有條件的計數MySQL

Question

我必須從表格中獲取每週報告，每個記錄的狀態如1表示爲開放狀態，2表示關閉狀態，所以我使用周作爲組的鍵和用於差異計數的使用情況（如開放，封閉，紅色等，但數據並不準確，因爲我想。 你會更好地支持下面提到的查詢。 每個響應都已撥出。

SELECT DISTINCT discussion_date,department,DAYNAME(discussion_date) AS DAY,CONCAT(YEAR(discussion_date),'/',WEEK(discussion_date)) AS week_name,YEAR(discussion_date) AS YEAR,CONCAT('WEEK -', WEEK(discussion_date)) AS week_no,COUNT(*) as total_queries,(CASE WHEN review_items.status = 1 AND due_date > curdate() THEN count(*) ELSE 0 END) as open,(CASE WHENreview_items.status = 2 THEN count(*) ELSE 0 END) as closed,(case when manager_closuredate > due_date THEN count(*) ELSE 0 END) as red,(CASE WHEN review_items.status = 3 THEN count(*) ELSE 0 END) as hold,(CASE WHEN review_items.status = 4 THEN count(*) ELSE 0 END) as discussion,DATE_ADD(discussion_date,INTERVAL(2 - DAYOFWEEK(discussion_date)) DAY) AS Week_start_date,DATE_ADD(discussion_date,INTERVAL(7 - DAYOFWEEK(discussion_date)) DAY) AS Week_end_date,DAYNAME(DATE_ADD(discussion_date,INTERVAL(2 - DAYOFWEEK(discussion_date)) DAY)) AS WeeK_Start_DAY FROM review_items 
WHERE department = 'Development' AND discussion_date BETWEEN '2017-08-01' AND '2017-08-31' 
GROUP BY week_name 
ORDER BY YEAR (discussion_date) ASC,WEEK(discussion_date) ASC 

Answer 1

您沒有正確使用條件聚合：

COUNT(CASE WHEN review_items.status = 1 AND due_date > curdate() THEN 1 END) as open, 
COUNT(CASE WHEN review_items.status = 2 THEN 1 END) as closed, 
COUNT(case when manager_closuredate > due_date THEN 1 END) as red, 
COUNT(CASE WHEN review_items.status = 3 THEN 1 END) as hold, 
COUNT(CASE WHEN review_items.status = 4 THEN 1 END) as discussion, 

可以既COUNT和使用的值，該列是任一種聚合溫控功能，或不是，不能同時使用。

所以我們的想法是隻有在條件滿足的情況下，纔會給計數器加1。有幾種方法可以做到這一點，其中一個是COUNT()，如果條件滿足，則評估爲COUNT(1)，即1。如果不滿足，則評估爲COUNT(NULL)，即0。

您還可以使用SUM()：

SUM(CASE WHEN <Condition> THEN 1 ELSE 0 END)