Question

这是我要问的上一个问题的第二部分。我有一张许可证号码表，我正在试图确定它们是否合法。我试图运行的其中一个测试是数字是顺序/计数模式。真正有用的是，如果我可以获得连续字符的最大数量，但我觉得这更难。例如：

 LICENSE_NUMBER   IS_COUNTING    NUM_SEQ_CHARS
 123456789        TRUE           9
 123455678        FALSE          5
 456789012        TRUE           9
 12345            TRUE           5
 123451234        FALSE          5

感谢您的帮助！

Answer 1

这是确定给定许可证号码中连续增加数字的最大数量的有效方法。我没有打扰YES / NO列 - 这可以通过比较MAX_COUNT到length(license_number)来获得。将其留作练习。

我们需要将许可证号码分解为各个字符，并跟踪位置。然后我们可以减去“位置”并取残差模10.因为Oracle认为mod(-1, 10)是-1而不是9（他们真的需要回到小学重新学习MODULO函数），我需要添加10，所以我得到了所有情况下的正确结果（解释下面的解决方案中的“奇怪”）。现在问题变成：为每个许可证号找到最长的“相等”残差模块10。

使用所谓的“Tabibitosan方法”（在我的解决方案中称为row_number()的CTE中两个prep的差异）最有效地查找序列中的连续“某事”。然后，这一切都变成了分组，计数和取max()的问题。

with
-- begin of test data; not part of the solution
     test_data ( license_number ) as (
       select '123456789' from dual union all
       select '123455678' from dual union all
       select '456789012' from dual union all
       select '12345'     from dual union all
       select '123451234' from dual union all
       select '402023488' from dual union all
       select '4189012'   from dual
     ),
-- end of test data; solution (SQL query) continues below this line
    tokenized ( license_number, idx, res ) as (
      select license_number, level,
             mod(10 + to_number(substr(license_number, level, 1)) - level, 10)
      from   test_data
      connect by level <= length(license_number)
          and prior license_number = license_number
          and prior sys_guid() is not null
    ),
    prep ( license_number, res, grp ) as (
      select license_number, res, 
             row_number() over (partition by license_number order by idx) -
               row_number() over (partition by license_number, res order by idx)
      from   tokenized
    ),
    grouped ( license_number, res, grp, ct ) as (
      select   license_number, res, grp, count(*)
      from     prep
      group by license_number, res, grp
    )
select   license_number, max(ct) as max_count
from     grouped
group by license_number
;

<强>输出：

LICENSE_NUMBER  MAX_COUNT
--------------  ---------
123455678               5
123456789               9
456789012               9
123451234               5
4189012                 5
12345                   5
402023488               3

Answer 2

^(01(2345678901)*(2(3(4(5(6(7(8(90?)?)?)?)?)?)?)?)?
 |12(3456789012)*(3(4(5(6(7(8(9(01?)?)?)?)?)?)?)?)?
 |23(4567890123)*(4(5(6(7(8(9(0(12?)?)?)?)?)?)?)?)?
 |34(5678901234)*(5(6(7(8(9(0(1(23?)?)?)?)?)?)?)?)?
 |45(6789012345)*(6(7(8(9(0(1(2(34?)?)?)?)?)?)?)?)?
 |56(7890123456)*(7(8(9(0(1(2(3(45?)?)?)?)?)?)?)?)?
 |67(8901234567)*(8(9(0(1(2(3(4(56?)?)?)?)?)?)?)?)?
 |78(9012345678)*(9(0(1(2(3(4(5(67?)?)?)?)?)?)?)?)?
 |89(0123456789)*(0(1(2(3(4(5(6(78?)?)?)?)?)?)?)?)?
 |90(1234567890)*(1(2(3(4(5(6(7(89?)?)?)?)?)?)?)?)?)$

它将匹配任何两个或更多数字的字符串，完全由连续数字组成。

123456789     MATCH
123455678     NOT MATCH
456789012     MATCH
12345         MATCH

演示：http://regexr.com/3enkb

Answer 3

此查询还计算连续字符的数量：

SELECT LICENSE_NUMBER,
       CASE length( LICENSE_NUMBER ) WHEN  max( cnt) THEN 'TRUE' ELSE 'FALSE' END As IS_COUNTING ,
       max( cnt) As NUM_SEQ_CHARS
FROM (
      SELECT LICENSE_NUMBER, p, count(*) As cnt
      FROM (
            SELECT LICENSE_NUMBER,
                   SUM( xx ) OVER ( Partition By LICENSE_NUMBER ORDER BY x ) As p
            FROM (
                  SELECT LICENSE_NUMBER,
                         x, qq,
                         CASE WHEN qq - 1 = LAG( qq ) OVER ( Partition By LICENSE_NUMBER ORDER BY x ) 
                              THEN 0 
                              WHEN qq = 0 AND 9 = LAG( qq ) OVER ( Partition By LICENSE_NUMBER ORDER BY x ) 
                              THEN 0
                              ELSE 1 
                          END As xx
                  FROM (
                      SELECT LICENSE_NUMBER, x, substr( LICENSE_NUMBER, x, 1 ) qq
                      FROM (
                          SELECT *
                          FROM table
                          CROSS JOIN (
                             SELECT level x FROM dual
                             CONNECT BY LEVEL <= ( SELECT MAX( length( LICENSE_NUMBER )) FROM table )
                          )
                      )
                  )
            )
       )
       GROUP BY LICENSE_NUMBER, p
)
GROUP BY LICENSE_NUMBER

如何在Oracle SQL中确定数字是否是连续的

3 个答案: