Question

我有一个MongoDB集合，组。每个组都有一个由成员的唯一ID引用的成员数组。用户可以是多个组的成员。

组数据如下：

{ _id: 1, members: ['1', '3', '5'], status: 'active' }
{ _id: 2, members: ['4', '1', '10', '11'], status: 'inactive' }
{ _id: 3, members: ['1', '2', '9'], status: 'active' }

我正在尝试将活动组的所有成员提取为单个成员ID数组，而不重复。我想使用mongoexport将它们导出到csv文件。

我可以导出相关项目的ID及其成员列表：

mongoexport -h localhost:3001 --db mydbname --collection groups --type=csv --fields _id,members --query '{"status": "active"}' --out mongotestoutput.txt

但是我不知道如何将成员列表合并到单个数组中。我一直在看Aggregation，但我只是迷失在所有不同的选择中，我看不到哪个可以满足我的需求。非常感谢您的帮助。

Answer 1

先使用Bank: The Enloe State Bank, City: Cooper, State: TX, Closing Date: May 31, 2019, Cert #: 10716, Acquiring Inst: Legend Bank, N. A. Bank: Washington Federal Bank for Savings, City: Chicago, State: IL, Closing Date: December 15, 2017, Cert #: 30570, Acquiring Inst: Royal Savings Bank Bank: The Farmers and Merchants State Bank of Argonia, City: Argonia, State: KS, Closing Date: October 13, 2017, Cert #: 17719, Acquiring Inst: Conway Bank Bank: Fayette County Bank, City: Saint Elmo, State: IL, Closing Date: May 26, 2017, Cert #: 1802, Acquiring Inst: United Fidelity Bank, fsb Bank: Guaranty Bank, (d/b/a BestBank in Georgia & Michigan) , City: Milwaukee, State: WI, Closing Date: May 5, 2017, Cert #: 30003, Acquiring Inst: First-Citizens Bank & Trust Company Bank: First NBC Bank, City: New Orleans, State: LA, Closing Date: April 28, 2017, Cert #: 58302, Acquiring Inst: Whitney Bank Bank: Proficio Bank, City: Cottonwood Heights, State: UT, Closing Date: March 3, 2017, Cert #: 35495, Acquiring Inst: Cache Valley Bank进行聚合，然后使用$unwind来创建看起来像您需要的新集合。然后将此新集合导出到CSV文件。

$out

db.test1.insertMany([ { _id: 1, members: ['1', '3', '5'], status: 'active' }, { _id: 2, members: ['4', '1', '10', '11'], status: 'inactive' }, { _id: 3, members: ['9'], status: 'active' } ])此处和下方用于禁止_id字段

{_id:0}

或者，如果您需要按数组中的状态获取成员，请添加另一个db.test1.aggregate([ {$unwind: "$members"}, {$project:{_id:0}}, {$out:"test2"} ]) db.test2.find({},{_id:0}) { "members" : "1", "status" : "active" } { "members" : "3", "status" : "active" } { "members" : "5", "status" : "active" } { "members" : "4", "status" : "inactive" } { "members" : "1", "status" : "inactive" } { "members" : "10", "status" : "inactive" } { "members" : "11", "status" : "inactive" } { "members" : "9", "status" : "active" }，$group阶段：

$addToSet

请参见MongoPlayground

Answer 2

This question显示了两种获得所需结果的方法。可接受的答案使用push，reduce和setUnion。较新的答案更简单，使用unwind和addToSet。两者都对我有用，但是我要使用更简单的版本：

db.collection.aggregate([
  { $match: { "status": "active" },
  { $unwind: "$members"},
  { $group:{
     _id: 0,
     selectedMembers: { $addToSet: '$members' }      
  }}
])

我正在从此表达式返回的JSON对象中提取所需的数组。

如何将多个文档中的数组字段合并到MongoDB中的单个输出数组中，以及如何将mongoexport合并到csv

2 个答案: