我试图在桌面上获得单个客户的最新记录。例如:
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
请注意,该表还包含created_at和updated_at字段。为简单起见,我省略了这些字段。
最后,我希望我的查询返回Cust001:
ID Customer City Amount
3 Cust001 City1 1
5 Cust001 City2 3
7 Cust001 City3 1
对于Cust002:
ID Customer City Amount
11 Cust002 City1 2
12 Cust002 City2 1
14 Cust002 City3 3
我试过了:
Table::where('Customer', 'Cust001')
->latest()
->groupBy('City')
->get()
以及
Table::select(DB::raw('t.*'))->from(DB::raw('(select * from table where Customer = \'Cust001\' order by created_at DESC) t'))
->groupBy('t.City')->get();
但它不断返回每组最老的记录(我想要最新的记录)。
我怎样才能做到这一点?如果对你们来说更容易,你可以在这里编写SQL查询,我会找到一种方法来翻译它#34;到Laravel语法。
答案 0 :(得分:1)
要根据created_at
获取每个城市的每位客户的最新记录,您可以使用自我加入
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();
在纯SQL中,它就像
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
自我内部联接的另一种方法是
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
答案 1 :(得分:0)
DB::table('table')->orderBy('id', 'desc')->get();
Laravel查询以获取id
的最新数据