在MongoDB中以“SQL UNION”的方式进行联合，可以在单个查询中使用聚合和查找。下面是我测试过的MongoDB 4.0的一个例子:

// Create employees data for testing the union.
db.getCollection('employees').insert({ name: "John", type: "employee", department: "sales" });
db.getCollection('employees').insert({ name: "Martha", type: "employee", department: "accounting" });
db.getCollection('employees').insert({ name: "Amy", type: "employee", department: "warehouse" });
db.getCollection('employees').insert({ name: "Mike", type: "employee", department: "warehouse"  });

// Create freelancers data for testing the union.
db.getCollection('freelancers').insert({ name: "Stephany", type: "freelancer", department: "accounting" });
db.getCollection('freelancers').insert({ name: "Martin", type: "freelancer", department: "sales" });
db.getCollection('freelancers').insert({ name: "Doug", type: "freelancer", department: "warehouse"  });
db.getCollection('freelancers').insert({ name: "Brenda", type: "freelancer", department: "sales"  });

// Here we do a union of the employees and freelancers using a single aggregation query.
db.getCollection('freelancers').aggregate( // 1. Use any collection containing at least one document.
  [
    { $limit: 1 }, // 2. Keep only one document of the collection.
    { $project: { _id: '$$REMOVE' } }, // 3. Remove everything from the document.

    // 4. Lookup collections to union together.
    { $lookup: { from: 'employees', pipeline: [{ $match: { department: 'sales' } }], as: 'employees' } },
    { $lookup: { from: 'freelancers', pipeline: [{ $match: { department: 'sales' } }], as: 'freelancers' } },

    // 5. Union the collections together with a projection.
    { $project: { union: { $concatArrays: ["$employees", "$freelancers"] } } },

    // 6. Unwind and replace root so you end up with a result set.
    { $unwind: '$union' },
    { $replaceRoot: { newRoot: '$union' } }
  ]);

下面是它的工作原理:

Instantiate an aggregate out of any collection of your database that has at least one document in it. If you can't guarantee any collection of your database will not be empty, you can workaround this issue by creating in your database some sort of 'dummy' collection containing a single empty document in it that will be there specifically for doing union queries. Make the first stage of your pipeline to be { $limit: 1 }. This will strip all the documents of the collection except the first one. Strip all the fields of the remaining document by using a $project stage: { $project: { _id: '$$REMOVE' } } Your aggregate now contains a single, empty document. It's time to add lookups for each collection you want to union together. You may use the pipeline field to do some specific filtering, or leave localField and foreignField as null to match the whole collection. { $lookup: { from: 'collectionToUnion1', pipeline: [...], as: 'Collection1' } }, { $lookup: { from: 'collectionToUnion2', pipeline: [...], as: 'Collection2' } }, { $lookup: { from: 'collectionToUnion3', pipeline: [...], as: 'Collection3' } } You now have an aggregate containing a single document that contains 3 arrays like this: { Collection1: [...], Collection2: [...], Collection3: [...] } You can then merge them together into a single array using a $project stage along with the $concatArrays aggregation operator: { "$project" : { "Union" : { $concatArrays: ["$Collection1", "$Collection2", "$Collection3"] } } } You now have an aggregate containing a single document, into which is located an array that contains your union of collections. What remains to be done is to add an $unwind and a $replaceRoot stage to split your array into separate documents: { $unwind: "$Union" }, { $replaceRoot: { newRoot: "$Union" } } Voilà. You now have a result set containing the collections you wanted to union together. You can then add more stages to filter it further, sort it, apply skip() and limit(). Pretty much anything you want.

Mongorestore有这样一个特性，即在数据库中已经存在的数据之上追加数据，所以这个行为可以用于组合两个集合:

mongodump文物 collection2.rename(文物) mongorestore

还没有尝试过，但它可能比map/reduce方法执行得更快。

代码片段。礼貌-关于堆栈溢出的多个帖子，包括这一篇。

 db.cust.drop();
 db.zip.drop();
 db.cust.insert({cust_id:1, zip_id: 101});
 db.cust.insert({cust_id:2, zip_id: 101});
 db.cust.insert({cust_id:3, zip_id: 101});
 db.cust.insert({cust_id:4, zip_id: 102});
 db.cust.insert({cust_id:5, zip_id: 102});

