Re: simple bash loop problem ...
* David selby <email@example.com> [030629 18:42]:
> 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
> kind of thing, but this is not BASIC !!
> My best guess is
> declare -i n=1
> while [ $n < 9 ]; do
> All i get is ...
> web@debian:/usr/local/myfiles/dave/websites/kcards$ ./gensite
> ./gensite: 9: No such file or directory
> I have defined it as an integer, used the less than operator for
I think you need to use -lt
> integers, ... errr ... I know its something stupid but I can't crack it ....
I would do something like:
while [ $i -lt $imax ] ; do
echo "i = $i"
i=`expr $i + 1`
> PS is there a more ellagent way to do a counted loop as well as a way
> that works ?
Well the above is not so elegant but does work with a standard bourne
shell, which is important for me.
Linux twofish 2.4.21-looxt93c #1 Thu Jun 26 15:38:09 JST 2003