甲骨文行至列轉換

select * from table

值返回被

login id | status_name | count 
=============================== 
admin | open  |  3 
admin | closed  |  5 
test  | inprogress | 10 
test  | open  | 10 
test  | closed  | 11 
user1 | closed  |  5 
user1 | pending  | 10

如何從行轉移該數據列？我想以這種方式

login_id | open | closed | inprogress | pending 
================================================ 
admin | 3 |  5 |   0 |  0 
test  | 10 |  10 |   10 |  0 
user1 | 0 |  5 |   0 |  10

來源

2012-03-06 user1251973

嘗試尋找到PIVOT還有：如http://stackoverflow.com/questions/4841718/oracle-sql-pivot-query – tbone 2012-03-06 12:12:43

select login_id 
    , sum(case when status_name='open' then count end) open 
    , sum(case when status_name='closed' then count end) closed 
    , sum(case when status_name='inprogress' then count end) inprogress 
    , sum(case when status_name='pending' then count end) pending 
from table 
group by login_id

來源

2012-03-06 10:58:00 turbanoff

喜...感謝提供查詢但計數沒有顯示出來....還有什麼辦法可以寫出動態的 – user1251973 2012-03-06 11:59:13

甲骨文行至列轉換

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