Question

我按照此处的说明操作：https://github.com/kriskowal/q#handling-errors

但我无法让我的promises返回处理程序抛出错误。它偷偷地失败了。这使得调试变得非常困难。谁能弄明白我做错了什么？

这是我的功能：

app.get('/api/recommend',function (req, res) {
    var results = require('./recommend').getOriginProfiles(req.query['origin'], req.query['target']).then(
        function getRecommendations(results) {
           // debugger;

            sasd / asdasdasdasdasdasd; // fail darn you!
            require('./recommend').matchProfile(results[0]);
            res.send(results);
        }).fail(function (error) { //never failed
            throw error;

        });
    logger.log('DEBUG', '/api/recommend response', results);
});

getOriginProfiles是一个返回promise的函数：

function getOriginProfiles(origin_qty, target_qty) {
    var deferred = q.defer();
    chooseRandomUsers(origin_qty).then(collectUserProfiles(origin_qty, target_qty, deferred), function (error) {
        logger.log('ERROR', 'recommending users', error);
        throw error;
    });
    return deferred.promise;
}

Answer 1

我会这样写：

var recommend = require('./recommend');

app.get('/api/recommend',function (req, res, next) {
    recommend.getOriginProfiles(req.query['origin'], req.query['target'])
        .then(function getRecommendations(results) {

            sasd / asdasdasdasdasdasd; // fail darn you!

            recommend.matchProfile(results[0]);
            res.send(results);

            logger.log('DEBUG', '/api/recommend response', results);
        })
        .fail(function (error) {
            next(error)
        })
        // Any function where you use a promise that isn't also returning a
        // promise should have .done on the chain to prevent it from swallowing
        // all exceptions.
        .done();
});

和

function getOriginProfiles(origin_qty, target_qty) {
    return chooseRandomUsers(origin_qty)
        .then(function(users){
             // Return the promise here will then wait for that promise to finish.
             return collectUserProfiles(users, origin_qty, target_qty);
        })
        .fail(function (error) {
            logger.log('ERROR', 'recommending users', error);
            throw error;
        });
}

无法得到我的q承诺来捕捉错误

1 个答案: