oracle-求sql 行列轉行解決方案

求sql 行列轉行解決方案

比如oralce查出一張表

要轉換成如圖需求的表 如何做

最佳回答：

差不多下面這樣

select (月份 || '月') 月份
  ,sum(decode(日期, 1, 行走公裡數,0)) as 1日
  ,sum(decode(日期, 2, 行走公裡數,0)) as 2日
  ,sum(decode(日期, 3, 行走公裡數,0)) as 3日
  ,sum(decode(日期, 4, 行走公裡數,0)) as 4日
  ,sum(decode(日期, 5, 行走公裡數,0)) as 5日
  ,sum(decode(日期, 6, 行走公裡數,0)) as 6日
  ,sum(decode(日期, 7, 行走公裡數,0)) as 7日
  ,sum(decode(日期, 8, 行走公裡數,0)) as 8日
  ,sum(decode(日期, 9, 行走公裡數,0)) as 9日
  ,sum(decode(日期, 10, 行走公裡數,0)) as 10日
  ,sum(decode(日期, 11, 行走公裡數,0)) as 11日
  ,sum(decode(日期, 12, 行走公裡數,0)) as 12日
  ,sum(decode(日期, 13, 行走公裡數,0)) as 13日
  ,sum(decode(日期, 14, 行走公裡數,0)) as 14日
  ,sum(decode(日期, 15, 行走公裡數,0)) as 15日
  ,sum(decode(日期, 16, 行走公裡數,0)) as 16日
  ,sum(decode(日期, 17, 行走公裡數,0)) as 17日
  ,sum(decode(日期, 18, 行走公裡數,0)) as 18日
  ,sum(decode(日期, 19, 行走公裡數,0)) as 19日
  ,sum(decode(日期, 20, 行走公裡數,0)) as 20日
  ,sum(decode(日期, 21, 行走公裡數,0)) as 21日
  ,sum(decode(日期, 22, 行走公裡數,0)) as 22日
  ,sum(decode(日期, 23, 行走公裡數,0)) as 23日
  ,sum(decode(日期, 24, 行走公裡數,0)) as 24日
  ,sum(decode(日期, 25, 行走公裡數,0)) as 25日
  ,sum(decode(日期, 26, 行走公裡數,0)) as 26日
  ,sum(decode(日期, 27, 行走公裡數,0)) as 27日
  ,sum(decode(日期, 28, 行走公裡數,0)) as 28日
  ,sum(decode(日期, 29, 行走公裡數,0)) as 29日
  ,sum(decode(日期, 30, 行走公裡數,0)) as 30日
  ,sum(decode(日期, 31, 行走公裡數,0)) as 31日

from 表 group by 月份

參考http://blog.csdn.net/danielinbiti/article/details/44977749