我假设卖单id和买单id是相关的，如果不是，那么在你的数据中sell和buy表之间就没有关系了。你可以尝试下面这样的方法来得到原始答案。首先cte为所有属性创建标记。其次CTE得到满足所有属性需要的所有sales id。然后final select获取采购订单符合销售属性需求且价格低于采购订单的情况。

with cte as ( 
select s.*
case
        when a.[attribute_id] = 'whatever attribute you choose' and a.value >= 'whatever the benchmark is' then 1
        when a.[attribute_id] = 'whatever 2nd attribute you choose' and a.value >= 'whatever the benchmark is' then 1
            else 0
            end as marker
    from [Sell Orders]  s
        inner join [Sell Order Attributes] a on s.id = a.sell_order_id
),

cte2 as (
select id , sum(marker)
from cte
group by id
having sum(marker) = 2 --however many attributes you are testing
)

select *
    from [Buy Orders] b on b.id = s.id
        inner join [Buy Order Attributes] c on b.id = c.buy_order_id
        inner join  cte2 t on t.id = b.id

        Where  t.price < b.price

因为你不把价格和报价存储为十进制或其他数字类型，所以你总是需要转换它，这将花费一些时间，而且不支持货币数据类型，所以你应该重新设计你的表。
剩下的就是许多带有正确ON子句的连接。
它会将价格与报价以及所有匹配的属性进行比较

SELECT DISTINCT s.*,b.*
FROM Sell_Orders s JOIN Sell_Order_Attributes sa ON sa.`sell_order_id` = s.`id`
JOIN Buy_Orders b ON CAST( REPLACE(REPLACE(REPLACE(b.`offer`,'€',''),'.',''),',','.') AS DECIMAL(20,2))  >= CAST( REPLACE(REPLACE(REPLACE(s.price,'€',''),'.',''),',','.') AS DECIMAL(20,2))  JOIN Buy_Order_Attributes ba ON ba.`buy_order_id` = b.`id` 
  and ba.`attribute_id` = sa.`attribute_id`and sa.`value` >= ba.`value`

| 身份证|标题|价格|身份证|要约|
| - ------|- ------|- ------|- ------|- ------|
| 1个|1号公寓|50万欧元|1个|600万欧元|
fiddle