Question

我正在尝试将SECTION_ID作为SECTION，并将ENROLL_DATE作为学生计算在这里，我从学生专栏获得正确的结果，但不是从专栏专栏获得。我只需要来自SECTION表的SECTION_ID而不是来自ENROLLMENT表，所以我做了减去设置操作。为什么它给我两张桌子的结果？我怎么才能从SECTION表中获得结果？

 SELECT s.LOCATION,COUNT(s.SECTION_ID) AS SECTIONS, COUNT(e.ENROLL_DATE) AS STUDENTS
 FROM SECTION s
 JOIN ENROLLMENT e
 ON s.SECTION_ID = e.SECTION_ID
 WHERE EXISTS (

           SELECT SECTION_ID
           FROM SECTION
            MINUS
           SELECT SECTION_ID
           FROM ENROLLMENT

            )
 AND s.SECTION_ID = e.SECTION_ID
 GROUP BY s.LOCATION
 ORDER BY s.LOCATION;

我收到了这个结果

 LOCATION  SECTIONS   STUDENTS
 ---------- ---------- ----------
   H310        1        1
   L206        8        8
   L210        29       29
   L211        10       10
   L214        36       36
   etc

我想要的结果

 LOCATION  SECTIONS   STUDENTS
 ---------- ---------- ----------
   H310        1        1
   L206        1        8
   L210        10       29
   L211        3        10
   L214        15       36
   etc

Answer 1

我最初的想法是你需要distinct关键字......

像这样：

 SELECT s.LOCATION,
        COUNT(DISTINCT s.SECTION_ID) AS SECTIONS, 
        COUNT(e.ENROLL_DATE) AS STUDENTS
 FROM SECTION s
 JOIN ENROLLMENT e
 ON s.SECTION_ID = e.SECTION_ID
 WHERE EXISTS (    
           SELECT SECTION_ID
           FROM SECTION
            MINUS
           SELECT SECTION_ID
           FROM ENROLLMENT    
            )
 AND s.SECTION_ID = e.SECTION_ID
 GROUP BY s.LOCATION
 ORDER BY s.LOCATION;

Answer 2

如果没有产生这些结果的示例数据，很难说，但我认为您可能希望计数不同：

Select
    s.location,
    count(distinct s.section_id) as sections,
    count(e.enroll_date) as students
from
   section s
       inner join
   enrollment e
       on s.section_id = e.section_id
group by
    s.location
order by
    s.location;

您存在的标准可能没有达到您的预期，因为内部和外部之间没有相关性。您还在join和where子句中重复了s.section_id = e.section_id。

设置运算符减去不显示结果

2 个答案: