NodeJs forEach请求承诺在返回之前等待所有诺言

Question

问题是我无法兑现任何承诺。他们..只是空了。

我在SO上看到的每个答案都告诉我要做这个，尽管出于某种原因这是行不通的。我不知所措，拉头发，砸键盘。有人可以指出我的愚蠢吗？

var q = require('q');
var request = require('request-promise'); // https://www.npmjs.com/package/request-promise

function findSynonym(searchList) {
    var defer = q.defer();
    var promises = [];
    var url = "http://thesaurus.altervista.org/service.php?word=%word%&language=en_US&output=json&key=awesomekeyisawesome";
    var wURL;
    searchList.forEach(function(word){
        wURL = url.replace('%word%',word);
        promises.push(request(wURL));
    });

    q.all(promises).then(function(data){
        console.log('after all->', data); // data is empty
        defer.resolve();
    });

    return defer;
}
var search = ['cookie', 'performance', 'danger'];

findSynonym(search).then(function(supposedDataFromAllPromises) { // TypeError: undefined is not a function [then is not a function]
    console.log('->',supposedDataFromAllPromises); // this never happens
});

Answer 1

您将返回defer没有.then方法的Deferred对象，而不是Promise对象defer.promise。

但是无论如何，这就是延迟的反模式，这里不需要使用延迟的。只要兑现Promise.all获得您的承诺：

function findSynonym(searchList) {
    var url = "http://thesaurus.altervista.org/service.php?word=%word%&language=en_US&output=json&key=awesomekeyisawesome";
    var promises = searchList.map(function(word) {
        return request(url.replace('%word%', word));
    });
    return q.all(promises).then(function(data){
        console.log('after all->', data); // data is empty
        return undefined; // that's what you were resolve()ing with
    });
}