Oracle PL/SQL我想要对应的MAX（O.CO_YEAR）列。怎么做？

下面是PL/SQL查询的代码，我必须为其获取有资格教授特定课程的讲师的姓名（如果提供了姓名），以及他们教授该课程的次数和最后一次教授该课程的时间（年份和学期）

我已经完成了问题的大部分，但不知道如何获得max（o.co_年）即期限的相应细节

declare
gname varchar2(20);
count_id number(2);
id varchar(20);
year1 number(4);

cursor abc   // cursor 1
     is
        SELECT  i.i_gname
          into gname
        FROM INSTRUCTOR I 
        WHERE i.i_id in (
        SELECT t.i_id  FROM TeachingQualification T
        WHERE t.c_id in (SELECT c.c_id  FROM  COURSE C
        WHERE c.c_title = 'Advanced Database App')) ;


cursor bcd // cursor two
    is
        select o.i_id, count(o.i_id), max(o.co_year)
          into id, count_id, year1
        from courseoffering o
        where (o.i_id = i_id and o.c_id = 1234567)
        group by o.i_id;


Begin
open abc;
open bcd; 

loop

FETCH abc into gname;
exit when abc%NOTFOUND;

FETCH bcd into id, count_id, year1;
exit when bcd%NOTFOUND;

DBMS_OUTPUT.PUT_LINE ('NAME: ' || gname || '  Number of times taught ' || count_id || ' Year ' || year1 || );  // want to output corresponding column details for the year1 attribute.

end loop;
close bcd;
close abc;
end;

---

PL/SQL我想要对应的MAX（O.CO_YEAR）列。怎么做

Oracle中的游标可以有参数。有关详细信息，请参阅

它在代码中的工作方式：

cursor bcd(p_c_id courseoffering.c_id%type) // cursor two
is
    select o.i_id, count(o.i_id), max(o.co_year)
      into id, count_id, year1
    from courseoffering o
    where (o.i_id = i_id and o.c_id = p_c_id)
    group by o.i_id;

---

是的，我知道，但这是一项大学任务，我想用PL/SQL实现

废话！我宁愿你们的大学教你们如何正确使用PLSQL和SQL，并给你们一个有意义的任务。如果您可以在SQL中执行某些操作，请在SQL中执行。
另外：为什么您的游标包含

到
关键字中？为什么你需要一个环？似乎您只期望1个返回值

我仔细分析了您的代码：

--instructors who are qualified to teach a course
SELECT i.i_gname --instructor name
  FROM INSTRUCTOR I 
 WHERE i.i_id IN ( SELECT t.i_id --instructor_id
                     FROM TeachingQualification T
                    WHERE t.c_id in (SELECT c.c_id --course_id
                                       FROM  COURSE C
                                      WHERE c.c_title = 'Advanced Database App'));


--instructors who taught a course, with amount and last time
SELECT o.i_id instructor_id, count(o.i_id) times_taught, max(o.co_year) last_time_taught
  FROM courseoffering o
 WHERE o.c_id = 1234567 --course_id
 GROUP BY o.i_id;

--all instructors (ID) who taught advanced database app, how many times, and last time
--consider that this may produce NO_DATA_FOUND
SELECT o.i_id instructor_id, count(o.i_id) times_taught, max(o.co_year) last_time_taught
  FROM course c
  JOIN courseoffering o 
    ON c.c_id = o.c_id
 WHERE c.c_title = 'Advanced Database App'
 GROUP BY o.i_id; 

-- i don't think teachingqualification is required. The last select providers instructor IDs.
-- There is no need to go through that table, unless it would contain extra data you'd want to
-- filter by. Since courseoffering is being queried, and it has instructors, it stands to reason
-- that those instructor are qualified to teach the course.
SELECT  (SELECT i_gname FROM instructor WHERE i_id = o.i_id) instructor
       ,count(o.i_id) times_taught
       , max(o.co_year) last_time_taught
  FROM course c
  JOIN courseoffering o 
    ON c.c_id = o.c_id
 WHERE c.c_title = 'Advanced Database App'
 GROUP BY o.i_id;

-- Look, if you do want a PLSQL block for this, go ahead.
BEGIN
   FOR r IN (SELECT  (SELECT i_gname FROM instructor WHERE i_id = o.i_id) instructor
                    ,count(o.i_id) times_taught
                    , max(o.co_year) last_time_taught
               FROM course c
               JOIN courseoffering o 
                 ON c.c_id = o.c_id
              WHERE c.c_title = 'Advanced Database App'
              GROUP BY o.i_id)
   LOOP
      DBMS_OUTPUT.PUT_LINE ('NAME: ' || r.instructor || '  Number of times taught ' || r.times_taught || ' Year ' || r.last_time_taught);
   END LOOP;
END;

哦，请给你的专栏起个有意义的名字。例如，将其称为
i\u id
，而不是
i\u id

我设置了一些示例数据：

create table course (id number(5,0), cname varchar2(50), constraint course_pk primary key (id))
/
create table courseoffering(id number(5,0), course_id number(5,0), instructor_id number(5,0), course_year number(5,0), constraint offering_pk primary key (id), constraint course_fk foreign key (course_id) references course (id))
/
insert into course values (1, 'Tech I');
insert into course values (2, 'Basic SQL');
insert into course values (3, 'Advanced SQL');
--Instructor 1
insert into courseoffering values (1, 1, 1, 2009); --Tech I
insert into courseoffering values (2, 1, 1, 2010); --Tech I
insert into courseoffering values (3, 1, 1, 2011); --Tech I
insert into courseoffering values (4, 2, 1, 2011); --Basic SQL
insert into courseoffering values (5, 2, 1, 2012); --Basic SQL
--Instructor 2
insert into courseoffering values (6, 2, 2, 2008); --Basic SQL
insert into courseoffering values (7, 2, 2, 2009); --Basic SQL
insert into courseoffering values (8, 2, 2, 2010); --Basic SQL
insert into courseoffering values (9, 3, 2, 2010); --Advanced SQL
insert into courseoffering values (10, 3, 2, 2011); --Advanced SQL
insert into courseoffering values (11, 3, 2, 2012); --Advanced SQL
insert into courseoffering values (12, 1, 2, 2009); --Tech I
insert into courseoffering values (13, 1, 2, 2010); --Tech I
commit;

运行此：

SELECT c.cname, o.instructor_id, count(o.instructor_id) times_taught, max(o.course_year) last_time_taught
  FROM course c
  JOIN courseoffering o 
    ON c.id = o.course_id
 GROUP BY c.cname, o.instructor_id
 ORDER BY c.cname, o.instructor_id; 

产生：

CNAME          INSTRUCTOR_ID  TIMES_TAUGHT   LAST_TIME_TAUGHT
Advanced SQL   2              3              2012
Basic SQL      1              2              2012
Basic SQL      2              3              2010
Tech I         1              3              2011
Tech I         2              2              2010

您甚至可以轻松地将所需的数据转换为视图。
不需要PLSQL。SQL中只有几行。如果您想在PLSQL中使用它，您仍然可以使用一个循环来覆盖每门课程的多个讲师，或者如果您将其缩小为一门课程和一个讲师，则可以使用一些变量。始终尽量减少SQL和PLSQL上下文之间的切换。
为什么要在PL/SQL中执行所有这些操作？你可以把这些都写成一条SQL语句。是的，我知道，但这是一项大学任务，我想用PL/SQL来实现。你说“想为year1属性输出相应的列详细信息”是什么意思？我的意思是在COURSEOFFERING表中有5列，即a、B、C、D、e，F…….当我们对D列进行最大值计算时，我想用解析函数输出最大值加上BLook列中相应的值。它们类似于聚合，但可以为每一行返回。请参阅示例。代码运行，但在实现上述更改后没有输出。首先：o.i\u id=i\u id始终为真，可以删除，但这不会神奇地使代码正常工作。阿门！对不起，我不得不说。非常感谢你，汤姆，这让我大开眼界。您构建查询的方式和解释对我非常有用。再次感谢！！！