rust - 从功能获得的mod中获取结构

Question

我如何从函数结果的mod中共享结构，以便每次我需要环境变量时都不会重新运行env::get()函数。
src/main:

mod env;

async fn main() {
    println!("{}", env::get().profile);
    println!("{}", env::get().redis);
}

src/env:

use serde::Deserialize;
extern crate dotenv;

#[derive(Deserialize, Debug)]
pub struct Config {
    pub redis: String,
    pub profile: String,
}

pub fn get() -> Config {
    dotenv::dotenv().expect("Failed to read .env file");

    let env = match envy::from_env::<Config>() {
        Ok(config) => config,
        Err(e) => panic!("{:#?}", e),
    };

    env
}

如果我按照mccarton的建议使用lazy_static，如何在main中获取ENV？
src/env:

 use serde::Deserialize;
    extern crate dotenv;
    
#[derive(Deserialize, Debug)]
pub struct Config {
    pub redis: String,
    pub profile: String,
}

pub fn get() -> Config {
    dotenv::dotenv().expect("Failed to read .env file");

    let env = match envy::from_env::<Config>() {
        Ok(config) => config,
        Err(e) => panic!("{:#?}", e),
    };

    env
}

lazy_static! {
    static ref ENV: Config = {
        dotenv::dotenv().expect("Failed to read .env file");

        let env = match envy::from_env::<Config>() {
            Ok(config) => config,
            Err(e) => panic!("{:#?}", e),
        };

        env
    };
}

Answer 1

这里是:

use serde::Deserialize;
extern crate dotenv;

#[derive(Deserialize, Debug)]
pub struct EnvConfig {
    pub redis: String,
    pub profile: String,
}

lazy_static! {
    pub static ref ENV: EnvConfig = {
        dotenv::dotenv().expect("Failed to read .env file");

        let env = match envy::from_env::<EnvConfig>() {
            Ok(config) => config,
            Err(e) => panic!("{:#?}", e),
        };

        env
    };
}

pub fn get() -> &'static ENV {
    &ENV
}