如何在 MongoDB 中仅返回唯一值(无重复项)?
您可以使用 distinct() 仅返回唯一值。语法如下 -
db.yourCollectionName.distinct("yourFieldName");为了理解这个概念,我们创建一个包含文档的集合。创建包含文档的集合的查询如下 -
> 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")
}使用 find() 方法显示集合中的所有文档。查询如下 -
> db.returnOnlyUniqueValuesDemo.find().pretty();
输出如下 -
{
"_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
}以下查询仅返回唯一值 -
> db.returnOnlyUniqueValuesDemo.distinct("CustomerAge");输出如下
[ 23, 21, 25, 22 ]
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