无法在MongoDB中实现$ addToSet以获取单个字段的值？
让我们创建一个包含文档的集合-
> db.demo533.insertOne({"ProjectName":"Online Hospital Management"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e8b4cfaef4dcbee04fbbbfc")
}
> db.demo533.insertOne({"ProjectName":"Online Library Management"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e8b4d02ef4dcbee04fbbbfd")
}
> db.demo533.insertOne({"ProjectName":"Online Hospital Management"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e8b4d04ef4dcbee04fbbbfe")
}
> db.demo533.insertOne({"ProjectName":"Online Customer Tracker"});{
"acknowledged" : true,
"insertedId" : ObjectId("5e8b4d0def4dcbee04fbbbff")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo533.find();
这将产生以下输出-
{ "_id" : ObjectId("5e8b4cfaef4dcbee04fbbbfc"), "ProjectName" : "Online Hospital Management" }
{ "_id" : ObjectId("5e8b4d02ef4dcbee04fbbbfd"), "ProjectName" : "Online Library Management" }
{ "_id" : ObjectId("5e8b4d04ef4dcbee04fbbbfe"), "ProjectName" : "Online Hospital Management" }
{ "_id" : ObjectId("5e8b4d0def4dcbee04fbbbff"), "ProjectName" : "Online Customer Tracker" }以下是实现$addToSet并获取字段ProjectName的值的查询-
> db.demo533.aggregate(
... [
... {
... $group:
... {
... _id:null,
... SetOfProject: { $addToSet: "$ProjectName" }
... }
... }
... ]
... )这将产生以下输出-
{ "_id" : null, "SetOfProject" : [ "Online Customer Tracker", "Online Library Management", "Online Hospital Management" ] }热门推荐
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