 db.zip.insert({zip_id:101, zip_cd:'AAA'});
 db.zip.insert({zip_id:102, zip_cd:'BBB'});
 db.zip.insert({zip_id:103, zip_cd:'CCC'});

mapCust = function() {
    var values = {
        cust_id: this.cust_id
    };
    emit(this.zip_id, values);
};

mapZip = function() {
    var values = {
    zip_cd: this.zip_cd
    };
    emit(this.zip_id, values);
};

reduceCustZip =  function(k, values) {
    var result = {};
    values.forEach(function(value) {
    var field;
        if ("cust_id" in value) {
            if (!("cust_ids" in result)) {
                result.cust_ids = [];
            }
            result.cust_ids.push(value);
        } else {
    for (field in value) {
        if (value.hasOwnProperty(field) ) {
                result[field] = value[field];
        }
         };  
       }
      });
       return result;
};


db.cust_zip.drop();
db.cust.mapReduce(mapCust, reduceCustZip, {"out": {"reduce": "cust_zip"}});
db.zip.mapReduce(mapZip, reduceCustZip, {"out": {"reduce": "cust_zip"}});
db.cust_zip.find();


mapCZ = function() {
    var that = this;
    if ("cust_ids" in this.value) {
        this.value.cust_ids.forEach(function(value) {
            emit(value.cust_id, {
                zip_id: that._id,
                zip_cd: that.value.zip_cd
            });
        });
    }
};

reduceCZ = function(k, values) {
    var result = {};
    values.forEach(function(value) {
        var field;
        for (field in value) {
            if (value.hasOwnProperty(field)) {
                result[field] = value[field];
            }
        }
    });
    return result;
};
db.cust_zip_joined.drop();
db.cust_zip.mapReduce(mapCZ, reduceCZ, {"out": "cust_zip_joined"}); 
db.cust_zip_joined.find().pretty();


var flattenMRCollection=function(dbName,collectionName) {
    var collection=db.getSiblingDB(dbName)[collectionName];

    var i=0;
    var bulk=collection.initializeUnorderedBulkOp();
    collection.find({ value: { $exists: true } }).addOption(16).forEach(function(result) {
        print((++i));
        //collection.update({_id: result._id},result.value);

        bulk.find({_id: result._id}).replaceOne(result.value);

        if(i%1000==0)
        {
            print("Executing bulk...");
            bulk.execute();
            bulk=collection.initializeUnorderedBulkOp();
        }
    });
    bulk.execute();
};


flattenMRCollection("mydb","cust_zip_joined");
db.cust_zip_joined.find().pretty();

非常基本的$lookup示例。

db.getCollection('users').aggregate([
    {
        $lookup: {
            from: "userinfo",
            localField: "userId",
            foreignField: "userId",
            as: "userInfoData"
        }
    },
    {
        $lookup: {
            from: "userrole",
            localField: "userId",
            foreignField: "userId",
            as: "userRoleData"
        }
    },
    { $unwind: { path: "$userInfoData", preserveNullAndEmptyArrays: true }},
    { $unwind: { path: "$userRoleData", preserveNullAndEmptyArrays: true }}
])

这里用到了

 { $unwind: { path: "$userInfoData", preserveNullAndEmptyArrays: true }}, 
 { $unwind: { path: "$userRoleData", preserveNullAndEmptyArrays: true }}

而不是

{ $unwind:"$userRoleData"} 
{ $unwind:"$userRoleData"}

因为{$unwind:"$userRoleData"}如果在$lookup中没有找到匹配的记录，将返回空或0结果。

虽然不能实时执行，但可以多次运行map-reduce，通过使用MongoDB 1.8+ map/reduce中的“reduce”out选项将数据合并在一起(参见http://www.mongodb.org/display/DOCS/MapReduce#MapReduce-Outputoptions)。您需要在两个集合中都有一些可以用作_id的键。

例如，假设您有一个用户集合和一个评论集合，并且您希望有一个新的集合，其中包含每个评论的一些用户统计信息。

让我们说users集合有以下字段:

_id firstName 姓 国家 性别 年龄

然后comments集合有以下字段:

_id 用户标识 评论 创建

你可以这样做:

var mapUsers, mapComments, reduce;
db.users_comments.remove();

