Question

我有一张桌子

id, location, status
1, london, 1
2, london, 0
3, boston, 1
4, boston, 1

我希望我的查询生成如下内容： -

location, status_1, status_0
london, 1, 1
boston, 2, 0

到目前为止，我有： -

select count(id) as freq, location from my_table
group by location order by freq desc;

我完全迷失在哪里。

Answer 1

在发出查询的客户端中，这种转换更好，但如果必须在数据库中执行，那么

select location,
    sum(status = 1) AS status_1,
    sum(status = 0) AS status_0
from my_table
group by location 

它有点hackish，但当status为1时'status = 1'将评估为布尔值true，MySQL将礼貌地转换为整数'1'，然后将其汇总。当status为0且status = 0的计算结果为true时也是如此。

Answer 2

所以你想计算每个城市每个州的记录吗？

在下面的查询中，我按位置分组（就像你的那样），然后为每个状态添加一个总和。总和内部是一种情况，如果记录与期望状态匹配则返回1，否则返回0。这样，您就可以有效地计算每个位置的每个州的记录数。

select
  a.location,
  sum(case when a.status = 1 then 1 else 0 end) as status_1,
  sum(case when a.status = 0 then 1 else 0 end) as status_0
from
  YourTable a
group by
  a.location

Answer 3

select location,
sum(case when status = 1 then 1 else 0 end) as status_1,
sum(case when status = 0 then 1 else 0 end) as status_0,
from my_table
group by location;