PHP MongoDB通过键选择字段（不同）

Question

I have a database with lottery games across the world. This is how each document look like:

Each game can appear in different country_code or state_code if country have states (Canada, USA).

Selecting all game_id's and then all the countries and/or states it belongs to is done like this:

// get all games
// $colCurrent = MongoCollection Object
$gamesRes = $colCurrent->distinct('game_id');
foreach($gamesRes as $gameId) {
    $disCountries = $colCurrent->distinct('country_code',array('game_id' => $gameId));
    $disStates = $colCurrent->distinct('state_code',array('game_id' => $gameId));
}

I believe this is inappropriate way to do it, as it does a lot of queries to database. I've tried using aggregate function, but it only select 1 field like distinct.

Anyone can help optimizing this query?

Thanks a lot!

Answer 1

Depending on what you are trying to achieve and the size of your data set, there are a few different approaches you can take.

Some examples using the Aggregation Framework in the mongo shell (MongoDB 2.2+):

1) Find all games and for each game create the set of unique country_code and state_code values:

db.games.aggregate(
    { $group: {
        _id: { gameId: "$game_id" },
        countries: { $addToSet: "$country_code" },
        states: { $addToSet: "$state_code" }
    }}
)

2) Find all games, and group by the unique combination of gameId, country_code, and state_code including a count:

db.games.aggregate(
    { $group: {
        _id: {
            gameId: "$game_id",
            country_code: "$country_code",
            state_code: "$state_code"
        },
        total: { $sum: 1 }
    }}
)

In this second example, note that the _id used for grouping can include multiple fields.

If you don't want to group on all the documents in the collection, you could make these aggregations more efficient by starting with the $match operator to limit the pipeline to the data you need ( $match can also take advantage of a suitable index).

Answer 2

Assuming you mean the total number of distinct countries and the total number of distinct states for each game_name (assuming one per game_id and that this is more readable [ interchange if needed ] )

Posting as mongo shell for general clarity, adapt to your driver and language as required:

db.lottery.aggregate([
    {$project: { country_code: 1, state_code: 1, game_name: 1 } },
    {$group: { 
        _id: "$game_name",
        countries: {$addToSet: "$country_code"},
        states: {$addToSet: "$state_code"}
    }},
    {$unwind: "$countries"},
    {$group:{ _id: { id: "$_id", states: "$states" }, country_count: {$sum: 1 } }},
    {$project: { _id: 0, game: "$_id.id", countries: "$country_count", states: "$_id.states" }},
    {$unwind: "$states"},
    {$group: { _id: { id: "$game", countries: "$countries" }, state_count: {$sum: 1 } }},
    {$project: { _id: 0, game: "$_id.id", countries: "$_id.countries", states: "$state_count" }},
    {$sort: { game: 1 }}
])

So there are a few fancy stages here:

1. Project the fields that are needed only
2. Group on the game and push each country and state into an array of its own
3. Now unwind the countries to get one record per country
4. Group a sum of the countries while retaining the game and states field array
5. Project --optional-- to make the records appear more natural. $group messes with _id
6. Unwind the states to get one record per state
7. Group a sum of the states while retaining game and countries count
8. Project into something more natural
9. Sorting --optional-- by what you like. In this case the name of the game

Phew! A reasonably hefty aggregate but it does show a way to work out the problem.

DISCLAIMER:

I have made the huge assumption here that your data does make some sense already and that there are not multiple records of games per country and/or per state. The additional "I didn't do it" part is that your code did not discern states within countries so "I didn't do it either" :-P

You can add in $group stages to do that though. Part of the fun of programming is learning and working out how to do things by yourself. So this should be a good place to start if not a perfect fit.

The reference is a really good place for learning how to apply all the operators used here. Apply one stage at a time ( data size permitting ) to get a good idea of what is going on in each step.