mysql - Laravel Eloquent集團最近的記錄
greatest-n-per-group (2)
我正試圖在桌面上獲得單個客戶的最新記錄。 例:
ID Customer City Amount
1 Cust001 City1 2
2 Cust001 City2 3
3 Cust001 City1 1
4 Cust001 City2 1
5 Cust001 City2 3
6 Cust001 City3 1
7 Cust001 City3 1
8 Cust002 City1 2
9 Cust002 City1 1
10 Cust002 City2 3
11 Cust002 City1 2
12 Cust002 City2 1
13 Cust002 City3 2
14 Cust002 City3 3
15 Cust003 City1 1
16 Cust003 City2 3
17 Cust003 City3 2
請注意,該表還包含created_at和updated_at字段。 為簡單起見我省略了這些字段。
最後,我希望我的查詢返回Cust001:
ID Customer City Amount
3 Cust001 City1 1
5 Cust001 City2 3
7 Cust001 City3 1
對於Cust002:
ID Customer City Amount
11 Cust002 City1 2
12 Cust002 City2 1
14 Cust002 City3 3
我試過了:
Table::where('Customer', 'Cust001')
->latest()
->groupBy('City')
->get()
並且
Table::select(DB::raw('t.*'))->from(DB::raw('(select * from table where Customer = \'Cust001\' order by created_at DESC) t'))
->groupBy('t.City')->get();
但它不斷返回每組最老的記錄(我想要最新的記錄)。
我怎樣才能做到這一點? 如果對你們來說更容易,你可以在這裡編寫SQL查詢,我會找到一種方法將其“翻譯”為Laravel語法。
要根據
created_at
獲取每個城市中每位客戶的最新記錄,您可以使用自聯接
DB::table('yourTable as t')
->select('t.*')
->leftJoin('yourTable as t1', function ($join) {
$join->on('t.Customer','=','t1.Customer')
->where('t.City', '=', 't1.City')
->whereRaw(DB::raw('t.created_at < t1.created_at'));
})
->whereNull('t1.id')
->get();
在簡單的SQL中它會是這樣的
select t.*
from yourTable t
left join yourTable t1
on t.Customer = t1.Customer
and t.City = t1.City
and t.created_at < t1.created_at
where t1.id is null
自我內部聯接的另一種方法是
select t.*
from yourTable t
join (
select Customer,City,max(ID) ID
from yourTable
group by Customer,City
) t1
on t.Customer = t1.Customer
and t.City = t1.City
and t.ID = t1.ID
DB::table('table')->orderBy('id', 'desc')->get();
Laravel查詢以獲取id的基礎上的最新數據