如何在Typescript中导出对象

我正在将现有的JavaScript项目转换为类型脚本.我不知道如何导出对象来获取此javscript输出

const path = require('path'),
      rootPath = path.normalize(__dirname + '/..'),
      env = process.env.NODE_ENV || 'development';

let config = {
  development: {
    amqpUrl: "amqp://localhost:15672",
    root: rootPath

  },
  test: {
    amqpUrl: "amqp://localhost:5672",
    root: rootPath

  },
  production: {
    amqpUrl: "amqp://localhost:5672",
    root: rootPath

  }
};
module.exports = config[env];

我写了如下的打字稿,但不清楚导出

import path = require("path")

    const rootPath = path.normalize(__dirname + '/..')
    const env = process.env.NODE_ENV || 'development'

    let config = {
      development: {
        amqpUrl: "amqp://localhost:15672",
        root: rootPath

      },
      test: {
        amqpUrl: "amqp://localhost:5672",
        root: rootPath

      },
      production: {
        amqpUrl: "amqp://localhost:5672",
        root: rootPath

      }
    };

   /* this is the line i'm having problem how can i export config object*/
    // export config[env];

当我导出我如何导出我试过的对象,export default config[env]但它产生一些从预期输出的顺序

在ES6中,您可以使用导出功能导出名称,或者默认情况下可以导出任何内容.该require格式是这样的:

let config = require('config')

它需要配置文件的默认导出.在你的情况下,你应该这样做:

export default config[env]

如果您想使用导出,您可以执行以下操作:

let Environment = config[env];
export {Environment}

区别在于:

import EnvirmentNameWhatever from "./config"

至

import {Environment} from "./config"

注意 - 默认导出时,您可以使用任何您喜欢的名称,而在导出时,您必须使用导出的名称.

使用export要导出的声明上的关键字应该完成这项工作,如下所示:

import path = require("path")

const rootPath = path.normalize(__dirname + '/..')
export const env = process.env.NODE_ENV || 'development'

export let config = {
    development: {
    amqpUrl: "amqp://localhost:15672",
    root: rootPath

    },
    test: {
    amqpUrl: "amqp://localhost:5672",
    root: rootPath

    },
    production: {
    amqpUrl: "amqp://localhost:5672",
    root: rootPath

    }
};