Question

GameScore个对象有一个名为Relation的{{1}}字段如何在查询中计算此关系中的所有对象：

Badges

我需要类似的东西：

var GameScore = Parse.Object.extend("GameScore");
var query = new Parse.Query(GameScore);
query.equalTo("playerName", "Dan Stemkoski");
query.find({
  success: function(results) {
    alert("Successfully retrieved " + results.length + " scores.");
    // Do something with the returned Parse.Object values
    for (var i = 0; i < results.length; i++) {
      var object = results[i];
      alert(object.id + ' - ' + object.get('playerName'));
    }
  },
  error: function(error) {
    alert("Error: " + error.code + " " + error.message);
  }
});

或

object.Badges.count

Answer 1

可悲的是，比一个人希望的更难。它需要通过数据进行另一次异步跳闸，因此您的回调表单可以执行此操作，但实际上并不适合这项工作。这是承诺......

// return a promise that's fulfilled with a badges count for the passed GameScore
function badgeCountOfGameScore(gameScore) {
    var relation = gameScore.get("Badges");  // it must have a relation col called Badges for this to work
    return relation.query.find().then(function(results) {
        return results.length;
    });
}

现在您的原始函数使用promises重做（并使用下划线来更好地处理数组）...

var _ = require("underscore");

function bunchOfBadgeCounts() {
    var GameScore = Parse.Object.extend("GameScore");
    var query = new Parse.Query(GameScore);
    query.equalTo("playerName", "Dan Stemkoski");
    return query.find().then(function(results) {
        alert("Successfully retrieved " + results.length + " scores.");
        var promises = _.map(results, function(object) {
            return badgeCountOfGameScore(object);
        });
        return Parse.Promise.when(promises);
    }).then(function() {
        alert("Counts = " + JSON.stringify(_.toArray(arguments)));
    }, function (error) {
        alert("Error = " + JSON.stringify(error));
    });
}

Answer 2

Parse.Relation对象实际上是一个查询描述，它将返回该关系中的对象，因此对于这种情况，您需要为每个GameScore运行另一个查询：

query.find().then(function (gameScores) {
    alert("Successfully retrieved " + gameScores.length + " scores.");
    var countPromises = gameScores.map(function (gs) {
        // the following executes a count() query, which is severely penalized by Parse
        return gs.get('Badges').query().count()
            .then(function (count) {
                // this extra step is to add the retrieved count as an extra property to the GameSccore object,
                // instead of returning only the counts
                gs.count = count; 
                return gs;
            });
    });
    return Parse.Promise.when(countPromises);
}).then(function () {
    var gameScoresWithBadgeCount = Array.prototype.slice.call(arguments);
}).fail(function(error) {
    alert("Error: " + error.code + " " + error.message);
});

这导致了很多额外的往返（我假设你因为alert()而在浏览器环境中），并且调用另外受Parse限制的count()个查询。

我建议您将计数缓存保留为GameScore类的额外字段，并通过CloudCode挂钩进行相应更新。或者，您可以尝试避免使用Relation并尽可能使等效using an Array field成为可能，通过该include，您可以根据需要始终https://jsfiddle.net/461huu5g/1/相关徽章，或者在不查询相关徽章的情况下进行计数！

如何在解析

2 个答案: