INTERSECT MySql

Question

我想獲得幾個SQL請求的交集。我嘗試了很多命令，但仍然卡住了。

以下是3個請求。

任何人都有想法來幫助我？

SELECT username 
FROM mdl_user 
INNER JOIN mdl_quiz_grades 
ON mdl_user.id=mdl_quiz_grades.userid 
WHERE mdl_quiz_grades.quiz = 17; 

SELECT username 
FROM mdl_user 
INNER JOIN mdl_quiz_grades 
ON mdl_user.id=mdl_quiz_grades.userid 
WHERE mdl_quiz_grades.quiz = 16; 

SELECT username 
FROM mdl_user 
INNER JOIN mdl_quiz_grades 
ON mdl_user.id=mdl_quiz_grades.userid 
WHERE mdl_quiz_grades.quiz = 15; 

Answer 1

INTERSECT可以使用容易地改寫連接：

SELECT t1.username 
from 
(
    SELECT username 
    FROM mdl_user 
    INNER JOIN mdl_quiz_grades 
    ON mdl_user.id=mdl_quiz_grades.userid 
    WHERE mdl_quiz_grades.quiz = 17 
) as t1 
join 
(
    SELECT username 
    FROM mdl_user 
    INNER JOIN mdl_quiz_grades 
    ON mdl_user.id=mdl_quiz_grades.userid 
    WHERE mdl_quiz_grades.quiz = 16 
) as t 
on t1.username = t2.username 
join 
(
    SELECT username 
    FROM mdl_user 
    INNER JOIN mdl_quiz_grades 
    ON mdl_user.id=mdl_quiz_grades.userid 
    WHERE mdl_quiz_grades.quiz = 15 
) as t3 
on t1.username = t3.username 

但在你的情況下，它幾乎是相同的查詢，只是不同的WHERE條件。可以將集合的邏輯：

SELECT username 
FROM mdl_user 
INNER JOIN mdl_quiz_grades 
ON mdl_user.id=mdl_quiz_grades.userid 
WHERE mdl_quiz_grades.quiz in (15,16,17) 
GROUP BY username 
HAVING COUNT(*) = 3 

如果同一用戶ID可能有你需要的應用獨特的同一測驗多次：

HAVING COUNT(DISTINCT quiz) = 3