. And for the reasons explained in that thread, I second the suggestion given by pixellany.

Thank you colucix for the welcome and the link to the other thread! Good information to know.

I like the solution provided by pixellany and will use if going forward.

Thanks to all!