在oracle中，如何将数据排序到右侧而不是底部？

假设我桌上有这些数据

select * from user.weekly_job;

得到：

但我需要员工提供的所有数据，其中包含每种工作类型的发生情况，如下所示：

Staff | 1 | 2 | 3 | 4 | 5
A     | 1 | 1 | 2 | 0 | 1
B     | 0 | 2 | 1 | 0 | 1

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如何使用oracle query执行此操作？

如果您有固定数量的JobType值：

select STAFF,
       SUM(CASE WHEN JobType=1 THEN 1 ELSE 0 END) as "1",
       SUM(CASE WHEN JobType=2 THEN 1 ELSE 0 END) as "2",
       SUM(CASE WHEN JobType=3 THEN 1 ELSE 0 END) as "3",
       SUM(CASE WHEN JobType=4 THEN 1 ELSE 0 END) as "4",
       SUM(CASE WHEN JobType=5 THEN 1 ELSE 0 END) as "5"
FROM t
GROUP BY STAFF

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按每种作业类型分组后，我有1万多个数据，但php需要很长时间才能完成这项工作，所以我需要先用oracle的方式进行比较

select STAFF,
       SUM(CASE WHEN JobType=1 THEN 1 ELSE 0 END) as "1",
       SUM(CASE WHEN JobType=2 THEN 1 ELSE 0 END) as "2",
       SUM(CASE WHEN JobType=3 THEN 1 ELSE 0 END) as "3",
       SUM(CASE WHEN JobType=4 THEN 1 ELSE 0 END) as "4",
       SUM(CASE WHEN JobType=5 THEN 1 ELSE 0 END) as "5"
FROM t
GROUP BY STAFF