Re: simple bash loop problem ...

[Shaun ONeil <shaun@dog-berry.com>]

On Sat, 2003-06-28 at 10:03, David selby wrote:
> Hello,
> 
> I am writing bash a bash & sed script, it has been going suprisingly 
> well. I need a loop to count 9 times & the variable n to the count ..
> 
> for n=1 to 9
> ....
> next
> 
> kind of thing, but this is not BASIC !!
> 
> My best guess is
> 
> declare -i n=1
> while [ $n < 9 ]; do
> .....
> n=$((n+=1))
> done
> 
> All i get is ...
> 
> web@debian:/usr/local/myfiles/dave/websites/kcards$ ./gensite
> ./gensite: 9: No such file or directory
> 
> I have defined it as an integer, used the less than operator for 
> integers, ... errr ... I know its something stupid but I can't crack it ....
> 
> Dave
> 
> PS is there a more ellagent way to do a counted loop as well as a way 
> that works ?
>  
> 
> 
> 

$ seq --help
Usage: seq [OPTION]... LAST
  or:  seq [OPTION]... FIRST LAST
  or:  seq [OPTION]... FIRST INCREMENT LAST
Print numbers from FIRST to LAST, in steps of INCREMENT.
<snip>

$ for n in `seq 1 9` ; do echo $n ; done
1
2
3
4
5
6
7
8
9

Note the backticks on seq and you should be good to go.

regards,
 Shaun.