我如何优化此查询以用于计算响应的 sql

Posted 2023-04-15

【中文标题】我如何优化此查询以用于计算响应的 sql

【英文标题】：How can i optimize this query for sql for counting response

【发布时间】：2019-12-21 10:17:41

【问题描述】：

我有一个问题响应表，当我尝试通过查询来计算问题的响应数以创建图表时，加载需要 65 秒

所以请指导我如何优化这个查询

SELECT
vr.question_id,
(SELECT COUNT(response) FROM visitors_response  WHERE question_id = vr.question_id AND response = 5 ) AS one_star,
(SELECT COUNT(response) FROM visitors_response  WHERE question_id = vr.question_id AND response = 4 ) AS two_star,
(SELECT COUNT(response) FROM visitors_response  WHERE question_id = vr.question_id AND response = 3 ) AS three_star,
(SELECT COUNT(response) FROM visitors_response  WHERE question_id = vr.question_id AND response = 2 ) AS four_star,
(SELECT COUNT(response) FROM visitors_response  WHERE question_id = vr.question_id AND response = 1 ) AS five_star,
(SELECT AVG(response)   FROM visitors_response  WHERE question_id = vr.question_id ) AS average 
FROM visitors_response vr
JOIN questions q ON q.id = vr.question_id 
JOIN survey s ON s.id = q.survey_id
WHERE s.user_id = 101 AND s.status = 'active' 
GROUP BY vr.question_id

【参考方案1】：

尝试条件聚合：

SELECT
vr.question_id,
COUNT(CASE WHEN response = 5 THEN response END) AS one_star,
COUNT(CASE WHEN response = 4 THEN response END) AS two_star,
COUNT(CASE WHEN response = 3 THEN response END) AS three_star,
COUNT(CASE WHEN response = 2 THEN response END) AS four_star,
COUNT(CASE WHEN response = 1 THEN response END) AS five_star,
AVG(response) AS average 
FROM visitors_response vr
JOIN questions q ON q.id = vr.question_id 
JOIN survey s ON s.id = q.survey_id
WHERE s.user_id = 101 AND s.status = 'active' 
GROUP BY vr.question_id

或者使用SUM代替COUNT：

SELECT
vr.question_id,
SUM(response = 5) AS one_star,
SUM(response = 4) AS two_star,
SUM(response = 3) AS three_star,
SUM(response = 2) AS four_star,
SUM(response = 1) AS five_star,
AVG(response) AS average 
FROM visitors_response vr
JOIN questions q ON q.id = vr.question_id 
JOIN survey s ON s.id = q.survey_id
WHERE s.user_id = 101 AND s.status = 'active' 
GROUP BY vr.question_id

【讨论】：

检查结果。 JOIN 可能会增加行数，然后计数发生在 之前 GROUP BY

【参考方案2】：

您可以在任何聚合函数中使用 IF() 函数。诀窍是 COUNT() 不仅计算空值，因此将 NULL 放入“else”部分。喜欢这里：

SELECT 
  COUNT(IF(response=1,1,NULL)) AS one_star,
  COUNT(IF(response=2,1,NULL)) AS two_star,
  COUNT(IF(response=3,1,NULL)) AS three_star,
  COUNT(IF(response=4,1,NULL)) AS four_star,
  COUNT(IF(response=5,1,NULL)) AS five_star,
  AVG(response) AS average
FROM visitors_response vr
JOIN questions q ON q.id = vr.question_id 
JOIN survey s ON s.id = q.survey_id
WHERE s.user_id = 101 AND s.status = 'active' 
GROUP BY vr.question_id

或者你也可以通过“OR”操作来做同样的事情：

  COUNT(response=1 OR NULL) AS one_star,

对我来说，这是最短且最容易理解的选项。

【参考方案3】：

请注意，对于任何查询优化问题，您都应该为查询中涉及的每个表提供SHOW CREATE TABLE tablename 语句。

也就是说，如果以下索引不存在，请将它们添加到您的表中：

survey: (user_id,status)
questions: (survey_id)
visitor_responses: (question_id,response)

以上索引假设survey和questions表上的id分别是每个表的主键。

报告如何提高性能，并为每个表添加最新的 SHOW CREATE TABLE tablename 语句，以便我们现在可以帮助确保您没有任何冗余索引。

如果性能不低于 1 秒，或您希望超过的任何其他阈值，还包括当前的EXPLAIN 计划。