Question

使用Oracle，我希望进行以下查询，但我想知道是否有更“智能”的方法。

ORDER  BY vote ASC, (id * -1) ASC

以下是我的示例数据的样子

Select * from Sales Sales1
left join Sales Sales2 on Sales2.val = Sales1.val
left join Sales Sales3 on Sales3.val = Sales2.val
left join Sales Sales4 on Sales4.val = Sales3.val
left join Sales Sales5 on Sales5.val = Sales4.val
...

结果集看起来像这样：

customer number | acct | start balance | open date | prev account 
       a            1         100         01-01-15       b-1     
       b            1          80         03-04-14       
       c            2         200         04-11-14       c-1
       c            1         150         06-12-15        
       d            1         600         08-16-15         
       e            3         400         12-19-15       e-2
       e            2         150         10-21-14       e-1
       e            1         100         01-18-13

正如您所看到的，我需要根据结果继续加入另一个销售记录，我需要继续这样做，直到我返回一个空的结果集。我当前的场景是运行查询并查看sales5，sales6，sales7等中是否有值。

Answer 1

每当你必须自己加入未知次数时，你应该考虑CONNECT BY。您的特殊需求并非如此简单，但CONNECT BY仍然是解决方案的关键要素。

在下面的SQL中，mockup_data子因子只是为了给我一些数据。你使用的是实际的桌子。

您的想法是搜索数据＆＃34; root＆＃34; - 不是任何其他记录prev_account的记录。然后，您从那些和CONNECT BY开始，以获得他们以前的所有帐户，尽可能多。然后你PIVOT将它们全部放入列中。

有一件事 - Oracle SQL语句不能具有任意（数据驱动）列数。解析SQL时必须知道该数字。因此，在PIVOT子句中，您需要指定＆＃34;级别的最大数量＆＃34;您将支持，以便Oracle知道结果集可以有多少列。

这是SQL。

WITH 
mockup_data as (
SELECT   
'a' customer_Number,           1 acct,         100 start_balance,        to_date('01-01-15','MM-DD-YY') open_date,       'b-1' prev_account from dual union all      
SELECT 'b'            ,1,          80,         to_date('03-04-14','MM-DD-YY'), null       from dual union all 
SELECT 'c'            ,2,         200,         to_date('04-11-14','MM-DD-YY'),       'c-1' from dual union all 
SELECT 'c'            ,1,         150,         to_date('06-12-15','MM-DD-YY'),        null from dual union all 
SELECT 'd'            ,1,         600,         to_date('08-16-15','MM-DD-YY'),        null from dual union all 
SELECT 'e'            ,3,         400,         to_date('12-19-15','MM-DD-YY'),       'e-2' from dual union all 
SELECT 'e'            ,2,         150,         to_date('10-21-14','MM-DD-YY'),     'e-1' from dual union all 
SELECT 'e'            ,1,         100,         to_date('01-18-13','MM-DD-YY'),     null  from dual ),
data_with_roots AS
       (SELECT d.*,
               CASE
                 WHEN (SELECT COUNT (*)
                       FROM   mockup_data d2
                       WHERE  d2.prev_account = d.customer_number || '-' || d.acct) = 0 THEN
                   'Y'
                 ELSE
                   'N'
               END
                 is_root
        FROM   mockup_data d),
     hierarchy AS
       (SELECT CONNECT_BY_ROOT (customer_number) customer_number,
               CONNECT_BY_ROOT (acct) acct,
               CONNECT_BY_ROOT (start_balance) start_balance,
               CONNECT_BY_ROOT (open_date) open_date,
               start_balance prev_start_balance,
               open_date prev_open_date,
               LEVEL - 1 lvl
        FROM   data_with_roots d
        CONNECT BY customer_number || '-' || acct = PRIOR prev_account
        START WITH is_root = 'Y'),
     previous_only AS
       (SELECT *
        FROM   hierarchy
        WHERE  lvl >= 1)
SELECT *
FROM   previous_only PIVOT (MAX (prev_start_balance) AS prev_start, MAX (prev_open_date) AS prev_open
                     FOR lvl
                     IN (1 AS "01", 2 AS "02", 3 AS "03", 4 AS "04", 5 AS "05" -- etc... as many levels as you need to support
                                                                              ));

简化冗余左连接

1 个答案: