这是我的查询结果：
WEEKENDDATE CCNAME PHNAME RATIO
08-Feb-15 Apple Line 1 - Day L&I Work 0.45
08-Feb-15 Apple Line 1 - Day Sorter 6.85
08-Feb-15 Apple Line 1 - Day Tray Fill 12.93
08-Feb-15 Apple Line 1 - Day WH General Labor 5.6
08-Feb-15 Apple Line 1 - Day WH Supervisor 1.48
15-Feb-15 Apple Line 1 - Day L&I Work 0.42
15-Feb-15 Apple Line 1 - Day Sorter 6.09
15-Feb-15 Apple Line 1 - Day Tray Fill 11.9
15-Feb-15 Apple Line 1 - Day WH General Labor 5.42
15-Feb-15 Apple Line 1 - Day WH Supervisor 3.46
22-Feb-15 Apple Line 1 - Day L&I Work 0.43
22-Feb-15 Apple Line 1 - Day Sorter 6.01
22-Feb-15 Apple Line 1 - Day Tray Fill 12.09
22-Feb-15 Apple Line 1 - Day WH General Labor 4.9
22-Feb-15 Apple Line 1 - Day WH Supervisor 1.71
我的老板想要数据横向 有没有办法让我的查询结果格式如下：
CCNAME PHNAME 08-Feb-15 15-Feb-15 22-Feb-15
Apple Line 1 - Day L&I Work 0.45 0.42 0.43
Apple Line 1 - Day Sorter 6.85 6.09 6.01
Apple Line 1 - Day Tray Fill 12.93 11.9 12.09
Apple Line 1 - Day WH General Labor 5.6 5.42 4.9
Apple Line 1 - Day WH Supervisor 1.48 3.46 1.71
或者像这样：
CCNAME PHNAME 08-FEB-15,15-FEB-15,22-FEB-15
Apple Line 1 - Day L&I Work .45,.42,.43
Apple Line 1 - Day Sorter 6.85,6.09,6.01
Apple Line 1 - Day Tray Fill 12.93,11.9,12.09
Apple Line 1 - Day WH General Labor 5.6,5.42,4.9
Apple Line 1 - Day WH Supervisor 1.48,3.46,1.71
以下是我能够弄清楚的问题：
请注意我事先不知道日期范围。
这取决于用户要求的周数。
select ccname, phname, listagg(weekenddate||','||ratio,'; ')
within group(order by ccname, phname) ratio
from
(...leaving out this code...)
group by ccname, phname
但是下面的结果还不够好：我可以使用＆＃39; listagg＆＃39;或者我应该研究＆＃39; pivot＆＃39;代替？
CCNAME PHNAME RATIO
Apple Line 1 - Day L&I Work 08-FEB-15,.45;15-FEB-15,.42;22-FEB-15,.43
Apple Line 1 - Day Sorter 08-FEB-15,6.85;15-FEB-15,6.09;22-FEB-15,6.01
Apple Line 1 - Day Tray Fill 08-FEB-15,12.93;15-FEB-15,11.9;22-FEB-15,12.09
Apple Line 1 - Day WH General Labor 08-FEB-15,5.6;15-FEB-15,5.42;22-FEB-15,4.9
Apple Line 1 - Day WH Supervisor 08-FEB-15,1.48;15-FEB-15,3.46;22-FEB-15,1.71