Javascript 在promise中设置一个变量，并在功能节点外部使用它

我有以下代码。 我试图在.then语句之外使用变量rsname、rslevel和rsexp，但我似乎找不到一种方法来实现这一点。（im使用的模块可在此处找到：）

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您无法使用当前设置。这就是它的本意。不管你对这些变量做什么，都不能离开这个函数

您可以使用如下函数将数据传递到其他地方：

const { hiscores } = require('runescape-api');

var rsn = "le me";

function doSomethingWithData(rsname, rslevel, rsexp) {
    console.log(rsname, rslevel, rsexp);
}

var rsnlookup = hiscores.getPlayer(rsn).then(data => {
    var rsname = data.name;
    var rslevel = data.skills.dungeoneering.level;
    var rsexp = data.skills.dungeoneering.experience;
    
    doSomethingWithData(rsname, rslevel, rsexp);
})

rsnlookup.catch((error) => {
  console.error("This username doesn\'t exist.");
});

这就是JavaScript的工作原理。无论何时向另一台服务器发出呼叫，JS都不会停止并等待请求返回。它实际上会继续执行，而请求仍然通过internet传递给您。这就是为什么您需要提供一个回调函数，以便它在完成后返回并执行

例如，这里有一个更简单的细分：

// The variables must be declared outside the function for them
// to be accessible from outside the function.
let rsname;
let rslevel;
let resexp;

// This call only BEGINS the request. It will not stay and wait for it.
hiscores.getPlayer(rsn).then(data => {
    // This code is only executed after the request comes back with the data,
    // so the data can only be used from here.
    // These variables do not get declared until this time,
    // so they won't exist outside this function until the request is done.
    // You need to pass them out to wherever they are needed from here,
    // or do all your logic with it right here.
    rsname = data.name;
    rslevel = data.skills.dungeoneering.level;
    rsexp = data.skills.dungeoneering.experience;
});
// This is executed the moment the request begins, it does not wait for it to return.
console.log(rsname); // The variables are still undefined here since the request is not back yet here.

---

本质上，所有的“异步”意味着代码将开始执行，并在不等待它完成的情况下继续执行。完成后，它会返回执行某种回调函数。

根据当前设置，您不能这样做。这就是它的本意。不管你对这些变量做什么，都不能离开这个函数

您可以使用如下函数将数据传递到其他地方：

const { hiscores } = require('runescape-api');

var rsn = "le me";

function doSomethingWithData(rsname, rslevel, rsexp) {
    console.log(rsname, rslevel, rsexp);
}

var rsnlookup = hiscores.getPlayer(rsn).then(data => {
    var rsname = data.name;
    var rslevel = data.skills.dungeoneering.level;
    var rsexp = data.skills.dungeoneering.experience;
    
    doSomethingWithData(rsname, rslevel, rsexp);
})

rsnlookup.catch((error) => {
  console.error("This username doesn\'t exist.");
});

这就是JavaScript的工作原理。无论何时向另一台服务器发出呼叫，JS都不会停止并等待请求返回。它实际上会继续执行，而请求仍然通过internet传递给您。这就是为什么您需要提供一个回调函数，以便它在完成后返回并执行

例如，这里有一个更简单的细分：

// The variables must be declared outside the function for them
// to be accessible from outside the function.
let rsname;
let rslevel;
let resexp;

// This call only BEGINS the request. It will not stay and wait for it.
hiscores.getPlayer(rsn).then(data => {
    // This code is only executed after the request comes back with the data,
    // so the data can only be used from here.
    // These variables do not get declared until this time,
    // so they won't exist outside this function until the request is done.
    // You need to pass them out to wherever they are needed from here,
    // or do all your logic with it right here.
    rsname = data.name;
    rslevel = data.skills.dungeoneering.level;
    rsexp = data.skills.dungeoneering.experience;
});
// This is executed the moment the request begins, it does not wait for it to return.
console.log(rsname); // The variables are still undefined here since the request is not back yet here.

---

本质上，所有的“异步”意味着代码将开始执行，并在不等待它完成的情况下继续执行。完成后，它会返回执行某种回调函数。

这是否回答了您的问题？同样相关：您可以在外部声明变量并分配它inside@AbhishekRanjan如果代码在不同的时间运行，则不起作用。@VLAZ我查看了您发送给我的链接，但我不确定如何在我的代码中使用此链接。我以前从未真正使用过异步调用。这能回答你的问题吗？同样相关：您可以在外部声明变量并分配它inside@AbhishekRanjan如果代码在不同的时间运行，则不起作用。@VLAZ我查看了您发送给我的链接，但我不确定如何在我的代码中使用此链接。我以前从未真正使用过异步调用。