Question

SQL>从标记中选择*;

 STUDENTID  SUBJECTID    MARK
---------- ---------- ----------
     1          1         **21**
     1          2         75
     1          3         87
     2          1         82
     2          2         64
     2          3         77
     3          1         82
     3          2         **23**
     3          3         67

已选择9行。

需要计算所有科目中谁及格（每门科目> 30）且总分> 170 这里的输出是这样的：

 STUDENTID  Status         Total_MARK

 ---------- ----------      ----------

  2           Passed              223

Answer 1

您可以将分组与具有having子句一起使用

 select student_id, 'Passed' as status, sum(mark) as total_mark 
   from marks
  group by student_id
 having sum(mark)>170 and min(mark)>30;

Demo

编辑 （由于您的最后评论）：使用

with marks( student_id, mark ) as
(
 select 1, 21 from dual union all
 select 1, 75 from dual union all
 select 1, 87 from dual union all
 select 2, 82 from dual union all
 select 2, 64 from dual union all
 select 2, 77 from dual union all
 select 3, 82 from dual union all
 select 3, 23 from dual union all
 select 3, 67 from dual
)
 select student_id as "Student ID", 
        case when sum(mark)>170 and min(mark)>30 then 
                  sum(mark)||' (Passed)' 
        else
                  sum(mark)||' (Failed)'         
        end as "Total Mark (Status)"
   from marks
  group by student_id;

Student ID  Total Mark (Status)
----------  -------------------
1           183 (Failed)
2           223 (Passed)
3           172 (Failed)

通过组合Total Mark和Status，

可能是一种明智的方法。如果您真的想分隔这些列，则需要以类似的方式编写单独的case when状态陈述。

在所有科目中及格且总分大于170的学生

1 个答案: