Question

我有3个人的黄金收入表。当用户获得一些金币时，新记录将通过e_user和f_value进入其中一个表格。

现在我要总结每个用户的所有金币，并选择那些收入低于1000的人。

select count(e_user) from (
    select count(e_user) from quest_gold union all
    select count(e_user) from monster_gold union all
    select count(e_user) from other_gold 
) AS money

where f_value < 1000
group by e_user

我不确定我的选择是否以正确的方式进行，但它现在给了我“未知的语法错误”。请告诉我这里有什么问题。

Answer 1

您可以使用having子句来过滤少于1000的用户，但不能使用count(f_user)来过滤sum(f_value)：

select e_user, sum(f_value) as sum_val
from (
    select e_user, f_value from quest_gold union all
    select e_user, f_value from monster_gold union all
    select e_user, f_value from other_gold 
) as money
group by e_user
having sum_val < 1000

Answer 2

赚取少于1000金币的用户应该包括那些一无所获的用户。因此：

select u.user, coalesce(earned.gold, 0) as earned_gold
from users u
left join
(
  select e_user, sum(f_value) as gold
  from
  (
    select e_user, f_value from quest_gold
    union all
    select e_user, f_value from monster_gold
    union all
    select e_user, f_value from other_gold
  ) all_gold
  group by e_user
) earned on earned.e_user = u.user and earned.gold < 1000
order by u.user;

如果您只想计算少于1000金的用户数：

select count(earned.e_user) as poor_users
from users u
left join
(
  select e_user, sum(f_value) as gold
  from
  (
    select e_user, f_value from quest_gold
    union all
    select e_user, f_value from monster_gold
    union all
    select e_user, f_value from other_gold
  ) all_gold
  group by e_user
) earned on earned.e_user = u.user and earned.gold < 1000;

选择黄金的总和值

2 个答案: