按计数分组的MySQL案例

我有三个问题

SELECT COUNT(question_id) AS correct_count, uID FROM mytable  WHERE id
>= 0 AND id <= 1000 AND correct = 1 AND answer_id IS NOT NULL GROUP BY user_id


SELECT COUNT(question_id) AS incorrect_count, uID FROM mytable  WHERE id >= 0 AND id <= 1000 AND correct !=1 AND answer_id IS NOT NULL GROUP BY user_id

SELECT COUNT(question_id) AS null_count, uID FROM mytable  WHERE id >= 0 AND id <= 1000 AND answer_id IS NULL GROUP BY user_id

如何使用CASE或IF将其作为单个查询连接起来

count只统计非null项，因此您可以将where子句转换为大小写表达式，当它们不匹配时返回null：

SELECT user_id, 
       sum(id >= 0 AND id <= 1000 AND correct = 1 AND answer_id IS NOT NULL) AS correctCnt, 
       sum(id >= 0 AND id <= 1000 AND correct !=1 AND answer_id IS NOT NULL) AS incorrectCnt,
       sum(id >= 0 AND id <= 1000 AND answer_id IS NULL) as nullCnt
FROM mytable  
GROUP BY user_id

SELECT   user_id,
         COUNT (CASE WHEN id >= 0 AND id <= 1000 AND correct = 1 AND answer_id IS NOT NULL THEN 1 ELSE NULL) AS correct_count, 
         COUNT (CASE WHEN id >= 0 AND id <= 1000 AND correct !=1 AND answer_id IS NOT NULL THEN 1 ELSE NULL) AS incorrect_count,
         COUNT (CASE WHEN WHERE id >= 0 AND id <= 1000 AND answer_id IS NULL) AS null_count
FROM     mytable  
GROUP BY user_id

---

这可能行得通，但它给了我一个超时时间，因为桌子太大了

SELECT   user_id,
         COUNT (CASE WHEN correct = 1 AND answer_id IS NOT NULL THEN 1 ELSE NULL) AS correct_count, 
         COUNT (CASE WHEN correct !=1 AND answer_id IS NOT NULL THEN 1 ELSE NULL) AS incorrect_count,
         COUNT (CASE WHEN WHERE answer_id IS NULL THEN 1 ELSE NULL) AS null_count
FROM     mytable
WHERE    id >= 0 AND id <= 1000  
GROUP BY user_id