Question

我正在尝试使用max seq_no记录的每个问题，但它返回了错误的数据。

这是查询：

SELECT id, attempt, seq_number, question
FROM question_states
WHERE attempt = 374
AND seq_number IN ( SELECT MAX(seq_number) FROM question_states WHERE attempt = 374 GROUP BY question)
GROUP BY question, id

我有一个表名作为问题，结构如下： -

id--seq_number--attempt--question
1 --0     --374  -- 130
2 --0     --374  -- 130
3 --1     --374  -- 130
4 --2     --374  -- 130
5 --1     --374  -- 131

为什么此查询会针对问题130返回2条记录？如果我删除GROUP BY id然后返回错误的数据。

Answer 1

试试这个：

SELECT id, attempt, seq_number, question
FROM question_states qs 
INNER JOIN (SELECT question, MAX(seq_number) seq_number 
            FROM question_states WHERE attempt = 374 GROUP BY question
           ) AS A ON qs.question = A.question AND qs.seq_number = A.seq_number 
WHERE attempt = 374

或

SELECT id, attempt, seq_number, question FROM (SELECT id, attempt, seq_number, question FROM question_states WHERE attempt = 374 ORDER BY question, seq_number DESC ) AS A GROUP BY question

Answer 2

在子查询中，由于您使用了group by，它将返回两个值“2”和“1”。并且在主查询中，因为您使用了'IN'语句，它将返回包含字段seq_number和attempt = 374的所有数据

尝试

选择max（seq_number），id，尝试，来自question_states组的问题by seq_number ORDER by seq_number DESC limit 1

SQL - MAX函数返回错误数据

2 个答案: