[英]Spring Data + MongoDB invalid reference error
这是我的代码(Spring数据):
MatchOperation matchOperation = Aggregation.match(new Criteria("stats.channelId").is(channelId));
UnwindOperation unwindOperation = Aggregation.unwind("stats");
GroupOperation groupOperation = Aggregation.group("id", "discordId")
.push("stats").as("stats");
AggregationOperation addFields = (AggregationOperationContext aggregationOperationContext) -> {
DBObject dbObject =
new BasicDBObject("allGamesOnChannel",
new BasicDBObject("$sum", "$stats.pickupsPlayed"));
return new BasicDBObject("$addFields", dbObject);
};
SortOperation sortOperation = Aggregation.sort(new Sort(Sort.Direction.DESC, "allGamesOnChannel"));
LimitOperation limitOperation = Aggregation.limit(maxElements);
ProjectionOperation projectionOperation = Aggregation.project("id", "discordId", "stats");
Aggregation aggregation = Aggregation.newAggregation(matchOperation,unwindOperation, matchOperation,groupOperation,addFields,
sortOperation,limitOperation,projectionOperation);
AggregationResults<UserSummaryChannel> userSummaries = mongoTemplate.aggregate(aggregation, "pickupUser", UserSummaryChannel.class);
我在上面执行时遇到此异常:
Caused by: java.lang.IllegalArgumentException: Invalid reference 'allGamesOnChannel'!
at org.springframework.data.mongodb.core.aggregation.ExposedFieldsAggregationOperationContext.getReference(ExposedFieldsAggregationOperationContext.java:99) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.aggregation.ExposedFieldsAggregationOperationContext.getReference(ExposedFieldsAggregationOperationContext.java:80) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.aggregation.SortOperation.toDBObject(SortOperation.java:73) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.aggregation.AggregationOperationRenderer.toDBObject(AggregationOperationRenderer.java:56) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.aggregation.Aggregation.toDbObject(Aggregation.java:580) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.MongoTemplate$BatchAggregationLoader.prepareAggregationCommand(MongoTemplate.java:2603) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.MongoTemplate$BatchAggregationLoader.aggregate(MongoTemplate.java:2585) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.MongoTemplate.aggregate(MongoTemplate.java:1570) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
at org.springframework.data.mongodb.core.MongoTemplate.aggregate(MongoTemplate.java:1511) ~[spring-data-mongodb-1.10.10.RELEASE.jar:na]
执行聚合而不分组:
Aggregation aggregation = Aggregation.newAggregation(matchOperation,addFields,
sortOperation,limitOperation,projectionOperation);
没有错误,所以我认为分组存在问题。 有任何想法我在这里做错了吗? 我还想做的是过滤嵌入元素的列表,计算那些元素的属性之一的总和,并基于它进行排序。 也许有类似的方法吗?
编辑所以这是我的文件:
@Document
public class PickupUser {
@Id
private String id;
private String discordId;
private List<UserModeStats> stats;
private String name;
}
嵌入式对象:
@Data
@EqualsAndHashCode(exclude = {"pickupsPlayed", "caps"})
public class UserModeStats {
public UserModeStats() {}
private String mode;
private String channelId;
private Long pickupsPlayed;
private Long caps;
}
我要创建的输出文档:
@Data
public class UserSummaryChannel {
public UserSummaryChannel() {}
private String id;
private String discordId;
private List<UserModeStats> stats;
//private Long allGamesOnChannel;
}
我不希望包括allGamesOnChannel
财产UserSummaryChannel
(聚集不具有或不具有现场工作)。
这不是真正的解决方案,但找到了实现我目标的另一种方法。 我本人“实现”了排序操作,以便对此感兴趣:
SortOperation sortOperation = Aggregation.sort(new Sort(Sort.Direction.DESC, "allGamesOnChannel"));
现在我有这个:
AggregationOperation sortOperation2 = (AggregationOperationContext aggregationOperationContext) -> {
DBObject dbObject =
new BasicDBObject("allGamesOnChannel", 1);
return new BasicDBObject("$sort", dbObject);
};
其余聚合操作保持不变。 没有异常被抛出。
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