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I'm having problems with a simple for loop on a newline

 
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# 1  
Old 10-02-2009
I'm having problems with a simple for loop on a newline
Code:
for i in `seq 1 10 ` ; do
        printf $i '\n';
done

gives me this:
Code:
1234567891064mbarch ~ $

(output followed by bash prompt) Smilie

I've tried so many ways to create a newline at the end. Does anyone have any ideas.. Thanks in advance. Sorry
# 2  
Old 10-02-2009
Hi.

Try this usage of printf:

Code:
printf "%d\n" $i

# 3  
Old 10-02-2009
Quote:
Originally Posted by scottn
Hi.

Try this usage of printf:

Code:
printf "%d\n" $i

That didn't do what I wanted. That gave the same effect as using:

Code:
for i in `seq 1 10 ` ; do
        echo $i;
done

I want
Code:
64mbarch ~ $
12345678910

# 4  
Old 10-02-2009
You want something like this:

Code:
for i in $(seq 10); do
    printf '%d' "$i"
    [ "$i" -eq 10 ] && printf '\n'
done

Or rather:

Code:
echo  $(seq 10)

... and with some shells even:
Code:
echo {1..10}

# 5  
Old 10-02-2009
Hi.

Sorry, misread your post.

Put an echo after the done...

Code:
for i in `seq 1 10 ` ; do
        printf $i
done
echo

Last edited by Scott; 10-02-2009 at 12:03 PM..
# 6  
Old 10-02-2009
Quote:
Originally Posted by radoulov
You want something like this:

Code:
for i in $(seq 10); do
    printf '%d' "$i"
    [ "$i" -eq 10 ] && printf '\n'
done

Or rather:

Code:
echo  $(seq 10)

... and with some shells even:
Code:
echo {1..10}

Thanks, that's what I was looking for.

---------- Post updated at 04:35 PM ---------- Previous update was at 03:59 PM ----------
Quote:
Originally Posted by scottn
Hi.

Sorry, misread your post.

Put an echo after the done...

Code:
for i in `seq 1 10 ` ; do
        printf $i
done
echo

That's ok. Sorry for not explaining my post well.

I prefer your solution because I understand it better. Thanks
# 7  
Old 10-03-2009
Quote:
Originally Posted by 64mb
Thanks, that's what I was looking for.

---------- Post updated at 04:35 PM ---------- Previous update was at 03:59 PM ----------



That's ok. Sorry for not explaining my post well.

I prefer your solution because I understand it better. Thanks
One more to go - why use printf?
Code:
for f in $(seq 1 10); do echo -n $f; done; echo

BR,

pen
 
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