Mongodb按复杂计算值对文档进行排序

本文介绍了Mongodb按复杂计算值对文档进行排序的处理方法，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 items = collection.aggregate([ {"$match": {}}, {"$project": { 'temp_score': { "$add": ["$total_score", 100], }, 'temp_votes': { "$add": ["$total_votes", 20], }, 'weight': { "$divide": ["$temp_score", "$temp_votes"] } } } ])

total_score 和 total_votes 已经存储在文档中，

The total_score and total_votes have stored in the document,

我可以按预期获得 temp_score 和 temp_votes，但无法获得体重，有什么建议吗?

I can get temp_score and temp_votes as expected, but can't get weight, any suggestion?

推荐答案

您的 $temp_score 和 $temp_votes 尚不存在于您的 $divide.

Your $temp_score and $temp_votes are not existing yet in your $divide.

你可以再做一个$project:

db.user.aggregate([{ "$project": { 'temp_score': { "$add": ["$total_score", 100], }, 'temp_votes': { "$add": ["$total_votes", 20], } } }, { "$project": { 'temp_score':1, 'temp_votes':1, 'weight': { "$divide": ["$temp_score", "$temp_votes"] } } }])

或重新计算 $divide 中的 temp_score 和 temp_votes :

or re-computing temp_score and temp_votes in $divide :

db.user.aggregate([{ "$project": { 'temp_score': { "$add": ["$total_score", 100], }, 'temp_votes': { "$add": ["$total_votes", 20], }, 'weight': { "$divide": [ { "$add": ["$total_score", 100] }, { "$add": ["$total_votes", 20] } ] } } }]);

您也可以在一个 $project 中使用 $let operator 将用于创建 2 个变量 temp_score 和 temp_votes.但是可以在单个字段下访问结果(此处为 total):

You can also do this in one single $project using the $let operator that will be used to create 2 variables temp_score and temp_votes. But the results will be accessible under a single field (here total) :

db.user.aggregate([{ $project: { total: { $let: { vars: { temp_score: { $add: ["$total_score", 100] }, temp_votes: { $add: ["$total_votes", 20] } }, in : { temp_score: "$$temp_score", temp_votes: "$$temp_votes", weight: { $divide: ["$$temp_score", "$$temp_votes"] } } } } } }])