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Dr_Death_UAE 03-01-2013 07:30 PM
Q: count inside a loop
 
hello

i am trying to write a for-loop that doesn't start from 1, but the loop count start from 1

Code:
for (( i=$a; i <= $b; i++ )); do echo "$x:$i"; done
where $a and $b always have different value
i want the result be like this:
Quote:
1: 10
2: 11
3: 12
4: 13
5: 14
6: 15
7: 16
8: 17
9: 18
10: 19
11: 20
any idea, thanks

kooru 03-02-2013 03:24 AM
If I understand well..

Code:
#!/bin/bash

x=1
a=10
b=20

for i in `seq $a $b`
do
echo "$x: $i"
x=$((x+1))
done

H_TeXMeX_H 03-02-2013 05:04 AM
Or even simpler:

Code:
bash-4.2$ for i in $(seq 1 11); do printf '%d: %d\n' "$i" "$(($i + 9))"; done
1: 10
2: 11
3: 12
4: 13
5: 14
6: 15
7: 16
8: 17
9: 18
10: 19
11: 20

Dr_Death_UAE 03-02-2013 07:04 AM
thanks kooru,
Quote:
x=$((x+1))
did the trick :)

chrism01 03-04-2013 12:15 AM
This is a simple way
Code:
for (( x=1,i=10 ; x <= 10, i<=20; x++, i++))
do
    echo "$x:$i"
done

# ./t.sh
1:10
2:11
3:12
4:13
5:14
6:15
7:16
8:17
9:18
10:19
11:20
:)

Dr_Death_UAE 03-04-2013 12:20 AM
Quote:
Originally Posted by chrism01 (Post 4904073)
This is a simple way
Code:
for (( x=1,i=10 ; x <= 10, i<=20; x++, i++))
do
    echo "$x:$i"
done

# ./t.sh
1:10
2:11
3:12
4:13
5:14
6:15
7:16
8:17
9:18
10:19
11:20
:)
thanks chrism01, i never thought it's possible to do it in this way

thanks again :hattip:

chrism01 03-04-2013 12:24 AM
No worries; glad to help :)


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