Laravel Implement greatest-n-per-group - php

Question

User table:
| id | name | age |
|----|------------|-----|
| 1 | Apple | 22 |
| 2 | Strawberry | 23 |
| 3 | Orange | 50 |
| 4 | Mango | 30 |
Memberships table:
| id | user_id | expire_at |
|----|---------|----------------------|
| 1 | 1 | 2019-08-17T11:19:30Z |
| 2 | 1 | 2019-08-10T11:20:10Z |
| 3 | 2 | 2019-08-29T11:20:19Z |
| 4 | 3 | 2019-08-02T11:20:30Z |
| 5 | 3 | 2019-08-28T11:20:40Z |
Problom
I want select users with the latest 'expire_at'.
After reference: https://stackoverflow.com/a/2111420/5588637,
I tried the following:
SELECT
u.*,
m1.*
FROM
users u
INNER JOIN memberships m1 ON u.id = m1.user_id
LEFT JOIN memberships m2 ON u.id = m2.user_id
AND (
m1.expire_at < m2.expire_at
OR m1.expire_at = m2.expire_at
AND m1.id < m2.id
)
WHERE
m2.id IS NULL;
Result
The id will appear twice because I used to join.
| id | name | age | id | user_id | expire_at |
|----|------------|-----|----|---------|----------------------|
| 1 | Apple | 22 | 1 | 1 | 2019-08-17T11:19:30Z |
| 2 | Strawberry | 23 | 3 | 2 | 2019-08-29T11:20:19Z |
| 3 | Orange | 50 | 5 | 3 | 2019-08-28T11:20:40Z |
After change m1.* to m1.expire_at. I got the result I want.
| id | name | age | expire_at |
|----|------------|-----|----------------------|
| 1 | Apple | 22 | 2019-08-17T11:19:30Z|
| 2 | Strawberry | 23 | 2019-08-29T11:20:19Z |
| 3 | Orange | 50 | 2019-08-28T11:20:40Z |
online try: http://sqlfiddle.com/#!9/27fa22/4
Implement in Lavavel
Laravel Framework version: 5.6.39
I am trying to convert the above SQL into Laravel using Database: Query Builder.
$users = DB::table('users as u')
->select('u.*', 'm1.*')
->join('memberships as m1','u.id','=','m1.user_id')
->leftJoin('memberships as m2', function($join){
$join->on('u.id', '=', 'm2.user_id')
->where(function ($query) {
$query->where('m1.expire_at','<','m2.expire_at')
->orWhere('m1.expire_at','=','m2.expire_at')
->where('m1.id','<','m2.id');
});
})
->whereNull('m2.id')
->toSQL();
I'm using toSql(). This will convert it to SQL first to make sure it's same of above SQL.
SELECT
`u`.*,
`m1`.*
FROM
`users` AS `u`
INNER JOIN `memberships` AS `m1` ON `u`.`id` = `m1`.`user_id`
LEFT JOIN `memberships` AS `m2` ON `u`.`id` = `m2`.`user_id`
AND (
`m1`.`expire_at` < ?
OR `m1`.`expire_at` = ?
AND `m1`.`id` < ?
)
WHERE
`m2`.`id` IS NULL
? seems to be the characteristic of laravel, I believe it is same of above SQL.
when i change toSQL() to get(), the result following:
Collection { ▼
#items: []
}
The above result is wrong, so i tried remove
whereNull('m2.id') in Laravel code (WHERE m2.id IS NULL in SQL), let’s see what happened.
Laravel result
Collection { ▼
#items: array:5 [▼
0 => { ▼
+"id": 1
+"name": "Apple"
+"age": "Eric Yiu SL"
+"user_id": 1
+"expire_at": "2019-08-10T11:20:10Z"
}
...
]
Ideal result
| id | name | age | id | user_id | expire_at |
|----|------------|-----|----|---------|----------------------|
| 1 | Apple | 22 | 2 | 1 | 2019-08-10T11:20:10Z |
| 3 | Orange | 50 | 4 | 3 | 2019-08-02T11:20:30Z |
| 1 | Apple | 22 | 1 | 1 | 2019-08-17T11:19:30Z |
| 2 | Strawberry | 23 | 3 | 2 | 2019-08-29T11:20:19Z |
| 3 | Orange | 50 | 5 | 3 | 2019-08-28T11:20:40Z |
Comparing results, Laravel result missing second id which is memberships table id, i guess this is the reason of incorrect results.
I have searched the Internet, seems is this problem.
https://github.com/laravel/framework/issues/4962
But I failed after various attempts...

Answer 1

You cannot select two rows with the same name in Laravel. The second one will override the first one. Use an alias instead.
$users = DB::table('users as u')
->select('u.*', 'm1.id as membership_id')
->join('memberships as m1','u.id','=','m1.user_id')
->leftJoin('memberships as m2', function($join){
$join->on('u.id', '=', 'm2.user_id')
->where(function ($query) {
$query->whereColumn('m1.expire_at','<','m2.expire_at')
->orWhere(function ($query) {
$query->whereColumn('m1.expire_at','=','m2.expire_at')
->whereColumn('m1.id','<','m2.id');
});
});
})
->whereNull('m2.id')
->get();
Note: I also encapsulated the orWhere() in the join to avoid confusion about the order of AND/OR.
What also works is using a different order in the select. You can for example use the following:
$query->select([
'm1.*',
'm1.id as membership_id',
'u.*'
])
It will return all columns of both tables plus the new membership_id column. But if there is a column on the users table which is named similarly to a column on the memberships table, only the users table column is returned (e.g. created_at). What comes last in your list is returned.

Answer 2

EDIT:
As #Namoshek mentioned, you should not select everything because you have a duplicate key problem in your SQL query. I modified my answer so that it would match #RaymondNijland answer. And by the way, even for the table user, you should select exactly what you need. And not only for a duplicate key problem but also for the speed of your SQL query. We don't think about it enough but it can quickly make the difference on a big set of results.
Less data to send from the database to your PHP server = faster
You should try this one :
DB::table('users as u')
->select('u.*', 'm1.id as membership_id')
->join('memberships as m1','u.id','=','m1.user_id')
->leftJoin('memberships as m2', function ($join) {
$join->on('u.id', '=', 'm2.user_id')
->on(function($join) {
$join->on('m1.id', '<', 'm2.id')
->on(function($join) {
$join->on('m1.expire_at', '<', 'm2.expire_at')
->orOn('m1.expire_at', '=', 'm2.expire_at');
});
});
})
->whereNull('m2.id')
->toSQL()
As mentioned in Laravel's documentation on this page: https://laravel.com/api/5.8/Illuminate/Database/Query/JoinClause.html#method_on
You can pass a closure to the on() method and there is the orOn() method that you can use in this closure.
I tested it and it gives the same result as your SQL query.