如何在oracle中获取特殊字符之间的数字

Question

你能帮助在 oracle 中获得以下输出吗

File_20170902_Name.txt
File200_Name_20170902_1.txt
File400_20170902_Name_1.txt
File1_name_20170902.txt

我期待像

这样的输出

20170902
20170902
20170902
20170902

提前致谢！！

Answer 1

假设文件名中的所有 8 位数字都是日期，并且前面是 _，后面是 _ 或 .txt，您可以使用

select REGEXP_SUBSTR(files,'_(\d{8})(_|.txt)',1,1,NULL,1) as dt FROM t;

Demo

但有些时间数字字符会变成 6 (YYMMDD) 字符

这将匹配_或_和.txt之间的6到8个字符的序列

REGEXP_SUBSTR(files,'_((\d{6,8}))(_|)',1,1,NULL,1)

Demo2

但是，这不是一个全面的正则表达式来匹配日期。即它可以匹配像20189999或180000这样的数字，如果它们是文件名的一部分。如果您只想检测日期，则可能必须将它们与此处指定的方法结合使用：Valid Date Checks in Oracle

Answer 2

试试这个演示：http://sqlfiddle.com/#!4/c361c/4

select
 regexp_replace(str,'.*(\d{8}).*','\1') as x
from table1

|        X |
|----------|
| 20170902 |
| 20170902 |
| 20170902 |
| 20170902 |

Answer 3

试试这个

create table tst (x varchar2(30));
insert into tst values ('File_20170902_Name.txt');
insert into tst values ('File200_Name_20170902_1.txt');
insert into tst values ('File400_20170902_Name_1.txt');
insert into tst values ('File1_name_20170902.txt');

Select x,       
      case 
          when substr(x, REGEXP_INSTR(x, '[^a-z|A-Z|0-9]') + 1, 1) between chr(48) and chr(57) Then 
            substr(x, REGEXP_INSTR(x, '[^a-z|A-Z|0-9]') + 1, REGEXP_INSTR(x, '[^a-z|A-Z|0-9]', 1, 2) - REGEXP_INSTR(x, '[^a-z|A-Z|0-9]') - 1) 
          else substr(x, REGEXP_INSTR(x, '[^a-z|A-Z|0-9]', 1, 2) + 1, REGEXP_INSTR(x, '[^a-z|A-Z|0-9]', 1, 3) - REGEXP_INSTR(x, '[^a-z|A-Z|0-9]', 1, 2) - 1) 
      end as t      
From tst  

注意：如果数字以 File_name_test_20170902.txt 等 2 个以上特殊字符开头，则此查询不支持

Answer 4

以下两者的工作方式大致相同 - 使用正则表达式进行基本选择，然后使用第二次调用 REGEXP_SUBSTR 或使用“普通”字符串函数来摆脱分隔符。我更喜欢对REGEXP_SUBSTR 的嵌套调用，而不是使用TRANSLATE 或REPLACE，但你可以选择：

对REGEXP_SUBSTR的嵌套调用：

WITH cteFiles AS (SELECT 'File_20170902_Name.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File200_Name_20170902_1.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File400_20170902_Name_1.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File1_name_20170902.txt' AS FILENAME FROM DUAL)
SELECT FILENAME, REGEXP_SUBSTR(REGEXP_SUBSTR(FILENAME, '_[0-9]+[_.]'), '[^_.]+') AS FINAL_NUMS
  FROM cteFiles;

使用REPLACE：

WITH cteFiles AS (SELECT 'File_20170902_Name.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File200_Name_20170902_1.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File400_20170902_Name_1.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File1_name_20170902.txt' AS FILENAME FROM DUAL)
SELECT FILENAME, REPLACE(REPLACE(REGEXP_SUBSTR(FILENAME, '_[0-9]+[_.]'), '_', NULL), '.', NULL) AS FINAL_NUMS
  FROM cteFiles;

使用TRANSLATE：

WITH cteFiles AS (SELECT 'File_20170902_Name.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File200_Name_20170902_1.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File400_20170902_Name_1.txt' AS FILENAME FROM DUAL UNION ALL
                  SELECT 'File1_name_20170902.txt' AS FILENAME FROM DUAL)
SELECT FILENAME, TRANSLATE(REGEXP_SUBSTR(FILENAME, '_[0-9]+[_.]'), '0123456789_.', '0123456789') AS FINAL_NUMS
  FROM cteFiles;

SQLFiddle here

祝你好运。