Question

所以我使用此查询来计算数据库中的身份验证错误数：

SELECT  (
SELECT COUNT(audit.server_response_code)
FROM   audit Where audit.server_response_code = '401' 
) AS Locked,
(
SELECT COUNT(audit.server_response_code)
FROM   audit Where audit.server_response_code = '403' 
) AS unlocked,
(
SELECT COUNT(audit.server_response_code)
FROM   audit Where audit.server_response_code = '490' 
) AS Passforget,
(
Select Count (audit.server_response_code)
From audit where audit.server_response_code = '491' 
) AS invalid

查询工作正常，但是我想在四个语句中添加一个聚合where语句。我该怎么办？

Answer 1

使用case表达式进行条件计数：

select count(case when audit.server_response_code = '401' then 1 end) AS Locked,
       count(case when audit.server_response_code = '403' then 1 end) AS unlocked,
       count(case when audit.server_response_code = '490' then 1 end) AS Passforget,
       count(case when audit.server_response_code = '491' then 1 end) AS invalid 
from audit
where audit.server_response_code in ('401','403','490','491')

可能不需要WHERE子句，但可能会加快速度（取决于数据和索引。）

更新：根据要求“我想添加一个总和列，它总结所有计数，但它现在正在运行，我该怎么办？”

select count(case when audit.server_response_code = '401' then 1 end) AS Locked,
       count(case when audit.server_response_code = '403' then 1 end) AS unlocked,
       count(case when audit.server_response_code = '490' then 1 end) AS Passforget,
       count(case when audit.server_response_code = '491' then 1 end) AS invalid, 
       count(*) as total_count
from audit
where audit.server_response_code in ('401','403','490','491')

Answer 2

简化你的逻辑！

select sum(case when a.server_response_code = '401' then 1 else 0 end) as locked,
       sum(case when a.server_response_code = '403' then 1 else 0 end) as unlocked,
       sum(case when a.server_response_code = '490' then 1 else 0 end) as Passforget,
       sum(case when a.server_response_code = '491' then 1 else 0 end) as Invalid
from audit a;

使用此结构，您可以轻松添加group by。

或者，更简单的方法是将值放在不同的行中：

select a.server_response_code, count(*)
from audit a
group by a.server_response_code;

注意：我没有过滤您提到的四个代码。如果还有其他代码，您可以添加该过滤器。

聚合多个选择查询sql

2 个答案: