我正在使用mongodb有一个集合,其中我正在为不同的用户存储项目的状态,它就像"userItemstatus" = 1 or 2
。这意味着1喜欢该项目,2意味着不喜欢该项目。
"_id" : NumberLong(104009),
"_class" : "com.samepinch.domain.registration.UserItemHistory",
"user_id" : NumberLong(85861),
"item_id" : NumberLong(103468),
"catagory" : "MISCELLANEOUS",
"userItemStatus" : 1,
"createdDate" : ISODate("2016-02-08T11:43:40.351Z"),
"updatedDate" : ISODate("2016-02-08T11:43:43.780Z")
}
{
"_id" : NumberLong(104018),
"_class" : "com.samepinch.domain.registration.UserItemHistory",
"user_id" : NumberLong(85861),
"item_id" : NumberLong(103352),
"catagory" : "MISCELLANEOUS",
"userItemStatus" : 2,
"createdDate" : ISODate("2016-02-08T11:44:26.803Z"),
"updatedDate" : ISODate("2016-03-17T19:34:07Z")
}
现在我想得到最受欢迎的项目。 i,具有最多"userItemStatus"= 1
个数的项目。我已经阅读了关于聚合但我有问题使用它与spring数据mongodb。我是使用spring数据mongodb的新手。请帮忙。
答案 0 :(得分:0)
public class ValueCount {
private String userid;
private int likes;
public String getValue() {
return value;
}
public void setValue(String value) {
this.value = value;
}
public int getCount() {
return count;
}
public void setCount(int count) {
this.count = count;
}
}
public List<ValueCount> agg(){
Criteria cr = Criteria.where("userItemStatus").is("1");
MatchOperation m = match(cr);
GroupOperation g = group("item_id").count().as("count");
ProjectionOperation p = project("count").and("item_id").previousOperation();
SortOperation s = sort(Sort.Direction.DESC,"count");
LimitOperation limit = limit(20);
Aggregation agg = newAggregation(m,g,p,s,limit);
AggregationResults<ValueCount> groupResults = mongoTemplate.aggregate(agg,"user_likedislike",ValueCount.class);
return groupResults.getMappedResults();
}
无论如何,您将从最喜欢的方式开始排序结果。您可以使用LimitOperation limit = limit(20);
将user_likedislike
替换为您的收藏名称来限制元素