Question

我有这样的查询

SELECT 
table1.question_id,
SUM(IF(table1.correct_answer = 1, 1, 0)) AS correct_answers,
SUM(IF(table1.correct_answer != 1, 1, 0)) AS incorrect_answers,
round((SUM(IF(table1.correct_answer = 1, 1, 0)) / (SUM(IF(table1.correct_answer = 1, 1, 0)) + SUM(IF(table1.correct_answer != 1, 1, 0))) * 100),
        2) AS percentage
FROM
    table2
        JOIN
    table1 ON table2.answer_id = table1.id
WHERE
    percentage BETWEEN 10 AND 20
group by table1.question_id;

我正在尝试获取question_id，正确答案的数量，错误答案的数量以及正确的百分比，即某个值之间的百分比。现在一切都在工作以外的地方。它显示未知的列＆＃39;百分比＆＃39;。

Answer 1

不仅where子句中不允许列别名，而且您需要的逻辑需要having子句。这是在聚合的结果：

SELECT table1.question_id,
       SUM(IF(table1.correct_answer = 1, 1, 0)) AS correct_answers,
       SUM(IF(table1.correct_answer != 1, 1, 0)) AS incorrect_answers,
       round((SUM(IF(table1.correct_answer = 1, 1, 0)) / (SUM(IF(table1.correct_answer = 1, 1, 0)) + SUM(IF(table1.correct_answer != 1, 1, 0))) * 100),
             2) AS percentage
FROM table2 JOIN
     table1
     ON table2.answer_id = table1.id
group by table1.question_id
HAVING percentage BETWEEN 10 AND 20;

Answer 2

要过滤聚合函数的结果，您需要使用HAVING子句，WHERE过滤器在这种情况下不起作用，如果您仍然希望在哪里执行，那么您必须重复整个where子句中的表达式

SELECT 
  table1.question_id,
  SUM(IF(table1.correct_answer = 1, 1, 0)) AS correct_answers,
  SUM(IF(table1.correct_answer != 1, 1, 0)) AS incorrect_answers,
  ROUND(
    (
      SUM(IF(table1.correct_answer = 1, 1, 0)) / (
        SUM(IF(table1.correct_answer = 1, 1, 0)) + SUM(IF(table1.correct_answer != 1, 1, 0))
      ) * 100
    ),
    2
  ) AS percentage 
FROM
  table2 
  JOIN table1 
    ON table2.answer_id = table1.id 
GROUP BY table1.question_id 
HAVING percentage BETWEEN 10 AND 20

WHERE条件别名不起作用

2 个答案: