在MongoDB中更新嵌套的嵌入式文档？
要更新MongDB中的最佳文档，请使用UPDATE()和positional（$）运算符。让我们创建一个包含文档的集合-
> db.demo643.insertOne({
... details : [
... {
... "CountryName":"US",
... StudentDetails:[{Name:"Chris"},{SubjectName:"MySQL"}]
... },
...
... {
... "CountryName":"UK",
... StudentDetails:[{Name:"Bob"},{SubjectName:"Java"}]
... }
... ]
... }
... )
{
"acknowledged" : true,
"insertedId" : ObjectId("5e9c737f6c954c74be91e6e3")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo643.find();
这将产生以下输出-
{ "_id" : ObjectId("5e9c737f6c954c74be91e6e3"), "details" : [ { "CountryName" : "US", "StudentDetails" : [ { "Name" : "Chris" }, { "SubjectName" : "MySQL" } ] }, { "CountryName" : "UK", "StudentDetails" : [ { "Name" : "Bob" }, { "SubjectName" : "Java" } ] } ] }以下是更新MongoDB中嵌套的嵌入式文档的查询-
> db.demo643.update({"details.CountryName": "UK"}, {"$push": {"details.$.StudentDetails": {Marks:78}}})
WriteResult({ "nMatched" : 1, "nUpserted" : 0, "nModified" : 1 })在find()方法的帮助下显示集合中的所有文档-
> db.demo643.find().pretty();
这将产生以下输出-
{
"_id" : ObjectId("5e9c737f6c954c74be91e6e3"),
"details" : [
{
"CountryName" : "US",
"StudentDetails" : [
{
"Name" : "Chris"
},
{
"SubjectName" : "MySQL"
}
]
},
{
"CountryName" : "UK",
"StudentDetails" : [
{
"Name" : "Bob"
},
{
"SubjectName" : "Java"
},
{
"Marks" : 78
}
]
}
]
}