Question

下表显示了课程与学生之间的多对多关系。

    CREATE Table CourseStudents
        (
          CourseId INT NOT NULL,
         StudentId INT NOT NULL,
         PRIMARY KEY (CourseId, StudentId)
        );

 INSERT INTO CourseStudents VALUES (1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (3, 2), 
 (4, 3), (4, 2), (5, 1)

示例数据

| CourseId | StudentId |
|----------|-----------|
|        1 |         1 |
|        1 |         2 |
|        2 |         1 |
|        2 |         2 |
|        3 |         2 |
|        3 |         3 |
|        4 |         2 |
|        4 |         3 |
|        5 |         1 |

我正在寻找一个返回所有具有完全相同学生的课程的查询。我能够提出下面显示的查询。

WITH CourseGroups AS
(
SELECT c.CourseId,
STUFF ((
SELECT ',' + CAST(c2.StudentId AS VARCHAR)
  FROM CourseStudents c2
  WHERE c2.CourseId = c.CourseId
  ORDER BY c2.StudentId
  FOR XML PATH ('')), 1, 1, '') AS StudentList
FROM CourseStudents c
GROUP BY c.CourseId)
SELECT cg.StudentList,
STUFF ((
SELECT ',' + CAST(cg2.CourseId AS VARCHAR(10))
  FROM CourseGroups cg2
  WHERE cg2.StudentList = cg.StudentList
  FOR XML PATH ('')), 1, 1, '') AS ExactMatchCourseList
FROM CourseGroups cg
GROUP BY cg.StudentList
HAVING COUNT(*) > 1

查询返回

| StudentList | ExactMatchCourseList |
|-------------|----------------------|
|         1,2 |                  1,2 |
|         2,3 |                  3,4 |

以上结果很好。但我只需要 ExactMatchCourseList 。我处理的表有超过十亿行，所以我需要一个有效的查询，可以在运行时间的几分钟内找到任何匹配的课程。感谢任何帮助。 SqlFiddle

Answer 1

这将为您提供课程对的列表，但如果您要获得一式三份（或更多），那么您最终会得到一些额外的结果。我没有时间再玩这个以纠正这个问题，但也许它指出了你正确的方向：

WITH CTE_CourseMatches AS (
    SELECT
        CS1.CourseId AS CourseId_1,
        CS2.CourseId AS CourseId_2,
        COUNT(*) AS cnt
    FROM
        CourseStudents CS1
    INNER JOIN CourseStudents CS2 ON CS2.StudentId = CS1.StudentId AND CS2.CourseId > CS1.CourseId
    GROUP BY
      CS1.CourseId,
      CS2.CourseId
 ),
 CTE_CourseCounts AS (SELECT CourseId, COUNT(*) AS cnt FROM CourseStudents GROUP BY CourseID)
 SELECT
     CM.CourseId_1,
     CM.CourseId_2
 FROM
     CTE_CourseMatches CM
 INNER JOIN CTE_CourseCounts CC1 ON CC1.CourseId = CM.CourseId_1 AND CC1.cnt = CM.cnt
 INNER JOIN CTE_CourseCounts CC2 ON CC2.CourseId = CM.CourseId_2 AND CC2.cnt = CM.cnt

Answer 2

这只会在您的CourseStudents表上运行2次，而不是您当前正在执行的4次。如果在CourseStudents表上的CourseId上添加索引，则第一次运行将只是索引扫描。它也只为每个课程运行一次原始STUFF，而不是每个学生运行一次，然后按课程分组。我遗漏了最后的STUFF，我不确定你是否想要或者它只是你计算它的副产品。

CREATE TABLE #Course
(
    CourseId INT NOT NULL PRIMARY KEY
);

INSERT INTO #Course
SELECT CourseId
FROM
CourseStudents s
GROUP BY
CourseId
ORDER BY
CourseId;

CREATE TABLE #CourseStudentList
(
CourseId INT NOT NULL PRIMARY KEY,
StudentList VARCHAR(MAX) NOT NULL
);

INSERT INTO #CourseStudentList
SELECT
c.CourseId,
STUFF ((
SELECT ',' + CAST(c2.StudentId AS VARCHAR)
  FROM CourseStudents c2
  WHERE c2.CourseId = c.CourseId
  ORDER BY c2.StudentId
  FOR XML PATH ('')), 1, 1, '') AS StudentList
FROM
#Course c
ORDER BY
c.CourseId;

SELECT *
FROM
(
    SELECT
    l.CourseId,
    l.StudentList,
    COUNT(*) OVER (PARTITION BY l.StudentList) AS [Count]
    FROM
    #CourseStudentList l
) l
WHERE
l.[Count] > 1
ORDER BY
l.StudentList;

在多对多关系表中查找完全匹配的组

2 个答案: