比较一行中的两个值

Question

select event_slots.id as event_slot_id, u.first_name, u.id AS user_id,

sum(case when next_round = 1 then 1 else 0 end) AS yes,
sum(case when next_round = 2 then 1 else 0 end) AS no

from event_slots

left join users as u on u.id = event_slots.selected_by_user_id 
left join audition_card_values as acv on acv.user_id = event_slots.selected_by_user_id 

where 
event_id = 1 and 
selected_by_user_id is not null

group by acv.user_id 

厄格布尼斯：

现在我只需要第一行“first_name”=“Eins”，因为“yes”等于“no”。 我怎样才能做到这一点？ WHERE 是 = 否...不起作用。

谢谢你的帮助

Answer 1

我会将查询表述为：

select 
    ev.id as event_slot_id, 
    u.first_name, 
    u.id AS user_id,
    sum(next_round = 1) AS yes,
    sum(next_round = 2) AS no
from event_slots as ev
inner join audition_card_values as acv 
    on acv.user_id = ev.selected_by_user_id 
left join users as u 
    on u.id = ev.selected_by_user_id 
where e.event_id = 1
group by ev.id, u.first_name, u.id
having sum(next_round = 1) = sum(next_round = 2)

理由：

• sum()中的条件表达式可以简化

• 您可以使用having子句进行过滤； MySQL 还支持在having子句中重用select子句中定义的别名，因此您可以拼写为：have having yes = no - 但我更喜欢重复条件表达式，因为这是标准 Z9778840A0100CB30C9828ZB7412

• audition_card_values上的left join后跟where... is not null可以重写为简单的inner join

• 我固定了你的group by条款； select子句中的每个非聚合列都必须出现在group by子句中