本文介绍了Mongo DB根据关键字段查找具有最高价值的所有记录的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想为集合中的每个用户获取具有最高价值的记录的注释.
I would like to get for each user in my collection the comment of the record with the highest value.
//myCol
[
{'user' : 1, 'value':20, "comment": "cloudy", "date":"2018-12-01"},
{'user' : 2, 'value':5, "comment": "sun","date":"2018-12-01"},
{'user' : 3, 'value':13, "comment": "rain","date":"2018-13-01"},
{'user' : 4, 'value':13, "comment": "cloudy","date":"2018-13-01"},
{'user' : 2, 'value':20, "comment": "rain","date":"2018-01-20"},
{'user' : 1, 'value':2, "comment": "cloudy","date":"2018-02-02"},
]
将给出以下结果:
//Result
[
{'user' : 1, 'value':20, "comment": "cloudy", "date":"2018-12-01"},
{'user' : 2, 'value':20, "comment": "rain","date":"2018-01-20"},
{'user' : 3, 'value':13, "comment": "rain","date":"2018-13-01"},
{'user' : 4, 'value':13, "comment": "cloudy","date":"2018-13-01"},
]
在MySQL中,我将使用一些联接逻辑
With MySQL I would use some join logic
SELECT user FROM myTable as t0
LEFT JOIN (SELECT user, max(value) as maxVal from myTable ) t1 on t0.user = t1.user and t0.value = t1.maxVal
WHERE maxVal IS NOT NULL
但是对于Mongo,这种逻辑似乎并不存在.
With Mongo however this logic does not seem to exist.
我尝试通过首先获取每个用户的最大值将代码分成两部分:
I tried to split the code in two by getting first the max values for each user:
max_list = myCol.aggregate(
[
{
"$group":
{
"_id": "$user",
"maxValue": { "$max": "$value" }
}
},
{
"$project" : {
"user" : "$_id",
"maxValue":"$maxValue",
"_id":0
}
}
]
)
==> [{'user':1, 'maxValue':20}...]
为此,我为使用find
函数的一种好方法scratch之以鼻,尤其是$in
的一种好用法,以便仅获取与my_list
中存在的两个值都匹配的结果
With this I'm scratching my head for a good way to use the find
function and especially the good use of $in
to get only the results which match both the values present in my_list
推荐答案
可以,但是方法略有不同:
It does, but the approach is slightly different:
db.myCol.aggregate([
{$sort: {value:-1}},
{$group:{
_id: "$user",
doc: {$first: "$$ROOT"}
}},
{$replaceRoot: {newRoot: "$doc"} }
])
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