Question

我有一个有列的表出席 date1，status（PRESENT或ABSENT），student_idstudent和student_subject_subjectid

我想查询student_idstudent和他们拥有的百分比出勤率，即每个学生对于特定主题的出勤率（当前/总计数）。基本上我只想要两个值的值。有没有简单的方法可以不使用视图，最好使用pl / sql？

下面是我尝试使用光标。

请告诉我使用以下代码是否保证为每个student_idstudent提供相应的答案（即现在的计数和总数）

还告诉我，在其他情况下是否会失败？

    DECLARE 
    s_id attendance.student_idstudent%type; 
    s_present number(3); 
    CURSOR student_present is select student_idstudent, count(student_idstudent) from  attendance 
    where status like 'Present' and student_subject_subjectid like 102 
    group by student_idstudent;

    s1_id attendence.student_idstudent%type; 
    s1_total number(3); 
    CURSOR student_total is select student_idstudent, count(student_idstudent) from attendance 
    where student_subject_subjectid like 102 
    group by student_idstudent;

    BEGIN 
    OPEN student_present; 
    OPEN student_total;
    LOOP 
    FETCH student_present into s_id, s_present;
    FETCH student_total into s1_id, s1_total;
    EXIT WHEN student_present%notfound;  
    dbms_output.put_line('student '|| s_id || ' has ' ||S_present/s1_total*100 || '% record' ); 
    END LOOP; 

    CLOSE student_present; 
    CLOSE student_total; 
    END; 
    /

Answer 1

我不确定这是否是一个家庭作业问题，所以我给你一个部分答案，你应该能够自己完成查询。

select student_idstudent
      ,count(case when status like 'Absent'  then 1 end) as absent_count
      ,count(case when status like 'Present' then 1 end) as present_count
      ,count(*)                                      as tot_count
  from attendance
 where student_subject_subjectid = 102
 group 
    by student_idstudent;

此外，使用不带任何外卡字符的LIKE等同于使用=。为清楚起见，条件应该写为status = 'Absent'和status = 'Present'。这也更清楚了。

like <number>也是如此。您的意思是student_subject_subjectid = 102，还是您还有其他想法？

Answer 2

随着@ronnis的帮助。
我想出了这个答案。

    SELECT student_idstudent,
    count(CASE WHEN  status = 'Present'  then 1 END)/count(*)*100 as percentage
    FROM attendence 
    where student_subject_subjectid = 102
    group by student_idstudent;

在单个列上执行两个COUNT而不在一个sql语句或pl / sql块中使用视图

2 个答案: