How to output a simple string that contains % symbol?

[el-hoshino]

For example, I need a string like this:

R.string.urls.url() → "http%3A%2F%2Fwww.abc.xyz"

The tricky point is that R.swift only uses String(fromat: ...) initializer while there are parameters in the given string, so:

1. If I apply the values like "url" = "http%%3A%%2F%%2Fwww.abc.xyz";, R.swift doesn't see any parameters, so treats it as a normal string, so the output is like "http%%3A%%2F%%2Fwww.abc.xyz"
2. If I apply the value like "url" = "http%3A%2F%2Fwww.abc.xyz";, R.swift takes strings like %3A as parameters, and asks me to use the method like R.string.urls.url(Double, Double, Double)

One of the workarounds that my friend came out with is to use a dummy parameter like "url" = "http%%3A%%2F%%2Fwww.abc.xyz%@";, so we can use R.string.urls.url("") and the output is just what I wanted. But obviously it's just a workaround.

Is there any better way?

[digitalby]

I ended up doing this workaround to escape the percent sign in a string:
%\U200B