Oracle SQL子查询

Question

我想选择工作时间少于任何人平均工作时间和一个营地的人

表是人和时间表

这是我到目前为止所拥有的：

 SELECT fname AS "First Name", lname AS "Last Name", 
 SUM((end_time - start_time) * 24)         
 FROM person JOIN schedule USING (person_ID)
 GROUP BY fname
 HAVING SUM((end_time - start_time) * 24) < (
 SELECT AVG(SUM((end_time - start_time) * 24)) FROM schedule);`

人员具有fname，lname和person_ID时间表具有sched_id，s_date，start_time，end_time和person_id

谢谢！

收到此错误：

Error starting at line 18 in command:
    SELECT fname AS "First Name", lname AS "Last Name", 
    SUM((end_time - start_time) * 24)     
    FROM person JOIN schedule USING (person_ID)
    GROUP BY fname
    HAVING SUM((end_time - start_time) * 24) < (
    SELECT AVG(SUM((end_time - start_time) * 24)) FROM schedule)
    Error at Command Line:18 Column:31
    Error report:
    SQL Error: ORA-00979: not a GROUP BY expression
    00979. 00000 -  "not a GROUP BY expression"
   *Cause:    
   *Action:

Answer 1

GROUP BY行需要在SELECT语句中包括所有未聚合的字段（因此，在这种情况下，除了SUM（）以外的所有内容）。

尝试将其更改为：

GROUP BY fname, lname

Answer 2

当使用聚合函数（例如SUM）时，您需要通过以下方式将所有非聚合属性包括在组中：

 SELECT fname AS "First Name", lname AS "Last Name", 
 SUM((end_time - start_time) * 24)         
 FROM person JOIN schedule USING (person_ID)
 GROUP BY fname, lname --this is changed
 HAVING SUM((end_time - start_time) * 24) < (
 SELECT AVG(SUM((end_time - start_time) * 24)) FROM schedule);`

Answer 3

我认为解决此问题的最佳方法是使用解析函数：

select "First Name", "Last Name", hours
from (SELECT fname AS "First Name", lname AS "Last Name",
             SUM((end_time - start_time) * 24) as hours,
             avg(SUM((end_time - start_time) * 24)) over () as avghours
      FROM person JOIN
           schedule
           USING (person_ID)
      GROUP BY fname, lname
     ) t
WHERE hours < avghours;

Oracle报告的特定问题是avg(sum())不允许group by （除非如上所述， avg()实际上是一个解析函数）。