如何在MongoDB中仅返回唯一值（无重复）？
您可以distinct()用来仅返回唯一值。语法如下-
db.yourCollectionName.distinct("yourFieldName");为了理解这个概念，让我们用文档创建一个集合。使用文档创建集合的查询如下-
> db.returnOnlyUniqueValuesDemo.insertOne({"CusomerName":"Larry","CustomerAge":23});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c8ed7262f684a30fbdfd580")
}
> db.returnOnlyUniqueValuesDemo.insertOne({"CusomerName":"Mike","CustomerAge":21});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c8ed72d2f684a30fbdfd581")
}
> db.returnOnlyUniqueValuesDemo.insertOne({"CusomerName":"Sam","CustomerAge":21});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c8ed7322f684a30fbdfd582")
}
> db.returnOnlyUniqueValuesDemo.insertOne({"CusomerName":"Carol","CustomerAge":25});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c8ed73a2f684a30fbdfd583")
}
> db.returnOnlyUniqueValuesDemo.insertOne({"CusomerName":"David","CustomerAge":22});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c8ed74a2f684a30fbdfd584")
}
> db.returnOnlyUniqueValuesDemo.insertOne({"CusomerName":"Chris","CustomerAge":23});
{
"acknowledged" : true,
"insertedId" : ObjectId("5c8ed7582f684a30fbdfd585")
}在find()method的帮助下显示集合中的所有文档。查询如下-
> db.returnOnlyUniqueValuesDemo.find().pretty();
以下是输出-
{
"_id" : ObjectId("5c8ed7262f684a30fbdfd580"),
"CusomerName" : "Larry",
"CustomerAge" : 23
}
{
"_id" : ObjectId("5c8ed72d2f684a30fbdfd581"),
"CusomerName" : "Mike",
"CustomerAge" : 21
}
{
"_id" : ObjectId("5c8ed7322f684a30fbdfd582"),
"CusomerName" : "Sam",
"CustomerAge" : 21
}
{
"_id" : ObjectId("5c8ed73a2f684a30fbdfd583"),
"CusomerName" : "Carol",
"CustomerAge" : 25
}
{
"_id" : ObjectId("5c8ed74a2f684a30fbdfd584"),
"CusomerName" : "David",
"CustomerAge" : 22
}
{
"_id" : ObjectId("5c8ed7582f684a30fbdfd585"),
"CusomerName" : "Chris",
"CustomerAge" : 23
}这是只返回唯一值的查询-
> db.returnOnlyUniqueValuesDemo.distinct("CustomerAge");以下是输出：
[ 23, 21, 25, 22 ]