As shown here, for
defaults to $@
if no in seq
is given. The for i
assigns your $i
variable.
"for i" without "in [sequence]" ending while using getopt
Question
I've found this example script for using getopt command in shell.
#!/bin/bash
args=$(getopt ab $*)
set -- $args
for i;
do
case "$i" in
-a)shift; echo "it was a";;
-b)shift; echo "it was b";;
esac;
done
It works well, but I don't understand where is variable $i assigned. How it knows that it must iterate through $arg. Can you explain this?
Solution
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