javascript - 避免带有 promise 的嵌套回调

Question

我对使用 Promise API 还是个新手，我正在努力避免深度嵌套的 Promise 链，据我所知，这是使用 Promise 的好处之一。以下面的伪代码为例，当后续的 Promise 依赖于先前的 context 时，如何避免嵌套 Promise？

function loadDependency1() {
    // return a promsise to load the first dependency
}

function loadDependency2(dependency1) {
    // return a promise to load the second dependency, which relies on the first dependency
}

function loadDependency3(dependency2) {
    // return a promise to load the third dependency, which relies on the second dependency
}

function doWork(dependency1, dependency2, dependency3) {
    // finally have all the things necessary to do work
}

// load all the dependencies and eventually doWork
loadDependency1().then(function(dependency1) {
    return loadDependency2(dependency1).then(function(dependency2) {
        return loadDependency3(dependency2).then(function(dependency3) {
            doWork(dependency1, dependency2, dependency3);
        });
    });
});

Answer 1

当您从 then 返回一个 promise 时，它​​将在该 promise 解析时解析:

所以，如果下一个只需要上一个:

loadDependency1().then(function(dependency1) {
    return loadDependency2(dependency1);
}).then(function(dependency2) {
     return loadDependency3(dependency2);
}).then(function(dependency3) {
    doWork(dependency3);
});

如果您需要第三个依赖项，则可以使用。

如果依赖关系不相互依赖:

Promise.all([loadDependency1(),loadDependency2(),loadDependency3])
.spread(function(dep1,dep2,dep3){
    doWork(dep1,dep2,dep3);
});

如果您想在整个 promise 链中保持“状态”并且正在使用现代 promise 库，例如 Bluebird你可以这样做:

loadDependency1().bind({}).then(function(dependency1) {
    this.dep1 = dependency1;
    return loadDependency2(dependency1);
}).then(function(dependency2) {
     this.dep2 = dependency2;
     return loadDependency3(dependency2);
}).then(function(dependency3) {
    doWork(this.dep1, this.dep2, dependency3);
});

如果你不是(你真的应该是:))你可以.all绕过它:

loadDependency1().then(function(dependency1) {
    return [loadDependency2(dependency1),dependency1];
}).spread(function(dependency2,dep1) {
     return [loadDependency3(dependency2),dependency2,dep1];
}).spread(function(dependency3,dependency2,dependency1) {
    doWork(dependency1, dependency2, dependency3);
});