如何将regex replace中捕获的模式转换为大写?

我想替换 /test/test1 具有 TEST1: .这就是我的出发点:

extern crate regex; // 1.0.1

use regex::Regex;

fn main() {
    let regex_path_without_dot = Regex::new(r#"/test/(\w+)/"#).unwrap();

    let input = "/test/test1/test2/";

    // Results in "test1:test2/"
    let result = regex_path_without_dot.replace_all(input, "$1:");
}

我试过用

let result = regex_path_without_dot.replace_all(&input, "$1:".to_uppercase());

但我有个错误:

error[E0277]: the trait bound `for<'r, 's> std::string::String: std::ops::FnMut<(&'r regex::Captures<'s>,)>` is not satisfied
  --> src/main.rs:10:41
   |
10 |     let result = regex_path_without_dot.replace_all(&input, "$1:".to_uppercase());
   |                                         ^^^^^^^^^^^ the trait `for<'r, 's> std::ops::FnMut<(&'r regex::Captures<'s>,)>` is not implemented for `std::string::String`
   |
   = note: required because of the requirements on the impl of `regex::Replacer` for `std::string::String`

我如何实现这一要求的特质?有什么简单的方法可以做到这一点吗?