使用 mongoDB 中的聚合删除集合中所有文档的特定字段

Posted 2023-02-16

【中文标题】使用 mongoDB 中的聚合删除集合中所有文档的特定字段

【英文标题】：Remove a particular field for all documents in a collection using aggregation in mongoDB

【发布时间】：2020-09-03 05:39:06

【问题描述】：

如何使用聚合删除集合中所有记录的特定值：

收集数据：

[
 
    _id: "bmasndvhjbcw",
    name: "lucas",
    occupation: "scientist",
    present_working:true,
    age: 55,
    location: "texas"

  ,
  
    _id: "bmasndvhjbcx",
    name: "mark",
    occupation: "scientist",
    age: 45,
    present_working:false,
    location: "texas"
  ,
  
    _id: "bmasndvhjbcq",
    name: "cooper",
    occupation: "physicist",
    age: 69,
    location: "texas",
    present_working:false
  
]

删除记录中存在present_working:false 的行。数据库中的数据不需要删除，只需要在聚合管道中修改即可

仅删除 present_working:false 和 present_working:false 后的预期输出应保留在数据库中。 ：

[
 
    _id: "bmasndvhjbcw",
    name: "lucas",
    occupation: "scientist",
    present_working:true,
    age: 55,
    location: "texas"
  ,
  
    _id: "bmasndvhjbcx",
    name: "mark",
    occupation: "scientist",
    age: 45,
    location: "texas"
  ,
  
    _id: "bmasndvhjbcq",
    name: "cooper",
    occupation: "physicist",
    age: 69,
    location: "texas"
  
]

MongoDB 版本：4.0

【参考方案1】：

你可以使用$$REMOVEas part of $project:

db.collection.aggregate([
    
        $project: 
            _id: 1,
            name: 1,
            occupation: 1,
            age: 1,
            location: 1,
            present_working:  $cond: [  $eq: [ "$present_working", false ] , "$$REMOVE", "$present_working" ] 
        
    
])

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