Mongodb按复杂的计算值对文档进行排序

本文介绍了Mongodb按复杂的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，但是没有重量，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$divide中还不存在您的$temp_score和$temp_votes.

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

您可以再做一个$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:

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] } ] } } }]);

您还可以使用 $let运算符，该运算符将用于创建2个变量temp_score和temp_votes.但是结果将可以在一个字段(此处为total)下访问:

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"] } } } } } }])