Question

我正在完成一项任务，但我已经接近完成任务了，我已经在一段时间内努力了。我需要显示一个具有SCHEDULE列的表和一个DAYS列，其中DAYS列显示基于SCHEDULE的天数列表。 SCHEDULE只有两个选项：'Weekend'和'Weekday。'

这是我想要达到的最终结果：

ID   Schedule     Days
001  Weekend      Saturday, Sunday

我创建了一个进程并使用了此代码：

BEGIN
UPDATE schedules
SET days =
        WHEN schedule = 'Weekend' THEN
        (SELECT LISTAGG(day, ', ') WITHIN GROUP (ORDER BY day_order)
        FROM days
        WHERE schedule = 'Weekend'
        )
        WHEN schedule = 'Weekday' THEN
        (SELECT LISTAGG(day, ', ') WITHIN GROUP (ORDER BY day_order)
        FROM days
        WHERE schedule = 'Weekday'
        )
    END

当我这样做时，我收到错误：

在遇到以下其中一项时遇到符号“文件结束”：begin function pragma procedure

如果有人能帮助我，我将不胜感激！

Answer 1

您可以使用MERGE声明：

MERGE INTO schedules dst
USING (
  SELECT schedule,
         LISTAGG( day, ', ' ) WITHIN GROUP ( ORDER BY day_order ) AS days
  FROM   days
  GROUP BY schedule
) src
ON ( dst.schedule = src.schedule )
WHEN MATCHED THEN
  UPDATE SET days = src.days;

但最好不要将值存储在分隔列表中，只需使用schedules.schedule和days.schedule之间的外键引用来连接这两个表。

Answer 2

您的更新等同于此。

UPDATE schedules s 
SET    days = (SELECT LISTAGG(day, ', ') 
                        within group ( ORDER BY day_order ) 
               FROM   days d 
               WHERE  d.schedule IN ( 'Weekend', 'Weekday' ) 
                      AND d.schedule = s.schedule 
               GROUP  BY d.schedule ) ;

但是，我会不推荐这个。将记录存储为逗号分隔值肯定会在将来导致问题。仅使用它来使用如下查询显示结果。这假设在您的日历表中，每天都有唯一的行。如果有重复项，则join将其设置为从天开始的不同结果集。

SELECT d.id, 
       d.schedule, 
       LISTAGG(s.day, ', ') 
         within GROUP ( ORDER BY d.day_order ) 
FROM   days d 
       join schedules s 
         ON ( d.schedule = s.schedule ) 
WHERE  d.schedule IN ( 'Weekend', 'Weekday' ) 
GROUP  BY d.id, 
          d.schedule

根据另一列的值显示单元格中的LISTAGG值

2 个答案: