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Re: simple bash loop problem ...



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 ?

You could mod this to increment:

#!/bin/sh
let count=10
echo -e "Colour test\n"
while [ $count -gt 0 ] ;
do
echo -ne \
"\e[40m\e[30m####"\
"\e[41m\e[31m####"\
"\e[42m\e[32m####"\
"\e[43m\e[33m####"\
"\e[44m\e[34m####"\
"\e[45m\e[35m####"\
"\e[46m\e[36m####"\
"\e[47m\e[37m####"\
"\e[49m\e[39m"\
"\e[1m"\
"\e[40m\e[30m####"\
"\e[41m\e[31m####"\
"\e[42m\e[32m####"\
"\e[43m\e[33m####"\
"\e[44m\e[34m####"\
"\e[45m\e[35m####"\
"\e[46m\e[36m####"\
"\e[47m\e[37m####"\
"\e[49m\e[39m"\
"\e[0m\n"
let count=count-1
done



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