如何获取 MongoDB 中子文档字段值的唯一列表?
若要获取子文档字段值的唯一列表,可使用点 (.)。语法如下:-
db.yourCollectionName.distinct("yourOuterFieldName.yourInnerFieldName");
为了理解这个概念,让我们使用文档创建一个集合。用于使用文档创建集合的查询如下:-
> db.getDistinctListOfSubDocumentFieldDemo.insertOne( ... { ... "StudentId": 101, ... "StudentPersonalDetails": [ ... { ... "StudentName": "John", ... "StudentAge":24 ... }, ... { ... "StudentName": "Carol", ... "StudentAge":21 ... } ... ] ... } ... ); { "acknowledged" : true, "insertedId" : ObjectId("5c90a9abb74d7cfe6392d7d8") } > db.getDistinctListOfSubDocumentFieldDemo.insertOne( ... ... { ... "StudentId": 102, ... "StudentPersonalDetails": [ ... { ... "StudentName": "Carol", ... "StudentAge":26 ... }, ... { ... "StudentName": "Bob", ... "StudentAge":21 ... } ... ] ... } ... ); { "acknowledged" : true, "insertedId" : ObjectId("5c90a9ceb74d7cfe6392d7d9") } > db.getDistinctListOfSubDocumentFieldDemo.insertOne( ... { ... "StudentId": 103, ... "StudentPersonalDetails": [ ... { ... "StudentName": "Bob", ... "StudentAge":25 ... }, ... { ... "StudentName": "David", ... "StudentAge":24 ... } ... ] ... } ... ); { "acknowledged" : true, "insertedId" : ObjectId("5c90a9e6b74d7cfe6392d7da") }
通过 find() 方法显示来自集合的所有文档。查询如下:-
> db.getDistinctListOfSubDocumentFieldDemo.find().pretty();
输出如下:-
{ "_id" : ObjectId("5c90a9abb74d7cfe6392d7d8"), "StudentId" : 101, "StudentPersonalDetails" : [ { "StudentName" : "John", "StudentAge" : 24 }, { "StudentName" : "Carol", "StudentAge" : 21 } ] } { "_id" : ObjectId("5c90a9ceb74d7cfe6392d7d9"), "StudentId" : 102, "StudentPersonalDetails" : [ { "StudentName" : "Carol", "StudentAge" : 26 }, { "StudentName" : "Bob", "StudentAge" : 21 } ] } { "_id" : ObjectId("5c90a9e6b74d7cfe6392d7da"), "StudentId" : 103, "StudentPersonalDetails" : [ { "StudentName" : "Bob", "StudentAge" : 25 }, { "StudentName" : "David", "StudentAge" : 24 } ] }
以下是获取子文档字段值唯一列表的查询:-
> db.getDistinctListOfSubDocumentFieldDemo.distinct("StudentPersonalDetails.StudentName");
输出如下:-
[ "Carol", "John", "Bob", "David" ]
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