mongodb java groupby查询转换

mongodb新手。我一直在玩MongoJava驱动程序，到目前为止一切顺利

我已成功使用DBCollection.aggregate–我通过了：

$match
$project
$group
$sort

例如，我已经制定了如下标准：

Map<String, Object> match_criteria;
match_criteria.put(“properties.type”, “monuments”);

DBObject match = new BasicDBObject();
match.putAll(match_criteria); //was just setting from Map
DBObject match_dbobj = new BasicDBObject("$match", match);
.
.
.
.
.in the same way was populating $project, $group and $sort

但现在使用这种方法转换以下mongo shell查询时遇到问题：

db.test.aggregate({
    $match:{             ---> still can pass with Map
        "properties.type":'monuments',
    }
},
{
    $project:{           ---> still can pass with Map
        "props.country":1,
        "attrs.time":1,
        "attrs.status":1
    }
},
{
    $group:{              ---> trying to find a way to handle $group from java code
        _id:{           
            status:'$attrs.status',
            country:'$props.country'
        },
        time:{
            $last:'$attrs.time'
        }
    }
}
)

如何转换$group-我不能再使用地图了。有什么想法吗？

在您的示例中，$group是一个嵌套的DBObject，因此创建它的方法是：

// Create the group sub documents
DBObject idDocument=new BasicDBObject("status","$attrs.status").append("country","$props.country");
DBObject timeDocument=new BasicDBObject("$last","$attrs.time");

// Create the group document
DBObject groupDocument=new BasicDBObject("_id",idDocument).append("time",timeDocument);

// Create the whole group statement
DBObject group=new BasicDBObject("$group", groupDocument);

---

使用计算出的into Aggregates.project。

// Create the group sub documents
DBObject idDocument=new BasicDBObject("status","$attrs.status").append("country","$props.country");
DBObject timeDocument=new BasicDBObject("$last","$attrs.time");

// Create the group document
DBObject groupDocument=new BasicDBObject("_id",idDocument).append("time",timeDocument);

// Create the whole group statement
DBObject group=new BasicDBObject("$group", groupDocument);