Promise.all()将所有结果作为未定义

时间:2016-11-27 13:14:03

标签: javascript node.js promise

我正在构建一系列来自Mongoose的Model.save()的承诺但是由于某种原因,来自Promise.all()的返回数据是一个未定义的数组

代码

const saveDataArray = [];
//go through results and save them to the db
Object.keys(passData.scrapeUniqueData).map((index) => {
    const result = gamesExport.addGame(passData.scrapeUniqueData[index], passData.leagueInfo.league_id, passData.leagueInfo.leagueID, passData.leagueInfo.year);
    if(result)
    saveDataArray.push(result);
});

//const -> saveDataArray 
[Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise,  Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise, Promise]

return Promise.all(saveDataArray)
.then((savedDBData) => {
    console.info('savedDBData', savedDBData);
    console.info('saveDataArray', saveDataArray);
    passData.savedDBGames = savedDBData.filter(function(n){ return n != undefined });
    return passData;
})

//const -> savedDBData 
[undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined, undefined]

所以imo savedDBData应该和`saveDataArray一样返回。为什么没有Promise.all返回保存的数据?

AddGame

addGame: (details, league_id, leagueID, year) => {
    "use strict";
    if(!details.result && details.result[0] === "P"){
        console.error('result skipped for postponded match');
        return;
    }
    if (details && details.notes !== "Postponed" && !details.postponed) {
        var result = new Games();
        result.fixtureID = details.fixtureID;

        if (details.date.toString().indexOf('z') === -1) {
            result.date = Date.parse(moment(details.date.toString(), "DD-MM-YYYY H:m").format());
        } else {
            result.date = details.date;
        }
        result.home_id = details.home_id._id;
        result.away_id = details.away_id._id;
        result.league_id = league_id;
        result.leagueID = leagueID;
        result.year = year;
        result.norsemen = details.home_id.name.isNorsemen() || details.away_id.name.isNorsemen();

        if (details.result) {
            if (details.result.home === "A" && details.result.away === "A") {
                result.result = ['A', 'A'];
                result.postponed = true;
            } else if (details.result.home === "H" && details.result.away === "W") {
                result.result = ['H', 'W'];
            } else if (details.result.home === "A" && details.result.away === "W") {
                result.result = ['A', 'W'];
            } else if (details.result.home === "C" && details.result.away === "C") {
                result.result = ['C', 'C'];
                result.cancelled = true;
            } else if (details.result.home === "P" && details.result.away === "P") {
                result.result = ['P', 'P'];
                result.postponed = true;
            } else {
                result.result = [parseInt(details.result.home), parseInt(details.result.away, 10)];
            }
        }
        return result.save(function (err, saveData) {
            if (err) console.error(err);
            return result;
        });
    } else {
        return null;
    }
},

1 个答案:

答案 0 :(得分:1)

当您将回调函数传递给Mongoose方法时,它会假定您不会使用它返回的承诺。

我不确定这是否是设计的(我猜是这样,因为否则会调用常规回调和promise处理程序,结果相同),但我可以使用代码重现您的问题看起来像这样:

let doc = new Model(...);
doc.save((err, res) => {
  console.log('CB RES', res);
}).then(res => {
  console.log('PR RES', res);
});

PR RES"承诺结果" )的记录结果为undefined,但CB RES&# 34;回调结果" )记录正确的文件。如果您未将回调函数传递给save(),则PR RES会记录正确的文档。

所以你应该重写addGame只返回一个promise,而不是传入一个回调函数:

return result.save().catch(function(err) {
  console.error(err);
  // XXX: be aware that if you don't do anything else here,
  //      the error will _just_ be logged, but not propagated.
  //      This is also what your original code does, so I assume
  //      it's intentional.
});