如何代碼從Postgres的轉換到Oracle

Question

我有一個源表（T1）：

ID1 | ID2 
---------- 
1 | 2 
1 | 5 
4 | 7 
7 | 8 
9 | 1 

我想將數據轉換成這（T2）：

ID1 | ID2 | LABEL 
------------------ 
1 | 2 | 1 
1 | 5 | 1 
4 | 7 | 2 
7 | 8 | 2 
9 | 1 | 1 

我找到了解決爲此在PostgreSQL中：

with 
recursive cte(id1, id2) as (
select id1, id2, 1 as level 
from t 
union all 
select t.id1, cte.id2, cte.level + 1 
from t join 
     cte 
     on t.id2 = cte.id1 
) 
select id1, id2, 
    dense_rank() over (order by grp) as label 
from (select id1, id2, 
     least(min(id2) over (partition by id1), min(id1) over (partition by id2)) as grp, 
     level 
    from cte 
) t 
where level = 1; 

我想將此代碼轉換爲Oracle。我如何將此代碼從Postgres轉換爲Oracle？

Answer 1

Oracle 11.2支持遞歸CTE。但它不符合標準，因爲recursive關鍵字不是必需的（實際上：不能使用）。因此，如果您刪除recursive關鍵字並正確定義CTE列，則以下內容應可用。您還需要使用與LEVEL不同的東西，因爲這是保留字。

with cte (id1, id2, lvl) as (
select id1, id2, 1 as lvl 
from t 
union all 
select t.id1, cte.id2, cte.lvl + 1 
from t 
    join cte on t.id2 = cte.id1 
) 
select id1, 
     id2, 
     dense_rank() over (order by grp) as label 
from (
    select id1, 
     id2, 
     least(min(id2) over (partition by id1), min(id1) over (partition by id2)) as grp, 
     lvl 
    from cte 
) t 
where lvl = 1; 

下面是一個SQLFiddle例如：http://sqlfiddle.com/#!4/deeb2/3

但是我懷疑，原來的查詢是正確的，因爲你沒有在遞歸CTE「起始條件」。該聯盟的第一部分檢索全部行的表格。除非我誤解了查詢的目的，否則應該有條件將其限制爲層次結構的「根」。

遞歸CTE也可以用CONNECT BY查詢替換，你的情況，這將是：

select id1, id2, level as lvl 
from t 
connect by prior id1 = id2; 

你可以結合，與原單查詢：

with cte (id1, id2, lvl) as (
    select id1, id2, level as lvl 
    from t 
    connect by prior id1 = id2 
) 
select id1, 
     id2, 
     dense_rank() over (order by grp) as label 
from (
    select id1, 
     id2, 
     least(min(id2) over (partition by id1), min(id1) over (partition by id2)) as grp, 
     lvl 
from cte 
) t 
where lvl = 1; 

雖然我認爲它應該是一樣的，似乎層次結構是以不同的順序遍歷的。可能是因爲遞歸CTE首先執行深度並且connect by執行了深度優先遞歸（或者相反方向）。對第二個版本

SQLFiddle例如：http://sqlfiddle.com/#!4/deeb2/4