我正在开发一个项目,我的mongo db里面有一个巨大的集合。现在我必须从mongo db中获取一些非常类似SQL group by子句的详细信息。我在mongo shell上成功执行了查询:
> db.votes.aggregate(
... { $group : {
... _id : "vote_post_id",
... votesPerId : { $sum : 1 }
... }}
... );
{
"result" : [
{
"_id" : "vote_post_id",
"votesPerId" : 27371750
}
],
"ok" : 1
}
>
现在,当从java程序执行查询时,我面临以下错误:
Exception in thread "main" com.mongodb.CommandResult$CommandFailure: command
failed [aggregate]: { "serverUsed" : "localhost/127.0.0.1:27017" ,
"errmsg" : "exception: aggregation result exceeds maximum document size (16MB)"
, "code" : 16389 , "ok" : 0.0}
at com.mongodb.CommandResult.getException(CommandResult.java:88)
at com.mongodb.CommandResult.throwOnError(CommandResult.java:134)
at com.mongodb.DBCollection.aggregate(DBCollection.java:1311)
at main.Connetion.CheckConnection.GetNoOfVotesForAType(CheckConnection.java:210)
at main.DAO.Votes.VoteDAO.findKeyWord(VoteDAO.java:109)
at main.DAO.Votes.VoteDAO.main(VoteDAO.java:139)
我调试了我的java程序,它正在生成我在mongo shell上写的确切查询:
{ "$group" : { "_id" : "$vote_post_id" , "count" : { "$sum" : 1}}}
我已经为这个问题找到了很多答案,似乎没有人解决这个问题。
我创建聚合函数的java代码:
//创建我们的管道操作,首先使用$ match DBObject match = new BasicDBObject(“$ match”,new BasicDBObject(“vote_type_id”,“1”));
// build the $projection operation
DBObject fields = new BasicDBObject("vote_post_id", 1);
fields.put("_id", 0);
DBObject project = new BasicDBObject("$project", fields );
// Now the $group operation
DBObject groupFields = new BasicDBObject( "_id", "$vote_post_id");
groupFields.put("count", new BasicDBObject( "$sum", 1));
DBObject group = new BasicDBObject("$group", groupFields);
// run aggregation
AggregationOutput output = votesCollection.aggregate( match, project, group );
System.out.println(output.getCommandResult());
答案 0 :(得分:1)
要使您的查询与shell中的查询相同,它将是:
// $group operation
DBObject groupFields = new BasicDBObject( "_id", "vote_post_id");
groupFields.put("votesPerId", new BasicDBObject( "$sum", 1));
DBObject group = new BasicDBObject("$group", groupFields);
// run aggregation
AggregationOutput output = votesCollection.aggregate( group );
System.out.println(output.getCommandResult());
答案 1 :(得分:0)
在你的shell查询中,你正在做
_id : "vote_post_id",
但是在你的Java查询中你正在做
"_id", "$vote_post_id"
区别在于美元符号。 $vote_post_id
替换该字段的值作为存储区键,而vote_post_id
只使用该字面值(因此只有一个存储区)。