Twitter-like app using MongoDB

问题

I'm making an app that uses the classic "follow" mechanism (the one used by Twitter and a lot of other apps around the web). I'm using MongoDB. My system has a difference, though: an user can follow groups of users. That means that if you follow a group, you'll automatically follow all the users who are members of that group. Of course users can belong to more than one group.

This is what I came up with:

when user A follows user B, id of user B gets added to an embedded array (called following) in user A's document
for unfollowing, I remove the id of the followed user from the following array
groups work in the same way: when user A follows group X, id of group X gets added to the following array. (I actually add a DBRef so I know if the connection is to an user or a group.)
when I have to check if user A follows group X, I just search for the group's id in user A's following array.
when I have to check if user A follows user B, things gets a little trickier. Each user's document has an embedded array listing all the groups the user belongs to. So I use an $or condition to check if user A is either following user B directly or via a group. Like this:

db.users.find({'$or':{'following.ref.$id':$user_id,'following.ref.$ref','users'},{'following.ref.$id':{'$in':$group_ids},'following.ref.$ref':'groups'}}})

This works fine, but I think I have a few issues. For example how do I show a list of followers for a particular user, including pagination? I can't use skip() and limit() on an embedded document.

I could change the design and use an userfollow collection, which would do the same job of the embedded following document. The problem with this approach, which I tried, is that with the $or condition I used earlier, users following two groups containing the same user would be listed twice. To avoid this I could use group or MapReduce, which I actually did and it works, but I'd love to avoid this to keep things simpler. Maybe I just need to think out of the box. Or maybe I took the wrong approach with both tries. Anyone already had to do a similar thing and came up with a better solution?

(This is actually a follow-up to this older question of mine. I decided to post a new question to explain my new situation better; I hope it's not a problem.)

回答1:

You have two possible ways in which a user can follow another user; either directly, or indirectly through a group, in which case the user directly follows the group. Let's begin with storing these direct relations between users and groups:

{
  _id: "userA",
  followingUsers: [ "userB", "userC" ],
  followingGroups: [ "groupX", "groupY" ]
}

Now, you'll want to be able to quickly find out which users user A is following, either directly or indirectly. To achieve this, you can denormalize the groups that user A is following. Let's say that group X and Y are defined as follows:

{
  _id: "groupX",
  members: [ "userC", "userD" ]
},
{
  _id: "groupY",
  members: [ "userD", "userE" ]
}

Based on these groups, and the direct relations user A has, you can generate subscriptions between users. The origin(s) of a subscription are stored with each subscription. For the example data the subscriptions would look like this:

// abusing exclamation mark to indicate a direct relation
{ ownerId: "userA", userId: "userB", origins: [ "!" ] },
{ ownerId: "userA", userId: "userC", origins: [ "!", "groupX" ] },
{ ownerId: "userA", userId: "userD", origins: [ "groupX", "groupY" ] },
{ ownerId: "userA", userId: "userE", origins: [ "groupY" ] }

You can generate these subscriptions pretty easily, using a map-reduce-finalize call for an individual user. If a group is updated, you only have to re-run the map-reduce for all users that are following the group and the subscriptions will be up-to-date again.

Map-reduce

The following map-reduce functions will generate the subscriptions for a single user.

map = function () {
  ownerId = this._id;

  this.followingUsers.forEach(function (userId) {
    emit({ ownerId: ownerId, userId: userId } , { origins: [ "!" ] });
  });

  this.followingGroups.forEach(function (groupId) {
    group = db.groups.findOne({ _id: groupId });

    group.members.forEach(function (userId) {
      emit({ ownerId: ownerId, userId: userId } , { origins: [ group._id ] });
    });
  });
}

reduce = function (key, values) {
  origins = [];

  values.forEach(function (value) {
    origins = origins.concat(value.origins);
  });

  return { origins: origins };
}

finalize = function (key, value) {
  db.subscriptions.update(key, { $set: { origins: value.origins }}, true);
}

You can then run the map-reduce for a single user, by specifying a query, in this case for userA.

db.users.mapReduce(map, reduce, { finalize: finalize, query: { _id: "userA" }})

A few notes:

You should delete the previous subscriptions of a user, before running map-reduce for that user.
If you update a group, you should run map-reduce for all the users that follow the group.

I should note that these map-reduce functions turned out to be more complex than what I had in mind, because MongoDB doesn't support arrays as return values of reduce functions. In theory, the functions could be much simpler, but wouldn't be compatible with MongoDB. However, this more complex solution can be used to map-reduce the entire users collection in a single call, if you ever have to.

来源：https://stackoverflow.com/questions/4042917/twitter-like-app-using-mongodb

标签

database-design

mongodb

software-design