// setup sample data - wouldn't actually use this in production
db.users.remove();
db.comments.remove();
db.users.save({firstName:"Rich",lastName:"S",gender:"M",country:"CA",age:"18"});
db.users.save({firstName:"Rob",lastName:"M",gender:"M",country:"US",age:"25"});
db.users.save({firstName:"Sarah",lastName:"T",gender:"F",country:"US",age:"13"});
var users = db.users.find();
db.comments.save({userId: users[0]._id, "comment": "Hey, what's up?", created: new ISODate()});
db.comments.save({userId: users[1]._id, "comment": "Not much", created: new ISODate()});
db.comments.save({userId: users[0]._id, "comment": "Cool", created: new ISODate()});
// end sample data setup

mapUsers = function() {
    var values = {
        country: this.country,
        gender: this.gender,
        age: this.age
    };
    emit(this._id, values);
};
mapComments = function() {
    var values = {
        commentId: this._id,
        comment: this.comment,
        created: this.created
    };
    emit(this.userId, values);
};
reduce = function(k, values) {
    var result = {}, commentFields = {
        "commentId": '', 
        "comment": '',
        "created": ''
    };
    values.forEach(function(value) {
        var field;
        if ("comment" in value) {
            if (!("comments" in result)) {
                result.comments = [];
            }
            result.comments.push(value);
        } else if ("comments" in value) {
            if (!("comments" in result)) {
                result.comments = [];
            }
            result.comments.push.apply(result.comments, value.comments);
        }
        for (field in value) {
            if (value.hasOwnProperty(field) && !(field in commentFields)) {
                result[field] = value[field];
            }
        }
    });
    return result;
};
db.users.mapReduce(mapUsers, reduce, {"out": {"reduce": "users_comments"}});
db.comments.mapReduce(mapComments, reduce, {"out": {"reduce": "users_comments"}});
db.users_comments.find().pretty(); // see the resulting collection

此时，您将拥有一个名为users_comments的新集合，其中包含合并的数据，您现在可以使用它了。这些简化的集合都有_id，这是你在map函数中发出的键，然后所有的值都是value键内的子对象-这些值不在这些简化文档的顶层。

这是一个比较简单的例子。您可以重复使用更多的集合，只要您想继续构建减少的集合。您还可以在该过程中对数据进行总结和聚合。随着聚合和保存现有字段的逻辑变得更加复杂，您可能会定义多个reduce函数。

您还会注意到，现在每个用户都有一个文档，数组中包含该用户的所有评论。如果我们合并的数据是一对一的关系，而不是一对多的关系，它将是平坦的，你可以简单地使用这样的reduce函数:

reduce = function(k, values) {
    var result = {};
    values.forEach(function(value) {
        var field;
        for (field in value) {
            if (value.hasOwnProperty(field)) {
                result[field] = value[field];
            }
        }
    });
    return result;
};

如果你想平铺users_comments集合，所以每个注释只有一个文档，另外运行这个:

var map, reduce;
map = function() {
    var debug = function(value) {
        var field;
        for (field in value) {
            print(field + ": " + value[field]);
        }
    };
    debug(this);
    var that = this;
    if ("comments" in this.value) {
        this.value.comments.forEach(function(value) {
            emit(value.commentId, {
                userId: that._id,
                country: that.value.country,
                age: that.value.age,
                comment: value.comment,
                created: value.created,
            });
        });
    }
};
reduce = function(k, values) {
    var result = {};
    values.forEach(function(value) {
        var field;
        for (field in value) {
            if (value.hasOwnProperty(field)) {
                result[field] = value[field];
            }
        }
    });
    return result;
};
db.users_comments.mapReduce(map, reduce, {"out": "comments_with_demographics"});

这个技巧绝对不应该在飞行中执行。它适用于定期更新合并数据的cron作业或类似的工作。您可能希望在新集合上运行ensureIndex，以确保对它执行的查询能够快速运行(请记住，您的数据仍然在值键中，因此如果您要在注释创建时间上索引comments_with_demographic，那么它将是db.comments_with_demographic .ensureIndex({"value.created": 1});

如果mongodb没有批量插入，我们循环small_collection中的所有对象，并将它们逐个插入到big_collection中:

db.small_collection.find().forEach(function(obj){ 
   db.big_collection.insert(obj)
});