如何仅在MongoDB中获取嵌套JSON对象的数据？
要获取MongoDB中嵌套JSON对象的数据，请使用findOne()。让我们创建一个包含文档的集合-
> db.demo109.insertOne(
... {
... "Name" : "Chris",
... "Subjects" : [
... {
... "Id" : "100",
... "Name":"MySQL",
... "InstructorDetails" : [
... {
... "Name" : "John"
... }
... ]
... },
... {
... "Id" : "101",
... "Name":"MongoDB",
... "InstructorDetails" : [
... {
... "Name" : "Mike"
... }
... ]
... }
... ]
... }
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5e2ee7df9fd5fd66da21447a")
}在find()方法的帮助下显示集合中的所有文档-
> db.demo109.find();
这将产生以下输出-
{
"_id" : ObjectId("5e2ee7df9fd5fd66da21447a"), "Name" : "Chris", "Subjects" : [
{ "Id" : "100", "Name" : "MySQL", "InstructorDetails" : [ { "Name" : "John" } ] },
{ "Id" : "101", "Name" : "MongoDB", "InstructorDetails" : [ { "Name" : "Mike" } ] }
]
}以下是仅获取MongoDB中嵌套JSON对象的数据的查询-
> db.demo109.findOne(
... { Name: "Chris"}
... , { 'Subjects': { $elemMatch:{'Id':"100"} } }
... , function (err, doc) { console.log(doc) });这将产生以下输出-
{
"_id" : ObjectId("5e2ee7df9fd5fd66da21447a"),
"Subjects" : [
{
"Id" : "100",
"Name" : "MySQL",
"InstructorDetails" : [
{
"Name" : "John"
}
]
}
]
}