Oracle 9i中字符之间的子字符串 [英] Substring between characters in Oracle 9i

问题描述

在更高版本中，我可以使用此regexp_substr:

In later versions I can use this regexp_substr:

SELECT 
    ID, 
    regexp_substr(ID, '[^.]+', 1, 2) DATA 1, 
    regexp_substr(ID, '[^.]+', 1, 3) DATA 2 
FROM employees

表格:员工

ID
--------------------------
2017.1.3001-ABC.01.01
2017.2.3002-ABCD.02.02
2017.303.3003-ABC.03.03
2017.404.3004-ABCD.04.04

预期输出:

ID                       DATA 1 DATA 2
------------------------ ------ ---------
2017.1.3001-ABC.01.01    1      3001-ABC
2017.2.3002-ABCD.02.02   2      3002-ABCD
2017.303.3003-ABC.03.03  303    3003-ABC
2017.404.3004-ABCD.04.04 404    3004-ABCD

请帮助我获取 SQL Oracle 9i 中ID列中.个字符之间的子字符串.

Please help me to get the sub-string between . characters in ID column in SQL Oracle 9i.

推荐答案

您不需要正则表达式-只需使用SUBSTR和INSTR:

You don't need regular expressions - just use SUBSTR and INSTR:

SELECT id,
       SUBSTR( id, dot1 + 1, dot2 - dot1 - 1) AS data1,
       SUBSTR( id, dot2 + 1, dot3 - dot2 - 1) AS data2
FROM   (
  SELECT id,
         INSTR( id, '.', 1, 1 ) AS dot1,
         INSTR( id, '.', 1, 2 ) AS dot2,
         INSTR( id, '.', 1, 3 ) AS dot3
  FROM   employees
);

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