如何在Java中实现此MongoDB聚合

Question

我有以下工作正常的MongoDB聚合shell命令：

db.followrequests.aggregate([{
    $match: {
        _id: ObjectId("551e78c6de5150da91c78ab9")
    }
}, {
    $unwind: "$requests"
}, {
    $group: {
        _id: "$_id",
        count: {
            $sum: 1
        }
    }
}]);

哪个返回：

{“ _id”：ObjectId（“ 551e78c6de5150da91c78ab9”），“计数”：7}

我需要用Java来实现，我正在尝试以下方法：

List<DBObject> aggregationInput = new ArrayList<DBObject>();

BasicDBObject match = new BasicDBObject();
match.put("$match", new BasicDBObject().put("_id",new ObjectId(clientId)));
aggregationInput.add(match);

BasicDBObject unwind = new BasicDBObject();
unwind.put("$unwind", "$requests");
aggregationInput.add(unwind);

BasicDBObject groupVal = new BasicDBObject();
groupVal.put("_id", "$_id");
groupVal.put("count", new BasicDBObject().put("$sum", 1));

BasicDBObject group = new BasicDBObject();
group.put("$group", groupVal);
aggregationInput.add(group);

AggregationOutput output = followRequestsCol.aggregate(aggregationInput);
for (DBObject result : output.results()) {
    System.out.println(result);
}

我有一个例外：

mongodb匹配过滤器必须是对象中的表达式。

您能帮我找出以上代码中的错误吗？ 谢谢！

Answer 1

尝试打印aggregationInput的值，您将意识到.put()不会返回BasicDBObject而只会返回与您更新的键相关联的先前值。 因此，当您执行以下操作时：

match.put("$match", new BasicDBObject().put("_id",new ObjectId(clientId)));

您实际上是将$match设置$match null ，因为new BasicDBObject().put("_id",new ObjectId(clientId))返回null 。

将您的代码更新为：

List <DBObject> aggregationInput = new ArrayList <DBObject> ();

BasicDBObject match = new BasicDBObject();
BasicDBObject matchQuery = new BasicDBObject();
matchQuery.put("_id", new ObjectId());
match.put("$match", matchQuery);
aggregationInput.add(match);

BasicDBObject unwind = new BasicDBObject();
unwind.put("$unwind", "$requests");
aggregationInput.add(unwind);

BasicDBObject groupVal = new BasicDBObject();
groupVal.put("_id", "$_id");
groupVal.put("count", new BasicDBObject().put("$sum", 1));

BasicDBObject group = new BasicDBObject();
group.put("$group", groupVal);
aggregationInput.add(group);

AggregationOutput output = followRequestsCol.aggregate(aggregationInput);
for (DBObject result : output.results()) {
    System.out.println(result);
}

或者，更通俗易懂的方法是使用流畅的BasicDBObjectBuilder ：

final DBObject match = BasicDBObjectBuilder.start()
                                           .push("$match")
                                               .add("_id", new ObjectId())
                                           .get();
aggregationInput.add(match);

它应该工作正常。

Answer 2

每个{}必须是新的DBObject 。 也可以使用.append(key,value)方法使.append(key,value)更优雅。

尝试这个：

List<DBObject> pipeline = new ArrayList<DBObject>(Arrays.asList(
    new BasicDBObject("$match", new BasicDBObject("_id", 
        new ObjectId("551e78c6de5150da91c78ab9"))), 
    new BasicDBObject("$unwind", "$requests"),
    new BasicDBObject("$group", 
        new BasicDBObject("_id","$_id").append("count", new BasicDBObject("$sum", 1)))));

AggregationOutput output = followRequestsCol.aggregate(pipeline);
for (DBObject result : output.results()) {
    System.out.println(result);
}

Answer 3

这是基于上述建议的最终工作版本。//使用mongodb聚合框架确定关注者的数量Integer returnCount = 0; 列表aggregationInput = new ArrayList（）;

    BasicDBObject match = new BasicDBObject();
    BasicDBObject matchQuery = new BasicDBObject();
    matchQuery.put("_id", new ObjectId(clientId));
    match.put("$match", matchQuery);
    aggregationInput.add(match);

    BasicDBObject unwind = new BasicDBObject();
    unwind.put("$unwind", "$requests");
    aggregationInput.add(unwind);

    BasicDBObject groupVal = new BasicDBObject();
    groupVal.put("_id", null);
    BasicDBObject sum = new BasicDBObject();
    sum.put("$sum", 1);
    groupVal.put("count", sum);
    BasicDBObject group = new BasicDBObject();
    group.put("$group", groupVal);
    aggregationInput.add(group);

    AggregationOutput output = followRequestsCol.aggregate(aggregationInput);
    for (DBObject result : output.results()) {
        returnCount = (Integer) result.get("count");
        break;
    }
    return returnCount;