问题描述
我正在尝试将 sql 查询转换为 Laravel 查询构建器, 但我很难理解哪种是 Laravel 中 FORM SUBQUERY 的最佳方式, 因为如果我没记错的话,这不是您通常可以使用查询生成器执行的操作,但您应该以原始方式实现。
这样对吗?
应该如何翻译?
SELECT t1.* FROM (//with out sub query it would be just products AS t1
SELECT SUM(amount)AS amount,products.*
FROM products,orders
WHERE orders.product_id = products.product_id
GROUP BY orders.product_id )AS t1
LEFT OUTER JOIN products as t2
ON(t1.name = t2.name AND t1.year_production< t2.year_production)
WHERE t2.name is null
$poductlist =Product::from('products AS t1')->select('t1.*')
->leftjoin('products AS t2',function($join){
$join->on('t1.name','=','t2.name')
->on('t1.year_production','<','t2.year_production')
->whereNull('t2.name')
->orderBy('t2.name','desc');})
->get();
解决方法
我相信您可以将您的查询重写为相关子查询,以过滤具有最大值为 year_production 且名称重复的产品
select p.product_id,p.name,coalesce(sum(o.amount),0) as amount
from products as p
left join orders o on p.product_id = o.product_id
where exists (
select 1
from products p1
where p1.name = p.name
having max(p1.year_production) = p.year_production
)
group by p.product_id,p.name
order by p.product_id
并且在查询构建器中您可以将其转换为
$products = DB::table('products as p')
->select(['p.product_id','p.name',DB::raw('coalesce(sum(o.amount),0) as amount')
])
->leftJoin('orders as o','p.product_id','=','o.product_id' )
->whereExists(function ($query) {
$query->select(DB::raw(1))
->from('products as p1')
->whereRaw(DB::raw('p1.name = p.name'))
->havingRaw('max(p1.year_production) = p.year_production')
;
})
->groupBy('p.product_id','p.name')
->orderBy('p.product_id')
->get();