bash : replace variable value inside ' '
shell, variables
Solution
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`)
Problem
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 ?