骨料凡多選擇查詢的SQL

所以我用這個查詢計數驗證錯誤的數目在我的數據庫：骨料凡多選擇查詢的SQL

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

查詢工作正常，但是我想補充一個聚合where語句這適用於四個陳述。我該怎麼辦？

來源

2016-11-21 Abdalla Ismail

你可以在'SELECT'外面添加一個'WHERE' - 這是你的意思嗎？ –

我嘗試使用where外部select語句，但它給了我一個錯誤，指出from子句丟失，我使用了from子句但它沒有工作 –

查看我想要添加的聚合的'filter（）'子句 –

使用case表達式進行條件計數：

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')

的WHERE子句也許不是必要的，但可以加快速度（取決於數據和索引）。

UPDATE：根據要求「我想添加一個總和列，總結所有的數字，但它現在工作，我該怎麼做？」

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')

來源

2016-11-21 14:57:43 jarlh

一個總和列，總結所有的計數，但它現在的工作，我怎麼能做到這一點？ –

簡化您的邏輯！

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;

有了這個結構，你可以隨時添加一個group by。

或者，一個更簡單的方法是將這些值在不同行：

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

注：我沒有過濾對於你提到的四個代碼。如果有其他代碼，則可以添加該過濾器。

來源

2016-11-21 14:58:31

骨料凡多選擇查詢的SQL

回答

相關問題