使用聚合将值添加到记录

本文介绍了使用聚合将值添加到记录的处理方法，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

我的文档样本

{ _id: "bmasndvhjbcw", name: "lucas", occupation: "scientist", present_working:true, age: 55, location: "texas", }, { _id: "bmasndvhjbcx", name: "mark", occupation: "scientist", age: 45, present_working:true, location: "texas", }, { _id: "bmasndvhjbca", name: "stuart", occupation: "lab assistant", age: 25, location: "texas", }, { _id: "bmasndvhjbcq", name: "cooper", occupation: "physicist", age: 69, location: "texas" } ]

对于没有present_working:true的记录，需要添加present_working:false

For the records which doesn't have present_working:true need to add present_working:false

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{ _id: "bmasndvhjbcw", name: "lucas", occupation: "scientist", present_working:true, age: 55, location: "texas", }, { _id: "bmasndvhjbcx", name: "mark", occupation: "scientist", age: 45, present_working:true, location: "texas", }, { _id: "bmasndvhjbca", name: "stuart", occupation: "lab assistant", age: 25, present_working:false location: "texas", }, { _id: "bmasndvhjbcq", name: "cooper", occupation: "physicist", age: 69, present_working:false, location: "texas" } ]

推荐答案

您可以根据需要使用其中之一:

You can use one of these as per your need:

db.collection.aggregate( [ { $match: { present_working: { $exists: false } } }, { $addFields: { present_working: false } } ] ) db.collection.aggregate( [ { $addFields: { present_working: { $ifNull: [ "$present_working", false ] } } } ] )

第一次聚合仅返回带有新添加字段的文档.第二个有文档，有和没有添加字段.

The first aggregation returns only the documents with the newly added field. The second has documents, with and without the added field.