mongoDB根据两个条件选择记录 [英] mongoDB selecting record based on two conditions

问题描述

我的数据库中的示例记录如下所示：

A sample record in my database looks like :

{
    "_id" : 2,
    "name" : "Corliss Zuk",
    "scores" : [
            {
                    "type" : "exam",
                    "score" : 87.53859552156015
            },
            {
                    "type" : "quiz",
                    "score" : 24.49132971110967
            },
            {
                    "type" : "homework",
                    "score" : 99.24881912510654
            }
    ]

}

我正在尝试选择所有家庭作业分数> 90或考试分数我的OR条件是这样的：

I am trying to select all records that have homework scores > 90 or exam scores < 50. My OR condition is like this:

 DBObject clause1 = new BasicDBObject("scores.type", "homework").append("scores.score", new BasicDBObject("$gt", 90)); 
    DBObject clause2 = new BasicDBObject("scores.type", "exam").append("scores.score", new BasicDBObject("$lt", 50));    
    BasicDBList or = new BasicDBList();
    or.add(clause1);
    or.add(clause2);
    DBObject query = new BasicDBObject("$or", or);

这将取得所有分数> 90或<1的记录。但是没有涉及被检查分数应该是家庭作业分数或考试分数的条件。
我错过了一个将分数与类型联系起来的条件，然后检查它们的价值。

This is picking up all records where the score is > 90 or < 50 but not relating the condition that the score being examined should be either a homework score or an exam score. I am missing a condition that will relate the scores to the type and then examine their value.

任何帮助都会非常感激。

Any help would be much appreciated.

谢谢，
John

Thanks, John

推荐答案

DBObject clause1 = new BasicDBObject("scores", new BasicDBObject("$elemMatch", new BasicDBObject("type", "homework").append("score", new BasicDBObject("$gt", 90))));
DBObject clause2 = new BasicDBObject("scores", new BasicDBObject("$elemMatch", new BasicDBObject("type", "exam").append("score", new BasicDBObject("$lt", 50))));

以这种方式使用。

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