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# 1  
Old 03-03-2009
for cycle question
i have a question how to modify below script to generate the expect result below :

test.sh
#!/bin/bash
for ((i=0; i < 25; i++))
do
echo $1$i
done

current result:
test.sh 20090101
200901010
200901011
200901012
200901013
200901014
200901015
200901016
200901017
200901018
200901019
2009010110
2009010111
2009010112
.
.
.
2009010124

expect result:
test.sh 20090101
2009010100
2009010101
2009010102
2009010103
2009010104
2009010105
2009010106
2009010107
2009010108
2009010109
2009010110
2009010111
2009010112
.
.
.

2009010124

Thanks
# 2  
Old 03-03-2009
Use printf to format the value and this loop should be more portable:

Code:
#!/bin/bash

i=0
while [ $i -le 25 ];
do
  printf "%s%02d\n" $1 $i 
  let "i=i+1"
done

Regards
# 3  
Old 03-03-2009
Quote:
Originally Posted by Franklin52
Use printf to format the value and this loop should be more portable:

Code:
#!/bin/bash

i=0
while [ $i -le 25 ];
do
  printf "%s%02d\n" $1 $i 
  let "i=i+1"


That is not portable. The portable syntax for arithmetic is:

Code:
i=$(( $i + 1 ))

Quote:
Code:
done

Regards
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