以下NoSQL查询返回特定用户的所有评论:
db.getCollection('catalog-review').aggregate([
{
$project: {
reviews: {
$filter: {
input: "$reviews",
as: "review",
cond: { $eq: [ "$$review.userId", 121 ] }
}
}
}
}
])
这个查询工作正常,但是如果我使用Mongo驱动程序在Java中实现此功能,则“ $$”不起作用。
List<CatalogReview> reviews = collection.aggregate(Arrays.asList(
new Document("$project", new Document("reviews", new Document("$filter", new Document("input", "$reviews")
.append("as", "review").append("cond", new Document("$eq", Arrays.asList(new Document("$$review.userId", 121)))))))
)).into(new ArrayList<>());
错误信息:
com.mongodb.MongoCommandException: Command failed with error 15999: 'invalid operator '$$review.userId'' on server localhost:27017. The full response is { "ok" : 0.0, "errmsg" : "invalid operator '$$review.userId'", "code" : 15999 }
Mongo驱动程序是否支持聚合函数?
最佳答案
问题是Arrays.asList(new Document("$$review.userId", 121))
。它应该是Arrays.asList("$$review.userId", 121)
。Arrays.asList(new Document("$$review.userId", 121))
= [ { "$$review.userId" : 121 } ]
Arrays.asList("$$review.userId", 121)
= [ "$$review.userId", 121 ]
该代码段应如下所示:
List<CatalogReview> reviews = collection.aggregate(Arrays.asList(
new Document("$project", new Document("reviews", new Document("$filter", new Document("input", "$reviews")
.append("as", "review").append("cond", new Document("$eq", Arrays.asList("$$review.userId", 121))))))
)).into(new ArrayList<>());
关于java - 在Mongo Java驱动程序中过滤嵌套数组,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/40003410/