[英]MongoDB Java Driver aggregation with regex filter
我正在使用MongoDB Java驱动程序3.6.3。 我想通过聚合创建正则表达式查询以检索不同的值。
假设我有json:
[{
"name": "John Snow",
"category": 1
},
{
"name": "Jason Statham",
"category": 2
},
{
"name": "John Lennon",
"category": 2
},
{
"name": "John Snow",
"category": 3
}]
我想创建正则表达式像“ John。*”的查询并按名称分组,这样就只有一个“ John Snow”
预期结果是:
[{
"name": "John Snow",
"category": 1
},
{
"name": "John Lennon",
"category": 2
}]
根据Mongo Shell命令, felix
提供的答案是正确的。 使用MongoDB Java驱动程序的该命令的等效表达式为:
MongoClient mongoClient = ...;
MongoCollection<Document> collection = mongoClient.getDatabase("...").getCollection("...");
AggregateIterable<Document> documents = collection.aggregate(Arrays.asList(
// Java equivalent of the $match stage
Aggregates.match(Filters.regex("name", "John")),
// Java equivalent of the $group stage
Aggregates.group("$name", Accumulators.first("category", "$category"))
));
for (Document document : documents) {
System.out.println(document.toJson());
}
上面的代码将打印出来:
{ "_id" : "John Lennon", "category" : 2 }
{ "_id" : "John Snow", "category" : 1 }
您可以在$match
阶段使用$regex
,然后再进行$group
阶段来实现此目的:
db.collection.aggregate([{
"$match": {
"name": {
"$regex": "john",
"$options": "i"
}
}
}, {
"$group": {
"_id": "$name",
"category": {
"$first": "$category"
}
}
}])
输出:
[
{
"_id": "John Lennon",
"category": 2
},
{
"_id": "John Snow",
"category": 1
}
]
您可以在这里尝试: mongoplayground.net/p/evw6DP_574r
您可以使用Spring Data Mongo
像这样
Aggregation agg = Aggregation.newAggregation(
ggregation.match(ctr.orOperator(Criteria.where("name").regex("john", "i")),
Aggregation.group("name", "category")
);
AggregationResults<CatalogNoArray> aggResults = mongoTemp.aggregate(agg, "demo",demo.class);
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