MongoDB noob here ...
好的，我有一个学生集合，每个学生都有一个如下所示的记录....我想按照降序排序'类型'：'作业'分数。
mongo shell上的咒语是什么样的？
> db.students.find({'_id': 1}).pretty()
{
"_id" : 1,
"name" : "Aurelia Menendez",
"scores" : [
{
"type" : "exam",
"score" : 60.06045071030959
},
{
"type" : "quiz",
"score" : 52.79790691903873
},
{
"type" : "homework",
"score" : 71.76133439165544
},
{
"type" : "homework",
"score" : 34.85718117893772
}
]
}
我正在尝试这个咒语......
doc = db.students.find()
for (_id,score) in doc.scores:
print _id,score
但它不起作用。
答案 0 :(得分：46)
您需要在应用程序代码中操作嵌入式数组，或在MongoDB 2.2中使用新的Aggregation Framework。
mongo shell中的示例聚合：
db.students.aggregate(
// Initial document match (uses index, if a suitable one is available)
{ $match: {
_id : 1
}},
// Expand the scores array into a stream of documents
{ $unwind: '$scores' },
// Filter to 'homework' scores
{ $match: {
'scores.type': 'homework'
}},
// Sort in descending order
{ $sort: {
'scores.score': -1
}}
)
示例输出：
{
"result" : [
{
"_id" : 1,
"name" : "Aurelia Menendez",
"scores" : {
"type" : "homework",
"score" : 71.76133439165544
}
},
{
"_id" : 1,
"name" : "Aurelia Menendez",
"scores" : {
"type" : "homework",
"score" : 34.85718117893772
}
}
],
"ok" : 1
}
答案 1 :(得分：8)
我们如何使用JS和mongo控制台来解决这个问题：
db.students.find({"scores.type": "homework"}).forEach(
function(s){
var sortedScores = s.scores.sort(
function(a, b){
return a.score<b.score && a.type=="homework";
}
);
var lowestHomeworkScore = sortedScores[sortedScores.length-1].score;
db.students.update({_id: s._id},{$pull: {scores: {score: lowestHomeworkScore}}}, {multi: true});
})
答案 2 :(得分：3)
这是java代码，可用于找出数组中的最低分数并将其删除。
public class sortArrayInsideDocument{
public static void main(String[] args) throws UnknownHostException {
MongoClient client = new MongoClient();
DB db = client.getDB("school");
DBCollection lines = db.getCollection("students");
DBCursor cursor = lines.find();
try {
while (cursor.hasNext()) {
DBObject cur = cursor.next();
BasicDBList dbObjectList = (BasicDBList) cur.get("scores");
Double lowestScore = new Double(0);
BasicDBObject dbObject = null;
for (Object doc : dbObjectList) {
BasicDBObject basicDBObject = (BasicDBObject) doc;
if (basicDBObject.get("type").equals("homework")) {
Double latestScore = (Double) basicDBObject
.get("score");
if (lowestScore.compareTo(Double.valueOf(0)) == 0) {
lowestScore = latestScore;
dbObject = basicDBObject;
} else if (lowestScore.compareTo(latestScore) > 0) {
lowestScore = latestScore;
dbObject = basicDBObject;
}
}
}
// remove the lowest score here.
System.out.println("object to be removed : " + dbObject + ":"
+ dbObjectList.remove(dbObject));
// update the collection
lines.update(new BasicDBObject("_id", cur.get("_id")), cur,
true, false);
}
} finally {
cursor.close();
}
}
}
答案 3 :(得分：2)
这很容易猜到，但无论如何，尽量不要与mongo大学课程作弊，因为你不会理解基础知识。
./sass/style.scss'答案 4 :(得分：2)
由于这个问题可以通过不同的方式进行管理，我想说另一个解决方案是＆＃34;插入和排序＆＃34;，这样你就可以在你创建一个Find（）时得到Ordered数组。
考虑这些数据：
<?xml version="1.0" encoding="utf-8"?>
<android.support.design.widget.CoordinatorLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:app="http://schemas.android.com/apk/res-auto"
android:id="@+id/main_content"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:background="@android:color/white"
android:fitsSystemWindows="true">
<android.support.design.widget.AppBarLayout
android:id="@+id/appbar"
android:layout_width="match_parent"
android:layout_height="@dimen/detail_backdrop_height"
android:fitsSystemWindows="true"
android:theme="@style/ThemeOverlay.AppCompat.Dark.ActionBar">
<android.support.design.widget.CollapsingToolbarLayout
android:id="@+id/collapsing_toolbar"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:fitsSystemWindows="true"
app:contentScrim="?attr/colorPrimary"
app:expandedTitleMarginEnd="64dp"
app:expandedTitleMarginStart="48dp"
app:expandedTitleTextAppearance="@android:color/transparent"
app:layout_scrollFlags="scroll|exitUntilCollapsed">
<RelativeLayout
android:layout_width="wrap_content"
android:layout_height="wrap_content">
<ImageView
android:id="@+id/backdrop"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:fitsSystemWindows="true"
android:scaleType="centerCrop"
app:layout_collapseMode="parallax" />
<LinearLayout
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_centerInParent="true"
android:gravity="center_horizontal"
android:orientation="vertical">
<TextView
android:id="@+id/love_music"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="@string/backdrop_title"
android:textColor="@android:color/white"
android:textSize="@dimen/backdrop_title" />
<TextView
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="@string/backdrop_subtitle"
android:textColor="@android:color/white"
android:textSize="@dimen/backdrop_subtitle" />
</LinearLayout>
</RelativeLayout>
<android.support.v7.widget.Toolbar
android:id="@+id/toolbar"
android:layout_width="match_parent"
android:layout_height="?attr/actionBarSize"
app:layout_collapseMode="pin"
app:popupTheme="@style/ThemeOverlay.AppCompat.Light" />
</android.support.design.widget.CollapsingToolbarLayout>
</android.support.design.widget.AppBarLayout>
<include layout="@layout/content_main" />
</android.support.design.widget.CoordinatorLayout>
在这里，我们将更新文档，进行排序。
{
"_id" : 5,
"quizzes" : [
{ "wk": 1, "score" : 10 },
{ "wk": 2, "score" : 8 },
{ "wk": 3, "score" : 5 },
{ "wk": 4, "score" : 6 }
]
}
结果是：
db.students.update(
{ _id: 5 },
{
$push: {
quizzes: {
$each: [ { wk: 5, score: 8 }, { wk: 6, score: 7 }, { wk: 7, score: 6 } ],
$sort: { score: -1 },
$slice: 3 // keep the first 3 values
}
}
}
)
文档： https://docs.mongodb.com/manual/reference/operator/update/sort/#up._S_sort
答案 5 :(得分：1)
我相信你正在做M101P: MongoDB for Developers，其中的家庭作业3.1是从两个家庭作业中删除较低的一个。由于到目前为止没有教授聚合，你可以这样做：
import pymongo
conn = pymongo.MongoClient('mongodb://localhost:27017')
db = conn.school
students = db.students
for student_data in students.find():
smaller_homework_score_seq = None
smaller_homework_score_val = None
for score_seq, score_data in enumerate(student_data['scores']):
if score_data['type'] == 'homework':
if smaller_homework_score_seq is None or smaller_homework_score_val > score_data['score']:
smaller_homework_score_seq = score_seq
smaller_homework_score_val = score_data['score']
students.update({'_id': student_data['_id']}, {'$pop': {'scores': smaller_homework_score_seq}})
答案 6 :(得分：0)
@Stennie的答案很好，也许一个$ group运算符对于保留原始文档很有用，而不会在许多文档中爆炸（一个得分）。
我只是在为您的应用程序使用javascript时添加另一种解决方案。
如果只查询一个文档，它有时更容易通过JS对嵌入式数组进行排序，而不是进行聚合。 当你的文档有很多字段时，它甚至比使用$ push操作符更好，否则你可以逐个推送所有字段，或者使用$$ ROOT操作符（我错了吗？）< / p>
我的示例代码使用 Mongoose.js ： 假设您已初始化学生模型。
// Sorting
function compare(a, b) {
return a.score - b.score;
}
Students.findById('1', function(err, foundDocument){
foundDocument.scores = foundDocument.scores.sort(compare);
// do what you want here...
// foundModel keeps all its fields
});
答案 7 :(得分：0)
这项工作对我来说，这是一个有点粗略的代码，但每个学生最低任务的结果是正确的。
var scores_homework = []
db.students.find({"scores.type": "homework"}).forEach(
function(s){
s.scores.forEach(
function(ss){
if(ss.type=="homework"){
ss.student_id = s._id
scores_homework.push(ss)
}
}
)
})
for(i = 0; i < scores_homework.length; i++)
{
var b = i+1;
var ss1 = scores_homework[i];
var ss2 = scores_homework[b];
var lowest_score = {};
if(ss1.score > ss2.score){
lowest_score.type = ss2.type;
lowest_score.score = ss2.score;
db.students.update({_id: ss2.student_id},{$pull: {scores: {score: lowest_score.score}}});
}else if(ss1.score < ss2.score){
lowest_score.type = ss1.type;
lowest_score.score = ss1.score;
db.students.update({_id: ss1.student_id},{$pull: {scores: {score: lowest_score.score}}});
}else{
lowest_score.type = ss1.type;
lowest_score.score = ss1.score;
db.students.update({_id: ss1.student_id},{$pull: {scores: {score: lowest_score.score}}});
}
i++
}
答案 8 :(得分：0)
这是我使用pyMongo的方法，它是MongoDB的Python驱动程序：
//Put this line in onCreate
getListView().setPadding(0, (int) getTypedValueInDP(50), 0, 0);
private float getTypedValueInDP(float value) {
return TypedValue.applyDimension(TypedValue.COMPLEX_UNIT_DIP, value, getResources().getDisplayMetrics());
}
答案 9 :(得分：0)
这就是我用Java实现的方法（保持简单，以便更容易理解） -
方法：
下面是Java代码：
public void removeLowestScore(){
//Create mongo client and database connection and get collection
MongoClient client = new MongoClient("localhost");
MongoDatabase database = client.getDatabase("school");
MongoCollection<Document> collection = database.getCollection("students");
FindIterable<Document> docs = collection.find();
for (Document document : docs) {
//Get scores array
ArrayList<Document> scores = document.get("scores", ArrayList.class);
//Create a list of scores where type = homework
List<Double> homeworkScores = new ArrayList<Double>();
for (Document score : scores) {
if(score.getString("type").equalsIgnoreCase("homework")){
homeworkScores.add(score.getDouble("score"));
}
}
//sort homework scores
Collections.sort(homeworkScores);
//Create a new list to update into student collection
List<Document> newScoresArray = new ArrayList<Document>();
Document scoreDoc = null;
//Below loop populates new score array with eliminating lowest score of "type" = "homework"
for (Document score : scores) {
if(score.getString("type").equalsIgnoreCase("homework") && homeworkScores.get(0) == score.getDouble("score")){
continue;
}else{
scoreDoc = new Document("type",score.getString("type"));
scoreDoc.append("score",score.getDouble("score"));
newScoresArray.add(scoreDoc);
}
}
//Update the scores array for every student using student _id
collection.updateOne(Filters.eq("_id", document.getInteger("_id")), new Document("$set",new Document("scores",newScoresArray)));
}
}
答案 10 :(得分：0)
当然已经晚了，但我只是想在Mongo Shell上贡献自己的解决方案：
var students = db.getCollection('students').find({});
for(i = 0 ; i < students.length(); i++) {
var scores = students[i].scores;
var tmp = [];
var min = -1 ;
var valueTmp = {};
for(j = 0 ; j < scores.length; j++) {
if(scores[j].type != 'homework') {
tmp.push(scores[j]);
} else {
if (min == -1) {
min = scores[j].score;
valueTmp = scores[j];
} else {
if (min > scores[j].score) {
min = scores[j].score;
tmp.push(valueTmp);
valueTmp = scores[j];
} else {
tmp.push(scores[j]);
}
}
}
}
db.students.updateOne({_id:students[i]._id},
{$set:{scores:tmp}});
}
答案 11 :(得分：0)
为了对数组排序，请按照下列步骤操作：
1）使用展开来遍历数组
2）排序数组
3）使用组将数组的对象合并到一个数组中
4），然后投射其他字段
查询
m_SDF答案 12 :(得分：0)
同时显示订单标题和数组标题，并返回整个集合数据集合名称为菜单
[
{
"_id": "5f27c5132160a22f005fd50d",
"title": "Gift By Category",
"children": [
{
"title": "Ethnic Gift Items",
"s": "/gift?by=Category&name=Ethnic"
},
{
"title": "Novelty Gift Items",
"link": "/gift?by=Category&name=Novelty"
}
],
"active": true
},
{
"_id": "5f2752fc2160a22f005fd50b",
"title": "Gift By Occasion",
"children": [
{
"title": "Gifts for Diwali",
"link": "/gift-for-diwali"
},
{
"title": "Gifts for Ganesh Chathurthi",
"link": "/gift-for-ganesh-chaturthi",
}
],
"active": true
}
]
查询如下
let menuList = await Menu.aggregate([
{
$unwind: '$children'
},
{
$sort:{"children.title":1}
},
{
$group : { _id : "$_id",
root: { $mergeObjects: '$$ROOT' },
children: { $push: "$children" }
}
},
{
$replaceRoot: {
newRoot: {
$mergeObjects: ['$root', '$$ROOT']
}
}
},
{
$project: {
root: 0
}
},
{
$match: {
$and:[{'active':true}],
}
},
{
$sort:{"title":1}
}
]);
答案 13 :(得分：-2)
按分数排序可以很简单：
db.students.find({_id:137}).sort({score:-1}).pretty()
但你需要找到一个类型：homework ...
答案 14 :(得分：-3)
它应该是这样的：
db.students.find().sort(scores: ({"score":-1}));