英文:
Find Latest Object in MongoDB
问题
我有一些以以下形式存储在MongoDB中的数据：
{
"_id" : 154,
"record_id" : "001280000033x54AAA",
"jsonData" : "",
"user_id" : 1,
"userName" : "abc@xyz.com",
"backup_no" : 1
}
{
"_id" : 155,
"record_id" : "001280000033x54AAA",
"jsonData" : "",
"user_id" : 1,
"userName" : "abc@xyz.com",
"backup_no" : 2
}
...
我想根据'user_id'和'userName'检索数据，但是如果较低的'backup_no'中存在具有相同'record_id'的记录，则我需要选择具有最高'backup_no'的记录。
我尝试了对'record_id'进行聚合，然后查询，但是我找不到解决方案。
谢谢。
英文:
I have data in MongoDB in the form of :
{
"_id" : 154,
"record_id" : "001280000033x54AAA",
"jsonData" : "",
"user_id" : 1,
"userName" : "abc@xyz.com",
"backup_no" : 1
}
{
"_id" : 155,
"record_id" : "001280000033x54AAA",
"jsonData" : "",
"user_id" : 1,
"userName" : "abc@xyz.com",
"backup_no" : 2
}
...
I want to retrieve data based on 'user_id','userName', but if a record with same record_id exists in lower 'backup_no' then i need to choose record with highest backup_no.
I have tried to aggregate record_id's and then query but i am unable to find a solution.
Thanks
答案1
得分: 1
使用Mongo的聚合操作，你可以得到以下查询结果：
db.collectionName.aggregate({
"$sort": {
"backup_no": -1 //首先按backup_no排序
}
}, {
"$group": {
"_id": "$record_id", //按record_id分组，只获取唯一的record_id
"userId": {
"$first": "$user_id"
},
"userName": {
"$first": "$userName"
},
"backup_no": {
"$first": "$backup_no"
}
}
}, {
"$project": {
"_id": 0,
"record_id": "$_id",
"user_id": "$user_id",
"userName": "$userName",
"backup_no": "$backup_no"
}
})
希望对你有帮助！
英文:
Using mongo aggregation you can get your results check below query :
db.collectionName.aggregate({
"$sort": {
"backup_no": -1 //first sort by backup_no
}
}, {
"$group": {
"_id": "$record_id", // group by record_id so here only get distinct record_id
"userId": {
"$first": "$user_id"
},
"userName": {
"$first": "$userName"
},
"backup_no": {
"$first": "$backup_no"
}
}
}, {
"$project": {
"_id": 0,
"record_id": "$_id",
"user_id": "$user_id",
"userName": "$userName",
"backup_no": "$backup_no"
}
})
答案2
得分: 0
以下是翻译好的内容：
db.collection.aggregate([
{
$group: {
_id: "$record_id",
user_id: {
$last: "$user_id"
},
userName: {
$last: "$userName"
},
backup_no: {
$max: "$backup_no"
}
}
}
])
或者
你可以通过按backup_no降序排序，然后忽略除了每个不同record_id的第一条记录之外的所有记录。
db.users.find({}).sort({ backup_no: -1 });
英文:
db.collection.aggregate([{
$group: {
_id: "$record_id",
user_id: {
$last: "$user_id"
},
userName: {
$last: "$userName"
},
backup_no: {
$max: "$backup_no"
}
}
}])
or
You could simply sort descending by backup_no and then ignore all but the first record for each distinct record_id?
MongoDB $orderby
db.users.find( { } ).sort( { backup_no: -1 } );
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论