Oracle正则表达式从最后一次出现时拆分字符串

Question

我仍然在oracle中学习regexp，我陷入了中间，下面是我的示例代码：

with t(val)
as
(
  --format: xyz_year_month_date
  select 'my_new_table_2015_06_31' from dual
  union all
  select 'my_new_table_temp_2016_06_31' from dual
 )
 select reverse(regexp_substr(reverse(val),'[^_]+',1,4)) col4,
 reverse(regexp_substr(reverse(val),'[^_]+',1,3)) col3,
 reverse(regexp_substr(reverse(val),'[^_]+',1,2)) col2,
 reverse(regexp_substr(reverse(val),'[^_]+',1,1)) col1
 from t;

Output:
COL4 COL3 COL2 COL1
table 2015 06 31
temp 2016 06 31

Expected output:
COL4 COL3 COL2 COL1
my_new_table 2015 06 31
my_new_table_temp 2016 06 31

提前致谢。

Answer 1

您可以通过提取不同的捕获组（用()圆括号括起来）来避免双重反转：

WITH t ( VAL ) AS (
  SELECT 'my_new_table_2015_06_31' FROM DUAL UNION ALL
  SELECT 'my_new_table_temp_2016_06_31' FROM DUAL
)
SELECT REGEXP_SUBSTR( val, '^(.*)_([^_]+)_([^_]+)_([^_]+)$', 1, 1, NULL, 1 ) AS COL4,
       REGEXP_SUBSTR( val, '^(.*)_([^_]+)_([^_]+)_([^_]+)$', 1, 1, NULL, 2 ) AS COL3,
       REGEXP_SUBSTR( val, '^(.*)_([^_]+)_([^_]+)_([^_]+)$', 1, 1, NULL, 3 ) AS COL2,
       REGEXP_SUBSTR( val, '^(.*)_([^_]+)_([^_]+)_([^_]+)$', 1, 1, NULL, 4 ) AS COL1
FROM   t

您甚至可以通过使用以下命令使正则表达式简单得多：

'^(.+)_(.+)_(.+)_(.+)$'

第一个.+是贪婪的，因此它将尽可能匹配，直到在第二个至第四个捕获组中只剩下足够的字符串用于最小匹配为止。

但是，您不需要正则表达式 ：

WITH t ( VAL ) AS (
  SELECT 'my_new_table_2015_06_31' FROM DUAL UNION ALL
  SELECT 'my_new_table_temp_2016_06_31' FROM DUAL
)
SELECT SUBSTR( val, 1,        pos1 - 1        ) AS col4,
       SUBSTR( val, pos1 + 1, pos2 - pos1 - 1 ) AS col3,
       SUBSTR( val, pos2 + 1, pos3 - pos2 - 1 ) AS col2,
       SUBSTR( val, pos3 + 1                  ) AS col1
FROM   (
  SELECT val,
         INSTR( val, '_', -1, 1 ) AS pos3,
         INSTR( val, '_', -1, 2 ) AS pos2,
         INSTR( val, '_', -1, 3 ) AS pos1
  FROM   t
);