MYSQL - 从多行，相同用户，不同值中选择

Question

下面是我的名为fittest的表格。 我需要找到哪些学生根据学生ID（studid）进行了前置栏中指定的前后测试。 因此，基于下面的简单表格，我需要返回studid 123456.如何为此编写SELECT查询？

SQL query: SELECT studid, prepost FROM `fittest` LIMIT 0, 30 ; 

    studid  prepost
    123456  pre
    123456  post
    1031460 pre

Answer 1

加入怎么样：

select s1.studid, s1.prepost, s2.prepost
from fittest s1
inner join fittest s2
    on s1.studid = s2.studid
where s1.prepost = 'pre' 
  and s2.prepost = 'post'

Answer 2

尝试

SELECT studid
  FROM fittest
 GROUP BY studid
HAVING COUNT(DISTINCT prepost) = 2

或者，如果你想确保只有有一个确切的行与IDS pre和一排与post那么你可以强制它像这样

SELECT studid
  FROM fittest
 GROUP BY studid
HAVING (MAX(prepost = 'pre' ) + 
        MAX(prepost = 'post')) = 2
   AND COUNT(DISTINCT prepost) = 2

输出：

| STUDID |
----------
| 123456 |

这是两个查询的SQLFiddle演示

Answer 3

尝试：

    select f1.studid
       from fittest f1
       inner join fittest f2
         on f1.studid = f2.studid
    where f1.prepost = 'pre' and f2.prepost = 'post'

这是SQL小提琴

Answer 4

根据您的真实主键，您可以使用以下内容：

SELECT studid
FROM fittest
WHERE prepost = 'pre' OR prepost = 'post'
GROUP BY studid
HAVING COUNT(DISTINCT prepost) = 2

Answer 5

SELECT t1.studid
FROM fittest t1
INNER JOIN t2 ON t1.studid=t2.studid
WHERE t1.prepost = 'pre'
AND t2.prepost='post'

Answer 6

如果此列中只能进行PRE和POST，请尝试

SELECT studid FROM fittest group by studid
HAVING Min(prepost)<>max(prepost)

Answer 7

SELECT 
studid 
FROM `fittest` 
where prepost = 'pre' 
or prepost = 'post'
group by studid
having count(distinct prepost)=2

SQL小提琴额外记录。

Answer 8

你可以使用JOIN

SELECT a.studid 
FROM fittest a 
JOIN fittest b 
    ON a.studid=b.studid 
        and a.prepost like 'pre' 
        and b.prepost like 'post'

Answer 9

选择.... where studid in（123456,123457，...）