你正在以一个固定值，字符串字面值'RM'进行循环，所以你真的没有做任何有用的枢轴 - 输出是相同的，自己运行'pivot_data'查询：

You are pivoting on a fixed value, the string literal 'RM', so you're really not doing anything useful in the pivot - the output is the same as you'd get from running the 'pivot_data' query on its own:

SELECT eNAME,workhrs,room, SCR from PRODUCTIVITY p,PRODUCTIVITYd d, emp e, ROOMS R
where p.PRODUCTIVITYID=d.PRODUCTIVITYID and e.empno=p.employeeid
AND R.ID=P.ROOMID;

ENAME    WORKHRS       ROOM        SCR
----- ---------- ---------- ----------
JONES        3.6        101         53
ALLEN       1.32        101         43
ALLEN          6        102         22

您希望每个员工的总额 workhrs 以及他们售出的房间的枢纽。如果您更改该查询以获得 workhrs 的分析和，以及房间/ scr值的排名（并使用现代连接语法），您将获得：

You want the aggregate workhrs for each employee, and a pivot of the rooms they sold. If you change that query to get the analytic sum of workhrs and a ranking of the room/scr values (and using modern join syntax) you get:

select e.ename, r.room, p.scr,
  sum(d.workhrs) over (partition by e.ename) as wrkhrs,
  rank() over (partition by e.ename order by r.room, p.scr) as rnk
from productivity p
join productivityd d on d.productivityid = p.productivityid
join emp e on e.empno=p.employeeid
join rooms r on r.id = p.roomid;

ENAME       ROOM        SCR     WRKHRS        RNK
----- ---------- ---------- ---------- ----------
ALLEN        101         43       7.32          1
ALLEN        102         22       7.32          2
JONES        101         53        3.6          1

然后，您可以关注生成的 rnk 号码：

You can then pivot on that generated rnk number:

with pivot_data as (
  select e.ename, r.room, p.scr,
    sum(d.workhrs) over (partition by e.ename) as wrkhrs,
    rank() over (partition by e.ename order by r.room, p.scr) as rnk
  from productivity p
  join productivityd d on d.productivityid = p.productivityid
  join emp e on e.empno=p.employeeid
  join rooms r on r.id = p.roomid
)
select *
from   pivot_data
pivot (
  min(room) as room, min(scr) as scr  --<-- pivot_clause
  for rnk                             --<-- pivot_for_clause        
  in  (1, 2, 3)                       --<-- pivot_in_clause         
);

ENAME     WRKHRS     1_ROOM      1_SCR     2_ROOM      2_SCR     3_ROOM      3_SCR
----- ---------- ---------- ---------- ---------- ---------- ---------- ----------
ALLEN       7.32        101         43        102         22                      
JONES        3.6        101         53                                            

您需要知道任何员工可能拥有的房间 - 即最高的 rnk 可能是 - 并且包括 in 这意味着您可能最终会出现空列，如本示例中没有 3_room 或 3_scr $ c $的数据c>。你不能避免这种情况，除非你得到一个XML结果或动态生成查询。

You need to know the maximum number of rooms any employee may have - i.e. the highest rnk could ever be - and include all of those in the in clause. Which means you're likely to end up with empty columns, as in this example where there is no data for 3_room or 3_scr. You can't avoid that though, unless you get an XML result or generate the query dynamically.