6

我有一个像下面这样的数据，其中有标签限制。我在这里代表了他们的观点

with t_view as (select '6-21  6-21  6-21  6-21  6-21  6-21  6-21  ' as col from dual
union
select '6-20  6-20  6-20  6-20  6-20  ' from dual
union
select '6-9  6-9  6-9  6-9  6-9  6-9  6-9  ' from dual)

我的预期输出是

Mon: 6-21, Tue: 6-21, Wed: 6-21, Thu: 6-21, Fri: 6-21, Sat: 6-21, Sun: 6-21
Mon: 6-20, Tue: 6-20, Wed: 6-20, Thu: 6-20, Fri: 6-20
Mon: 6-9, Tue: 6-9, Wed: 6-9, Thu: 6-9, Fri: 6-9, Sat: 6-9, Sun: 6-9

我想用一些像这样的独特模式替换所有这些水平表，然后根据索引用 Mon, Tue 替换该模式

$1(6-20 )$2(6-20 )$3(6-20 )$4(6-20 )$5(6-20 )

我已尝试以下查询，但无法完成

select regexp_replace(col, '([[:digit:]]-[[:digit:]]{2}[[:space:]]+)','$(\1)') from t_view;
4

5 回答 5

3

您需要CASE表达式、REGEXP_COUNT和 REGEXP_REPLACE的组合，因为您对所有行都没有相同的表达式。根据数据，您可以在 case 表达式中有尽可能多的条件。

正则表达式模式是(\d-\d+ ).

例如，

SQL> WITH t_view(col) AS
  2    ( SELECT '6-21  6-21  6-21  6-21  6-21  6-21  6-21  ' FROM dual
  3    UNION
  4    SELECT '6-20  6-20  6-20  6-20  6-20  ' FROM dual
  5    UNION
  6    SELECT '6-9  6-9  6-9  6-9  6-9  6-9  6-9  ' FROM dual
  7    )
  8  SELECT REPLACE(new_col, '  ','') new_col
  9  FROM (
 10    SELECT
 11      CASE
 12        WHEN regexp_count(col, '\d+\-\d+') = 5
 13        THEN regexp_replace(col,
 14                            '(\d-\d+  )(\d-\d+  )(\d-\d+  )(\d-\d+  )(\d-\d+  )',
 15                            'Mon: \1, Tue: \2,Wed: \3,Thu: \4,Fri: \5')
 16        WHEN regexp_count(col, '\d+\-\d+') = 7
 17        THEN regexp_replace(col,
 18                            '(\d-\d+  )(\d-\d+  )(\d-\d+  )(\d-\d+  )(\d-\d+  )(\d-\d+  )(\d-\d+  )',
 19                            'Mon: \1, Tue: \2,Wed: \3,Thu: \4,Fri: \5,Sat: \6,Sun: \7')
 20      END new_col
 21    FROM t_view
 22  );

NEW_COL
----------------------------------------------------------------------------------------------------
Mon: 6-20, Tue: 6-20,Wed: 6-20,Thu: 6-20,Fri: 6-20
Mon: 6-21, Tue: 6-21,Wed: 6-21,Thu: 6-21,Fri: 6-21,Sat: 6-21,Sun: 6-21
Mon: 6-9, Tue: 6-9,Wed: 6-9,Thu: 6-9,Fri: 6-9,Sat: 6-9,Sun: 6-9

SQL>
于 2015-09-07T08:58:28.617 回答
2

SQL小提琴

Oracle 11g R2 模式设置：

查询 1：

with t_view ( col ) as (
      select '6-21  6-21  6-21  6-21  6-21  6-21  6-21  ' from dual
union select '6-20  6-20  6-20  6-20  6-20  ' from dual
union select '6-9  6-9  6-9  6-9  6-9  6-9  6-9  6-9  6-9' from dual
union select '6-1' from dual
union select '6-1 6-2' from dual
),
days ( id, day ) AS (
            SELECT 1, 'Mon' FROM DUAL
  UNION ALL SELECT 2, 'Tue' FROM DUAL
  UNION ALL SELECT 3, 'Wed' FROM DUAL
  UNION ALL SELECT 4, 'Thu' FROM DUAL
  UNION ALL SELECT 5, 'Fri' FROM DUAL
  UNION ALL SELECT 6, 'Sat' FROM DUAL
  UNION ALL SELECT 0, 'Sun' FROM DUAL
),
matches ( col, idx, day ) AS (
  SELECT col,
         COLUMN_VALUE,
         day || ': ' || REGEXP_SUBSTR( t.col, '\d+-\d+', 1, COLUMN_VALUE )
  FROM   t_view t,
         TABLE(
           CAST(
             MULTISET(
               SELECT LEVEL
               FROM   DUAL
               CONNECT BY LEVEL <= REGEXP_COUNT( t.col, '\d+-\d+' )
             )
             AS SYS.ODCINUMBERLIST
           )
         ) l
         INNER JOIN days d
         ON ( MOD( l.COLUMN_VALUE, 7 ) = d.id )
)
SELECT LISTAGG( day, ', ' ) WITHIN GROUP ( ORDER BY IDX ) AS col
FROM   matches
GROUP BY col

