#!/usr/bin/ksh
i=1
while [ "$i" -lt 121 ]
do
if [ $i -lt 100 ]
then
if [ $i -lt 10 ]
then
i=00$i
else
i=0$1
fi
fi
echo "fla${i}"
i=' expr $i+1 '
done
exit 0
Why does this script result in an error "Too many arguments"?
i=' expr $i+1 ' is not incrementing i, it assigns i a value of ' expr $i+1 '.
On the next iteration of the loop you run [ $i -lt 100 ]. Since i is not embraced in double quotes, this expands to [ expr '$i+1' -lt 100 ]. [ is actually a command, and you have given it too many arguments due to the reason above.
If you replace i=' expr $i+1 ' with i=$(($i + 1)), your code should work.
Edit:
It seems that at least Bash will have issues with the number 008, it interprets it as octal. You need to assign 00$i, 0$i and $i to another variable (or echo "fla00$i" etc) if you get some error after 008.
In this case, you'll need to do something when i >= 100.
This is what I would do:
i=1
while [ $i -lt 121 ]; do
if [ $i -ge 100 ]; then
echo fla$i
elif [ $i -ge 10 ]; then
echo fla0$i
else
echo fla00$i
fi
i=$(($i + 1))
done
elif instead of else if
Your entire script intention is to do exactly this: seq -f 'fla%03.0f' 1 128
As to the shell programming:
i=0$1 should probably become i="0$i"
' expr $i+1 ' should become $( expr "$i" + 1 )
Note the spaces required by expr. This won't work $( expr "$i"+1 )
printf 'fla%03d\n' {1..121} (or in bash zsh just printf '%s\n' fla{001..121} )
i=$((i+1)) or ((i++)) and let ksh handle the addition
[and[[set -xto the shebang line to enable verbose debug mode to see what exactly is happening while executing the script.for i in {1..128}; do printf "fla%03d\n" $i; doneprintf "fla%03d\n" {1..128}