我有一个简单的mongoDB集合,名为usersCollection,其常规结构如下:
[
{
"username" : "john",
"sex" : "m",
"email" : "john@gmail.com",
"courses" : [
{
"title" : "medicine",
"grade": 25,
},
{
"title" : "art",
"grade": 29,
},
{
"title" : "history",
"grade": 21,
}
]
},
{
"username" : "jane",
"sex" : "f",
"email" : "jane@gmail.com",
"courses" : [
{
"title" : "math",
"grade": 20,
},
{
"title" : "medicine",
"grade": 30,
}
]
},
{
"username" : "sarah",
"sex" : "f",
"email" : "sarah@gmail.com",
"courses" : [ ]
},
{
"username" : "josh",
"sex" : "f",
"email" : "josh@gmail.com",
"courses" : [
{
"title" : "english",
"grade": 28,
}
]
},
{
"username" : "mark",
"sex" : "m",
"email" : "mark@gmail.com",
"courses" : [
{
"title" : "history",
"grade": 30,
},
{
"title" : "medicine",
"grade": 19,
},
{
"title" : "math",
"grade": 22,
}
]
}
]
每个用户都是usersCollection的成员,并且具有一些常规信息以及他/她已经完成的相对等级的课程列表。
谁能告诉我如何查询usersCollection以获得包含所有已完成特定课程的用户的对象的降序排序数组?此数组的每个对象都应包含相对用户的“名称”,“电子邮件”和“等级”。
例如,围绕课程名称为“ medicine”的usersCollection发起查询,我将获得以下数组:
[
{
"username": "jane",
"email": "jane@gmail.com",
"grade": 30
},
{
"username": "john",
"email": "john@gmail.com",
"grade": 25
},
{
"username": "mark",
"email": "mark@gmail.com",
"grade": 19
}
]
答案 0 :(得分:2)
您可以使用mongoDB的排序和组管道进行管理。 如果您想获得任何特定的课程,可以在排序管道之前应用匹配管道。
db.users.aggregate([
{
$unwind: "$courses"
},
{
$match: {
'courses.title': 'medicine'
}
},
{
$sort: {
'courses.title': -1,
'courses.grade': -1,
}
},
{
$group: {
_id: "$courses.title",
records: {
$push: {
"_id" : "$_id",
"username" : "$username",
"sex" : "$sex",
"email" : "$email",
"grade":"$courses.grade"
}
}
}
}
])