结果：

|                                                                                      COL |
|------------------------------------------------------------------------------------------|
|                                                                                 Mon: 6-1 |
|                                                                       Mon: 6-1, Tue: 6-2 |
|                                    Mon: 6-20, Tue: 6-20, Wed: 6-20, Thu: 6-20, Fri: 6-20 |
|              Mon: 6-21, Tue: 6-21, Wed: 6-21, Thu: 6-21, Fri: 6-21, Sat: 6-21, Sun: 6-21 |
| Mon: 6-9, Tue: 6-9, Wed: 6-9, Thu: 6-9, Fri: 6-9, Sat: 6-9, Sun: 6-9, Mon: 6-9, Tue: 6-9 |
于 2015-09-07T09:02:13.043 回答
1

考虑Space作为分隔符，标记为行（使用级别）并使用重新加入LISTAGG()，使用级别作为日生成器（TO_CHAR(TRUNC(SYSDATE,'D')+level）

with t_view as
(
  select '6-21  6-21  6-21  6-21  6-21  6-21  6-21  ' as col from dual
  union all
  select '6-20  6-20  6-20  6-20  6-20  ' from dual
  union all
  select '6-9  6-9  6-9  6-9  6-9  6-9  6-9 6-9 ' from dual
)
SELECT LISTAGG(TO_CHAR(TRUNC(SYSDATE,'D')+level1,'Dy')||': '||
               REGEXP_SUBSTR(col,'[^ ]+',1,LEVEL1),', ')
        WITHIN GROUP (ORDER BY level1 )
from
(
SELECT  col,level level1
  FROM t_view
CONNECT BY REGEXP_SUBSTR(col,'[^ ]+',1,LEVEL) IS NOT NULL
   AND PRIOR col = col
   AND PRIOR sys_guid() IS NOT NULL
)
group by col;
于 2015-09-07T10:51:12.900 回答
1

为什么我们不能使用这种简单的方法？对我来说看起来不错

SELECT 'Mon: '||regexp_substr(col,'\d+\-\d+',1,1) ||
       ', Tue: '||regexp_substr(col,'\d+\-\d+',1,2) ||
       ', Wed: '||regexp_substr(col,'\d+\-\d+',1,3) ||
       ', Thu: '||regexp_substr(col,'\d+\-\d+',1,4) ||
       ', Fri: '||regexp_substr(col,'\d+\-\d+',1,5) ||  
       ', Sat: '||regexp_substr(col,'\d+\-\d+',1,6) ||
       ', Sun: '||regexp_substr(col,'\d+\-\d+',1,7)
 FROM t_view

显然，消除空的 Sat Sun 很容易，例如使用 nvl2：

SELECT 'Mon: '||regexp_substr(col,'\d+\-\d+',1,1) ||
       ', Tue: '||regexp_substr(col,'\d+\-\d+',1,2) ||
       ', Wed: '||regexp_substr(col,'\d+\-\d+',1,3) ||
       ', Thu: '||regexp_substr(col,'\d+\-\d+',1,4) ||
       ', Fri: '||regexp_substr(col,'\d+\-\d+',1,5) ||
       nvl2(regexp_substr(col,'\d+\-\d+',1,6), 
       ', Sat: '||regexp_substr(col,'\d+\-\d+',1,6) ||
       ', Sun: '||regexp_substr(col,'\d+\-\d+',1,7),null)
 FROM t_view

你应该记住，这只是一个例子，如果你能得到错过任意天数的数据，你应该在更多地方使用 nvl2

于 2015-09-07T08:54:35.493 回答
1

这是我的尝试。它与 MTo 的版本没有太大区别。想法是一样的：将字符串转换为行，添加有关日期的信息，然后重新组合记录。

with week as (
  select 1 day_num, 'Mon' day_name from dual union all
  select 2 day_num, 'Tue' day_name from dual union all
  select 3 day_num, 'Wed' day_name from dual union all
  select 4 day_num, 'Thu' day_name from dual union all
  select 5 day_num, 'Fri' day_name from dual union all
  select 6 day_num, 'Sat' day_name from dual union all
  select 7 day_num, 'Sun' day_name from dual
),
t_view as (select '6-21  6-21  6-21  6-21  6-21  6-21  6-21  ' as col from dual
                union all
                select '6-20  6-20  6-20  6-20  6-20  ' from dual
                union all
                select '6-9  6-9  6-9  6-9  6-9  6-9  6-9  ' from dual
                ),
lines as(
  select 
    col, WEEK.DAY_NAME, l, trim(regexp_substr(col, '[^,]+', 1, L)) elem
  from (
      select regexp_replace(col,'([[:digit:]]-[[:digit:]]{1,2}[[:space:]]+)','\1,')  col
      from t_view
      )
      join (select level l from dual connect by level < 10)
  on instr(col, ',', 1, L ) > 0
  join week on WEEK.DAY_NUM = l
  order by col,l
  )
select listagg(day_name||':'||elem,' ')  within group (order by l)
from lines
group by col;

结果：

Mon:6-20 Tue:6-20 Wed:6-20 Thu:6-20 Fri:6-20
Mon:6-21 Tue:6-21 Wed:6-21 Thu:6-21 Fri:6-21 Sat:6-21 Sun:6-21
Mon:6-9 Tue:6-9 Wed:6-9 Thu:6-9 Fri:6-9 Sat:6-9 Sun:6-9
于 2015-09-07T11:13:39.943 回答