如何在Typescript中导出对象 [英] How to export object in Typescript
问题描述
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这个javscript输出
I'm converting existing JavaScript project into type script. im not sure how to export object to get this javscript output
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];
我写的打字稿如下,但不清楚导出
i wrote typescript as follow but not clear with exporting
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];
当我导出时如何导出我试过的对象导出默认配置[env]
但它会从预期输出中产生一些差异
when i exporting how can i export just object i tried with export default config[env]
but it generates something deference from expected output
推荐答案
在ES6中,您可以导出使用导出功能的名称,或者默认情况下您可以导出任何内容。
require
格式如下:
In ES6 you are allowed to export names using the export function, or for default you can export anything.
The require
format goes like this:
let config = require('config')
它需要默认导出配置文件。
在你的情况下,你应该这样做:
And it takes the default export of config file. In your case, you should do:
export default config[env]
如果你想使用导出,你可以这样做:
If you want to use the export, you would do something like:
let Environment = config[env];
export {Environment}
区别在于:
import EnvirmentNameWhatever from "./config"
到
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。
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