在使用多个promise和Promise.all时,如何避免promise构造函数反模式？

说我有以下代码：

getFoo = function() {
    return new Promise(function(resolve, reject) {
        var promises = [];
        promises.push(new Promise(function(resolve, reject) => {
            getBar1().then(function(bar1) {
                processBar1(bar1); 
                resolve(bar1);
            });
        }));
        promises.push(new Promise(function(resolve, reject) => {
            getBar2().then(function(bar2) {
                processBar2(bar2); 
                resolve(bar2);
            });
        }));
        Promise.all(promises).spread(function(bar1, bar2) {
            var result = processBothBars(bar1, bar2);
            resolve(result);
        });
    });
}

它提出了反模式的一些基本问题,错误被吞噬,以及厄运的金字塔.

我正在使用蓝鸟BTW.

解决方法:

你可以一起摆脱新的Promise.

getFoo = function() {
    var promises = [];
    promises.push(getBar1().then(function(bar1) {
        processBar1(bar1);
        return bar1;
    }));
    promises.push(getBar2().then(function(bar2) {
        processBar2(bar2);
        return bar2;
    }));
    return Promise.all(promises).spread(function(bar1, bar2) {
        var result = processBothBars(bar1, bar2);
        return result;
    });
}

// start mock
function getBar1() {
    return Promise.resolve({name:'bar1',processed: false});
}
function getBar2() {
    return Promise.resolve({name:'bar2',processed: false});
}
function processBar1(bar1) {
  bar1.processed = true;
}
function processBar2(bar2) {
  bar2.processed = true;
}
function processBothBars (bar1, bar2) {
  return [bar1, bar2].filter(function (bar) {
    return bar.processed;
  }).map(function (bar) {
    return bar.name;
  });
}
Promise.prototype.spread = function (fn) {
  return this.then(function (arr) {
      return fn.apply(this, arr);
  });
};
// end mock

var getFoo = function (fail) {
    var promises = [];
    promises.push(getBar1().then(function (bar1) {
        processBar1(bar1);
        if (fail) {
          throw 'getBar1 Failed!';
        }
        return bar1;
    }));
    promises.push(getBar2().then(function (bar2) {
        processBar2(bar2);
        return bar2;
    }));
    return Promise.all(promises).spread(function (bar1, bar2) {
        var result = processBothBars(bar1, bar2);
        return result;
    });
}
getFoo().then(function (result) {
    console.log(result); // ['bar1', 'bar2']
});
getFoo(true).then(function (result) {
    console.log(result); // doesn't happen
}).catch(function (e) {
    console.error(e); // Error: getBar1 Failed!
});

.then返回一个promise,因此除非你想防止错误到达外部promise,否则不需要创建一个包装它的新的promise.