Question

我们有以下表格：

表1：Student_Records
StudentID | CourseID |期间|等级
12 6010 P1 90

23 6020 P1 80

12 6030 P2''空白，没有成绩

15 6010 P1 70

12 6020 P1 80

15 6020 P1 90

表2：Course_Records
CourseID CourseDec Credits
6010数学3

6020 Biology 3

6030英语3

表3：Student_Info
StudentID FirstName LastName ClassYear
12 Joe Smith 2013

15 Chak Li 2013

23 Pete Vo 2013

结果欲望：
ClassYear LastName FirstName StudentId数学生物学
2013史密斯乔12 90 80
2013 Li Chak 15 70 90

如何使用pivot命令实现此结果？

Answer 1

使用联接查询数字和课程，以便最终得到

StudentID CourseDec Grade
1         Math      20
1         Woodwork  82

透视你最终

StudentID Math WoodWork
1         20   82

然后加入Back to student获取名字Alst Name等。

Answer 2

您可以使用PIVOT，但这需要您知道您感兴趣的课程说明。

SELECT p.classyear, 
       p.lastname, 
       p.firstname, 
       p.studentid, 
       pvt.math, 
       pvt.biology 
FROM   (SELECT sr.grade, 
               si.classyear, 
               si.studentid, 
               si.firstname, 
               silastname 
        FROM   student_info si 
               INNER JOIN student_records sr 
                 ON si.studentid = sr.studentid 
               INNER JOIN course_records cr 
                 ON sr.courseid = cr.courseid) p PIVOT ( AVG (grade) FOR 
       coursedec IN ( 
       [Math], [Biology]) ) AS pvt 
ORDER  BY pvt.classyear;

Answer 3

http://data.stackexchange.com/stackoverflow/query/60493/http-stackoverflow-com-questions-9068600-sql-server-pivot-mulitple-tables

DECLARE @Student_Records AS TABLE (
  studentid INT,
  courseid  INT,
  period    VARCHAR(2),
  grade     INT);

INSERT INTO @Student_Records
VALUES      (12,
             6010,
             'P1',
             90),
            (23,
             6020,
             'P1',
             80),
            (12,
             6030,
             'P2',
             NULL),
            (15,
             6010,
             'P1',
             70),
            (12,
             6020,
             'P1',
             80),
            (15,
             6020,
             'P1',
             90);

DECLARE @Course_Records AS TABLE (
  courseid  INT,
  coursedec VARCHAR(50),
  credits   INT);

INSERT INTO @Course_Records
VALUES      ( 6010,
              'Math',
              3),
            ( 6020,
              'Biology',
              3),
            ( 6030,
              'English',
              3);

DECLARE @Student_Info AS TABLE (
  studentid INT,
  firstname VARCHAR(50),
  lastname  VARCHAR(50),
  classyear INT);

INSERT INTO @Student_Info
VALUES      (12,
             'Joe',
             'Smith',
             2013),
            (15,
             'Chak',
             'Li',
             2013),
            (23,
             'Pete',
             'Vo',
             2013);

SELECT DISTINCT coursedec
FROM   @Course_Records AS cr
       INNER JOIN @Student_Records sr
         ON sr.courseid = cr.courseid
WHERE  sr.grade IS NOT NULL;

SELECT classyear,
       lastname,
       firstname,
       summary.studentid,
       summary.math,
       summary.biology
FROM   (SELECT *
        FROM   (SELECT si.studentid,
                       coursedec,
                       grade
                FROM   @Course_Records AS cr
                       INNER JOIN @Student_Records sr
                         ON sr.courseid = cr.courseid
                       INNER JOIN @Student_Info si
                         ON si.studentid = sr.studentid
                WHERE  sr.grade IS NOT NULL) AS results PIVOT (AVG(grade) FOR
               coursedec
               IN (
               [Math], [Biology])) AS pvt) AS summary
       INNER JOIN @Student_Info si
         ON summary.studentid = si.studentid

请注意，您可以使用动态HTML来调整查询，因为添加了更多课程：

Pivot Table and Concatenate Columns

SQL Server透视多个表

3 个答案: