Question

Sorry if the question is very straight forward but am a newbie to shell scripting.
I am trying to write something like this :

for i in {1..20}
do
   curl "something $i ........ -d  'something "$i" something' "
done

The problem is that the second $i inside the single quotes part ” is not being replaced. What should be done to get it working ?

Answer 1

As said above, parameters are not expanded inside single quotes, you have to use double quotes. The only point is that since it occurs in a already double-quoted string, you have to escape them with a backslash (\), like this:

$ foo=bar
$ eval "echo \"something \\\"$foo\\\"\""
something "bar"

Note that there are three \ before the innermost ", as this will be expanded twice (once when evaluating the argument of eval and once when evaluating the argument of echo)