我有一个带有键的表(即数字),如org_key和par_org_key
Org_key | Par_Org_key 52 26 23 89 26 14 14 8
在上面的表格中,org_key 52的父级是26,org_key 26的父级是14,依此类推 上述系列的根父级是8(因为它未在org_key列中列出)。 我需要找到这样的根,并在单行上有完整的链接,例如:8 | 14 | 26 | 52。
此SQL适用于Oracle:
select
substr(sys_connect_by_path(org_nbr,'|'),2) spa
,substr(sys_connect_by_path(org_key,'|'),2) org_key_line
,connect_by_root(org_key) org_key_root
,level lvl
,org.*
from org
start with par_org_key is null
connect by par_org_key = prior org_key
我的问题是,如何使用Netezza SQL执行相同的查询?
答案 0 :(得分:1)
Netezza不支持递归公用表表达式,否则这将是更优雅和可扩展的解决方案。我发现最好的解决方法是在同一个表中使用Left Outer Joins。
创建一个测试表。我们使用-1来表示根节点。
CREATE TABLE t1 AS
SELECT 52 AS Org_key, 26 AS Par_Org_key
UNION
SELECT 23 AS Org_key, 89 AS Par_Org_key
UNION
SELECT 26 AS Org_key, 14 AS Par_Org_key
UNION
SELECT 14 AS Org_key, 8 AS Par_Org_key
UNION
SELECT 8 AS Org_key, -1 AS Par_Org_key;
返回8 | 14 | 26 | 52
SELECT NVL(a.Org_key,'0') || '|' || NVL(b.Org_key,'0') || '|' || NVL(c.Org_key,'0') || '|' || NVL(d.Org_key,'0')
FROM t1 a
LEFT OUTER JOIN t1 b ON a.Org_key = b.Par_Org_key
LEFT OUTER JOIN t1 c ON b.Org_key = c.Par_Org_key
LEFT OUTER JOIN t1 d ON c.Org_key = d.Par_Org_key
LEFT OUTER JOIN t1 e ON d.Org_key = e.Par_Org_key
WHERE a.Par_Org_key = -1;
如果你想添加更多左外连接以支持未知数量的级别,这可以帮助处理产生的空值
SELECT NVL(CAST(a.Org_key AS VARCHAR(10)),'') || '|' || NVL(CAST(b.Org_key AS VARCHAR(10)),'') || '|' || NVL(CAST(c.Org_key AS VARCHAR(10)),'') || '|' || NVL(CAST(d.Org_key AS VARCHAR(10)),'')