javascript - promise 不返回 [node.js]

Question

这个问题已经有答案了:

How do I return the response from an asynchronous call? (42 个回答)

已关闭 4 年前。

我现在正在开发 Dominos Pizza API，当我从主模块 (bot.js) 中 console.log() 时，一切似乎都正常。如果我在测试文件(testnearby.js)中创建一个函数来告诉它console.log(localStore)，它不会打印任何内容。代码如下。 /image/ovJMD.png

定义的不起作用的函数(上面提到的)是 getLocalStore

如有任何帮助，我们将不胜感激。代码:

bot.js(主模块):

//START 

// TABLE OF THINGS ADDED TO BASKET ETC.

var store = "";
var localStore = null;


// FIND LOCAL STORE

exports.getLocalStore = function(postcode) {
    // get the local store
    // get the api and find with the postcode provided
    // fetch the api
    fetch('https://www.dominos.co.uk/storefindermap/storesearch?SearchText=' + postcode)
        // json it
        .then(res => res.json())
        // return it
        .then(function(json) {
            // return console.log(json.localStore)
            // makes it a variable so we can return it elsewhere
            localStore = json.localStore
        })
    return localStore
}
// END


testnearby.js

// START

const ukdomino = require("./bot.js")
const postcode = "L129JH"

ukdomino.getLocalStore(postcode).then(localStore => console.log(localStore))

// END

Answer 1

您需要返回 fetch ...

exports.getLocalStore = function(postcode) {
    // get the local store
    // get the api and find with the postcode provided
    // fetch the api
    return fetch('https://www.dominos.co.uk/storefindermap/storesearch?SearchText=' + postcode)
        // json it
        .then(res => res.json())

}

当您在函数中返回 localStore 时，它​​仍然为 null，因为它在 fetch 设置其值之前返回