Question

通常我会选择最小的代码，但我目前正在构建一个支付网关，它可能每天处理数百万个连接，因此每次调整代码都可以帮助我建立性能，但是我有一个问题只能通过mysql gurus ..

使用union，最好用where子句过滤union的每一边，或者整理记录然后对union set进行过滤..

哪一个最快：

1）

select * from (
    select * from payment_routes 
    where channel_id is null 
    and currency_id != v_currency_id
    and card_id = v_card_id 
    and enabled = 1
  union 
    select * from payment_routes 
    where channel_id = p_channel_id 
    and currency_id = v_currency_id
    and card_id = v_card_id 
    and enabled = 1
) t1
join gateways g on t1.gateway_id = g.id
where t1.currency_id = v_currency_id
;

2）或者是这个较小的代码，快速/快速（但不慢）

select * from (
    select * from payment_routes 
    where channel_id is null 
    and currency_id != v_currency_id
  union 
    select * from payment_routes 
    where channel_id = p_channel_id 
    and currency_id = v_currency_id
) t1
join gateways g on t1.gateway_id = g.id
where t1.currency_id = v_currency_id
and t1.card_id = v_card_id 
and t1.enabled = 1
;

从逻辑上讲，我会说（1）因为每个联合集都被过滤得更快，所以需要合并更少的记录。

更新

我找到了一种更好的方法来编写没有联合的

select r.*, g.`name` 'gateway_name' 
from payment_routes r
join gateways g on r.gateway_id = g.id
where (
    (channel_id is null and currency_id != v_sell_currency_id)
    or
    (channel_id = p_channel_id and currency_id = v_sell_currency_id)
)
and card_id = v_card_id
and currency_id = v_pay_currency_id
and enabled = 1

MySQL UNION：哪种方法最快

更新

0 个答案: