我有以下聚合管道來計算嵌入評論的手機集合中評分最高的品牌。
public Document findTopRatedBrands(int minReviews, int results) {
UnwindOperation unwindOperation = unwind("reviews");
GroupOperation groupOperation = group("$brand").avg("$reviews.rating")
.as("avgRating").count().as("numReviews");
MatchOperation matchOperation = match(new Criteria("numReviews").gte(minReviews));
SortOperation sortOperation = sort(Sort.by(Sort.Direction.DESC, "avgRating",
"numReviews"));
LimitOperation limitOperation = limit(results);
ProjectionOperation projectionOperation = project().andExpression("_id").as
("brand").andExpression("avgRating").as("rating").andExclude("_id")
.andExpression("numReviews").as("reviews");
Aggregation aggregation = newAggregation(unwindOperation, groupOperation, matchOperation,
sortOperation, limitOperation, projectionOperation);
AggregationResults<Phone> result = mongoOperations
.aggregate(aggregation, "phones", Phone.class);
return result.getRawResults();
}
電話集合中的檔案示例如下：
{
"_id": {
"$oid": "61e1cc8f452d0aef89d9125f"
},
"brand": "Samsung",
"name": "Samsung Galaxy S7",
"releaseYear": 2016,
"reviews": [{
"_id": {
"$oid": "61d4403b86913bee0245c171"
},
"rating": 2,
"dateOfReview": {
"$date": "2019-12-24T00:00:00.000Z"
},
"title": "Won't do that again.",
"body": "I could not use with my carrier. Sent it back.",
"username": "bigrabbit324"
}]
}
我想按 avgRating（四舍五入到小數點后一位）排序，其次按評論數排序。現在平均評分它沒有四舍五入，所以它總是給出不同的值，所以我也不能按評論數量排序。我已經看過 ArithmeticOperators.Round 類，但如果可能的話，我不明白如何在此處包含它。
結果示例如下：
[Document{{brand=Nokia, rating=3.25, reviews=4}}]
我想有 3.2 作為評級。
這適用于 Mongo Compass：
$project: {
_id: 0,
brand: '$_id',
rating: { $round: ['$avgRating', 1] }
}
uj5u.com熱心網友回復：
嘗試
ProjectionOperation roundAverageRating = Aggregation.project("avgRating", "numReviews")
.and(ArithmeticOperators.Round.roundValueOf("avgRating").place(1))
.as("avgRatingRounded");
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