Question

我想按照描述的顺序使用两列来排序我的表：

当前余额将从下一行中的上一个余额开始。

下面是我的数据

ID_DATE     PREV_BAL    CURR_BAL
20200201,   157,        192
20200201,   192,        195
20200201,   123,        124
20200201,   124,        157
20200201,   125,        123

我希望按照以下顺序先订购

然后选择最上面的一行作为我的最终结果，即

ID_DATE     PREV_BAL    CURR_BAL
20200201,   192,        195

任何帮助

with da as (
    select 20200201 id_date, 157 prev_bal, 192 curr_bal from dual
    union all select 20200201 id_date, 192 prev_bal, 195 curr_bal from dual
    union all select 20200201 id_date, 123 prev_bal, 124 curr_bal from dual
    union all select 20200201 id_date, 124 prev_bal, 157 curr_bal from dual
    union all select 20200201 id_date, 125 prev_bal, 123 curr_bal from dual
)
SELECT * FROM da

Answer 1

这是一个hierarchy，您可以看到运行此查询的整个周期：

select da.*, sys_connect_by_path(curr_bal, ' - ') path
  from da 
  connect by prior prev_bal = curr_bal 
  start with not exists (select 1 from da t where t.prev_bal = da.curr_bal)      

 ID_DATE   PREV_BAL   CURR_BAL PATH
-------- ---------- ---------- -------------------------------
20200201        192        195  - 195
20200201        157        192  - 195 - 192
20200201        124        157  - 195 - 192 - 157
20200201        123        124  - 195 - 192 - 157 - 124
20200201        125        123  - 195 - 192 - 157 - 124 - 123

但是，如果只想让大多数父行使用Tejash的查询或not in或not exists：

select * from da 
  where not exists (select 1 from da t where t.prev_bal = da.curr_bal)

如果您要按照问题中的说明在整个树中朝其他方向移动，请更改connect by clause并选择叶行：

select da.*, sys_connect_by_path(curr_bal, ' - ') path
  from da 
  where connect_by_isleaf = 1
  connect by prev_bal = prior curr_bal 
  start with not exists (select 1 from da t where da.prev_bal = t.curr_bal)

^dbfiddle

Answer 2

我建议您进行self outer join并在无法找到任何记录的地方进行记录CURR_BAL包含与PREV_BAL相同的记录。

我不知道它是否对您有帮助，但是它给出了所需的答案

类似以下内容：

with da as (
      select 20200201 id_date, 157 prev_bal, 192 curr_bal from dual
      union all select 20200201 id_date, 192 prev_bal, 195 curr_bal from dual
      union all select 20200201 id_date, 123 prev_bal, 124 curr_bal from dual
      union all select 20200201 id_date, 124 prev_bal, 157 curr_bal from dual
      union all select 20200201 id_date, 125 prev_bal, 123 curr_bal from dual
 )
 -- Your query starts from here
 SELECT
      ID_DATE, PREV_BAL, CURR_BAL
 FROM
     (
          SELECT
              D.*, P.ID_DATE   AS P_ID_DATE
          FROM
              DA D LEFT JOIN DA P ON D.CURR_BAL = P.PREV_BAL
      )
 WHERE P_ID_DATE IS NULL;

结果

        ID_DATE   PREV_BAL   CURR_BAL
     ---------- ---------- ----------
       20200201        192        195

干杯！

Answer 3

这是一个简单的问题，只需使用下面的查询

SELECT * FROM ( 
with da as (
    select 20200201 id_date, 157 prev_bal, 192 curr_bal from dual
    union all select 20200201 id_date, 192 prev_bal, 195 curr_bal from dual
    union all select 20200201 id_date, 123 prev_bal, 124 curr_bal from dual
    union all select 20200201 id_date, 124 prev_bal, 157 curr_bal from dual
    union all select 20200201 id_date, 125 prev_bal, 123 curr_bal from dual
)
SELECT * FROM da ORDER BY prev_bal DESC  )
WHERE rownum=1 ORDER BY curr_bal DESC

根据顺序对列进行排序后获取最新记录

3 个答案: