Question

+------------+---------+------------+--------------+------+
| dt         | status  | code       | arena        | cnt  |
+------------+---------+------------+--------------+------+
| 2016-01-01 | failure | AB         | kingdom1     |    2 |
| 2016-01-01 | success | AB         | kingdom1     |   16 |
| 2016-01-01 | failure | CD         | kingdom1     |   50 |
| 2016-01-01 | success | CD         | kingdom1     |  662 |
| 2016-01-01 | failure | EF         | kingdom1     |  131 |
| 2016-01-01 | success | EF         | kingdom1     |  622 |

SQL查询：

select DATE(created) as dt, status, code, arena,count(status) as cnt
from game_table
where some_condition
group by dt, code, status)

要求：

我需要找到code，arena明智的失败百分比，但我尝试的却是100％失败（ baaaaad编码器）。

+------------+---------+------------+--------------+------+
| dt         | status  | code       | arena        |failed|
+------------+---------+------------+--------------+------+
| 2016-01-01 | failure | AB         | kingdom1     | 11.1 |
| 2016-01-01 | failure | CD         | kingdom1     | 7.02 |
| 2016-01-01 | failure | EF         | kingdom1     | 17.4 |

我试过了：

select dt, cnt/sum(cnt) as p, status, code, arena 
from (
  select DATE(created) as dt, status, code, arena,count(status) as cnt
  from game_table
  where some_condition
  group by dt, code, status
 ) as inner_t
group by dt, code, status;

Answer 1

您可以通过以下代码获得失败或成功组的百分比。

 select DATE(created) as dt, count(case when status = "failure" then 1 end) * 100/count(status) as failure, code, arena, cnt
 from percent_issue group by code

Answer 2

主要问题似乎是要正确理解GROUP BY。它只表示每个______＆＃34;显示一个结果行。您希望每个code有一个结果行，因此您group by code。如果您想要每code和arena行，group by code, arena。对于不在GROUP BY中的字段决定要显示哪个汇总，例如最小日期。

select 
  min(created)
  'failure' as status,
  code,
  min(arena) as arena,
  count(case when status = 'failure' then 1 end) / count(*) * 100 as failed
from game_table
where some_condition
group by code;

获取列的失败百分比

2 个答案: