Question

我有两个集合，都有这样的结构：

id    trips_in
1     5
2     10

id    trips_out
1     6
2     8

我的问题是如何将它们组合成一个像这样的集合：

id    trips_in    trips_out
1     5           6
2     10          8

我发现了mapReduce，但它的功能看起来比我需要的更多。我写了以下查询：

tripsInMap = function() {
    var values = {
        trips_in: this.trips_in
    };
    emit(this._id, values);
};
tripsOutMap = function() {
    var values = {
        trips_out: this.trips_out
    };
    emit(this._id, values);
};
reduce = function(key, values) {
  var result = {
      "trips_out" : "",
      "trips_in" : ""
    };

    values.forEach(function(value) {
      if(value.trips_out !== null) {result.trips_out = value.trips_out;}

      if(value.trips_in !== null) {result.trips_in = value.trips_in;}
    });

    return result;
}
db.tripsIn.mapReduce(tripsInMap, reduce, {"out": {"reduce": "joined"}});
db.tripsOut.mapReduce(tripsOutMap, reduce, {"out": {"reduce": "joined"}});

但是我最终得到了"trips_in": undefined。我想知道是否有更好的方法。

Answer 1

虽然这可能不是最快的方法，但你可以尝试这样的事情：

// Create the new collection with data from the tripsIn collection
db.tripsIn.find().forEach( function(trip) { 
    db.tripsJoined.insert({ _id: trip._id, trips_in: trip.trips_in, trips_out: 0 }); 
})

// Update the trips_out field in the tripsJoined collection
// using upsert:true to insert records that are not found
db.tripsOut.find().forEach( function(trip) {
    db.tripsJoined.update(
        { _id: trip._id}, 
        {$inc: {trips_in: 0, trips_out: trip.trips_out}}, 
        {upsert: true});
})

第一行将遍历tripsIn集合中的每个文档，并在tripsJoined集合中插入相应的文档并设置trips_out字段。

第二行将遍历tripsOut集合，对于每个文档，它将使用tripsJoined值更新相应的trips_out文档。

请注意，我添加了{$inc: {trips_in: 0 ...和upsert:true。这样做是为了如果tripsOut集合中存在任何在tripsIn集合中没有相应_id值的行程文档，则会插入文档并初始化trips_in字段到0。

将收藏集附加到

1 个答案: