kriskowal/qnode.jsq.all和spread-kriskowal/qnode.jsq.allandspread

7  

Spread calls q.all internally.

在内部传播q.all。

Here is the code for spread from q.js:

这是从q.js传播的代码：

Q.spread = spread;
function spread(promise, fulfilled, rejected) {
    return when(promise, function (valuesOrPromises) {
        return all(valuesOrPromises).then(function (values) {
            return fulfilled.apply(void 0, values);
        }, rejected);
    }, rejected);
}

Note that it expects a promise that resolves to an array or an array as the first argument.

请注意，它期望一个解析为数组或数组的promise作为第一个参数。

Therefore your call should look like this:

因此，您的通话应如下所示：

var p1 = doWork(data);
var p2 = p1.then(doMoreWork);
var p3 = doConcurrentWork(data);

return q.spread([p1,p2,p3], funcWith3params, function(err) {
        console.log(err):
    });

However, your original call should work as well. Not sure why it does not.

但是，您的原始呼叫也应该有效。不知道为什么不这样做。

3  

The most succinct way to express the example you’ve provided is:

表达您提供的示例的最简洁方式是：

var p1 = doWork(data);
var p2 = p1.then(doMoreWork);
var p3 = doConcurrentWork(data);

return Q.spread([p1, p2, p3], funcWith3params)
    .done();

However, the original is correct as written.

但是，原件正如所写的那样。

I suspect that the problem is that one or more of the input promises is never resolving. Try using timeouts to isolate the problem.

我怀疑问题是一个或多个输入承诺永远不会解决。尝试使用超时来隔离问题。

var p1 = doWork(data).timeout(1000, 'p1 timed out');
var p2 = p1.then(doMoreWork).timeout(1000, 'p2 timed out');
var p3 = doConcurrentWork(data).timeout(1000, 'p3 timed out');

return Q.spread([p1, p2, p3], funcWith3params)
   .done();