Question

我需要将一个0和1的字符串转换为表示1的整数序列，类似于打印对话框中的页面选择序列。

e.g。 '0011001110101' - ＆gt; '3-4,7-9,11,13'

是否可以在单个SQL select（在Oracle 11g中）中执行此操作？

我可以使用以下内容获取页码的单独列表：

with data as (
    select 'K1' KEY,   '0011001110101' VAL from dual
    union select 'K2', '0101000110'   from dual
    union select 'K3', '011100011010' from dual
)
select
    KEY,
    listagg(ords.column_value, ',') within group (
        order by ords.column_value
    ) PAGES
from
    data
cross join (
    table(cast(multiset(
        select level
        from   dual
        connect by level <= length(VAL)
    ) as sys.OdciNumberList)) ords
)
where
    substr(VAL, ords.column_value, 1) = '1'
group by
    KEY

但是这不会进行分组（例如，为第一个值返回“3,4,7,8,9,11,13”）。

如果我可以在每次值更改时分配组编号，那么我可以使用分析函数来获取每个组的最小值和最大值。即如果我可以生成以下内容，那么我将被设置：

Key     Page    Val     Group
K1      1       0       1
K1      2       0       1
K1      3       1       2
K1      4       1       2
K1      5       0       3
K1      6       0       3
K1      7       1       4
K1      8       1       4
K1      9       1       4
K1      10      0       5
K1      11      1       6
K1      12      0       7
K1      13      1       8

但我坚持这一点。

任何人都有任何想法，或其他方法来获得这个？

Answer 1

首先让我们平衡它：

select regexp_instr('0011001110101', '1+', 1, LEVEL) istr,
       regexp_substr('0011001110101', '1+', 1, LEVEL) strlen
FROM dual
CONNECT BY regexp_substr('0011001110101', '1+', 1, LEVEL) is not null

然后使用listagg轻松完成剩下的工作：

with data as
 (
    select 'K1' KEY,   '0011001110101' VAL from dual
    union select 'K2', '0101000110'   from dual
    union select 'K3', '011100011010' from dual
 )
SELECT key,
       (SELECT listagg(CASE
                         WHEN length(regexp_substr(val, '1+', 1, LEVEL)) = 1 THEN
                          to_char(regexp_instr(val, '1+', 1, LEVEL))
                         ELSE
                          regexp_instr(val, '1+', 1, LEVEL) || '-' ||
                          to_char(regexp_instr(val, '1+', 1, LEVEL) +
                                  length(regexp_substr(val, '1+', 1, LEVEL)) - 1)
                       END,
                       ' ,') within GROUP(ORDER BY regexp_instr(val, '1+', 1, LEVEL))
          from dual
        CONNECT BY regexp_substr(data.val, '1+', 1, LEVEL) IS NOT NULL) val
  FROM data

Answer 2

使用没有正则表达式的递归子查询分解子句：

Oracle安装程序：

CREATE TABLE data ( key, val ) AS
  SELECT 'K1', '0011001110101' FROM DUAL UNION ALL
  SELECT 'K2', '0101000110' FROM DUAL UNION ALL
  SELECT 'K3', '011100011010' FROM DUAL UNION ALL
  SELECT 'K4', '000000000000' FROM DUAL UNION ALL
  SELECT 'K5', '000000000001' FROM DUAL;

<强>查询：

WITH ranges ( key, val, pos, rng ) AS (
  SELECT key,
         val,
         INSTR( val, '1', 1 ), -- Position of the first 1
         NULL
  FROM   data
UNION ALL
  SELECT key,
         val,
         INSTR( val, '1', INSTR( val, '0', pos ) ), -- Position of the next 1

         rng || ',' || CASE
           WHEN pos = LENGTH( val )                 -- Single 1 at end-of-string
           OR   pos = INSTR( val, '0', pos ) - 1    -- 1 immediately followed by 0
           THEN TO_CHAR( pos )
           WHEN INSTR( val, '0', pos ) = 0          -- Multiple 1s until end-of-string
           THEN pos || '-' || LENGTH( val )
           ELSE pos || '-' || ( INSTR( val, '0', pos ) - 1 ) -- Normal range
         END
  FROM   ranges
  WHERE  pos > 0
)
SELECT KEY,
       VAL,
       SUBSTR( rng, 2 ) AS rng -- Strip the leading comma
FROM   ranges
WHERE  pos = 0 OR val IS NULL
ORDER BY KEY;

<强>输出

KEY VAL           RNG
--- ------------- -------------
K1  0011001110101 3-4,7-9,11,13
K2  0101000110    2,4,8-9
K3  011100011010  2-4,8-9,11
K4  000000000000  
K5  000000000001  12

Answer 3

这是一个稍微高效的Isalamon解决方案版本（使用分层查询）。它稍微有点效率，因为我使用单个分层查询而不是多个（在相关子查询中），并且我在内部查询中只计算一次1的每个序列的长度。（实际上它只计算一次，但函数调用本身有一些开销。）

此版本还会正确处理'00000'和NULL等输入。 Isalamon的解决方案没有，当输入值为NULL时，MT0的解决方案不会返回一行。目前尚不清楚输入数据中是否有NULL，如果可能，所需的结果是什么;我假设应该返回一行，并使用page_list NULL。

此版本的优化器成本为17，Isalamon解决方案为18，MT0为33。但是，与标准字符串函数相比，优化器成本没有考虑到正则表达式的处理速度明显变慢;如果执行速度很重要，那么应该尝试MT0的解决方案，因为它可能会更快。

with data ( key, val ) as (
       select 'K1', '0011001110101' from dual union all
       select 'K2', '0101000110'    from dual union all
       select 'K3', '011100011010'  from dual union all
       select 'K4', '000000000000'  from dual union all
       select 'K5', '000000000001'  from dual union all
       select 'K6', null            from dual union all
       select 'K7', '1111111'       from dual union all
       select 'K8', '1'             from dual
     )
--  End of test data (not part of the solution); SQL query begins below this line.
select key, val,
       listagg(case when len = 1 then to_char(s_pos)
                    when len > 1 then to_char(s_pos) || '-' || to_char(s_pos + len - 1) 
               end, ',') within group (order by lvl) as page_list
from   ( select key, level as lvl, val,
                regexp_instr(val, '1+', 1, level)          as s_pos,
                length(regexp_substr(val, '1+', 1, level)) as len
         from   data
         connect by regexp_substr(val, '1+', 1, level) is not null
                and prior key = key
                and prior sys_guid() is not null
       )
group by key, val
order by key
;

<强>输出：

KEY  VAL            PAGE_LIST
---  -------------  -------------
K1   0011001110101  3-4,7-9,11,13
K2   0101000110     2,4,8-9
K3   011100011010   2-4,8-9,11
K4   000000000000
K5   000000000001   12
K6
K7   1111111        1-7
K8   1              1

将0和1的序列转换为打印样式的页面列表

3 个答案: