如何删除mongodb中聚合查询返回的文档 [英] How to delete documents returned by an aggregation query in mongodb

问题描述

我正在尝试删除 Mongodb 中的聚合返回的所有文档.

I am attempting to delete all the documents returned by an aggregation in Mongodb.

我的查询如下:

db.getCollection("Collection")
  .aggregate([
    {
      $match: { status: { $in: ["inserted", "done", "duplicated", "error"] } }
    },
    {
      $project: {
        yearMonthDay: { $dateToString: { format: "%Y-%m-%d", date: "$date" } }
      }
    },  
    { $match: { yearMonthDay: { $eq: "2019-08-06" } } }
  ])
  .forEach(function(doc) {
    db.getCollection("Collection").remove({});
  });

我试过这个查询，但它删除了数据库中的所有数据，请问有什么建议吗?

I tried this query but it removes all the data in the database, any suggestions please?

推荐答案

由于 remove 没有查询条件，它会与所有文档匹配并删除，而不管聚合结果如何.

Since the remove doesn't have a query condition its going to match with all the documents and delete irrespective of the aggregation result.

解决方案(匹配当前光标文档的id):

Solution (match the ids of the current cursor doc):

db.getCollection("Collection")
  .aggregate([
    {
      $match: { status: { $in: ["inserted", "done", "duplicated", "error"] } }
    },
    {
      $project: {
        yearMonthDay: { $dateToString: { format: "%Y-%m-%d", date: "$date" } }
      }
    },
    { $match: { yearMonthDay: { $eq: "2019-08-06" } } }
  ])
  .forEach(function(doc) {
    db.getCollection("Collection").remove({ "_id": doc._id });
  });

另一个更好的解决方案是对db进行单次往返，而删除是从聚合中获取id列表cursor() 通过 cursor.map()

Another better solution would be to have single round trip to db while deletion is get a list of ids from the aggregation cursor() via cursor.map()

var idsList = db
  .getCollection("Collection")
  .aggregate([
    {
      $match: { status: { $in: ["inserted", "done", "duplicated", "error"] } }
    },
    {
      $project: {
        yearMonthDay: { $dateToString: { format: "%Y-%m-%d", date: "$date" } }
      }
    },
    { $match: { yearMonthDay: { $eq: "2019-08-06" } } }
  ])
  .map(function(d) {
    return d._id;
  });

//now delete those documents via $in operator
db.getCollection("Collection").remove({ _id: { $in: idsList } });

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