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Laravel雄辩地选择所有具有max created_at最大值的行

[英]Laravel Eloquent select all rows with max created_at

I have a table that contains: 我有一个包含以下内容的表:

id  seller_id   amount   created_at
1   10          100      2017-06-01 00:00:00
2   15          250      2017-06-01 00:00:00
....
154 10          10000    2017-12-24 00:00:00
255 15          25000    2017-12-24 00:00:00

I want to get all the latest rows for each individual seller_id. 我想获取每个单个Seller_id的所有最新行。 I can get the latest row for one like this: 我可以得到这样的最新行:

$sales = Snapshot::where('seller_id', '=', 15)
    ->orderBy('created_at', 'DESC')
    ->first();

How do I get only the latest row for each seller? 如何仅获得每个卖方的最新行?

To get latest record for each seller_id you can use following query 要获取每个Seller_id的最新记录,您可以使用以下查询

select s.*
from snapshot s
left join snapshot s1 on s.seller_id = s1.seller_id
and s.created_at < s1.created_at
where s1.seller_id is null

Using query builder you might rewrite it as 使用查询生成器,您可以将其重写为

DB::table('snapshot as s')
  ->select('s.*')
  ->leftJoin('snapshot as s1', function ($join) {
        $join->on('s.seller_id', '=', 's1.seller_id')
             ->whereRaw(DB::raw('s.created_at < s1.created_at'));
   })
  ->whereNull('s1.seller_id')
  ->get();

This worked: 这工作:

DB::table('snapshot as s')
  ->select('s.*')
  ->leftJoin('snapshot as s1', function ($join) {
        $join->on('s.seller_id', '=', 's1.seller_id');
        $join->on('s.created_at', '<', 's1.created_at');
   })
  ->whereNull('s1.seller_id')
  ->get();

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