从Oracle中的String中提取特定数据集

Question

I have the string ' 1_A_B_C_D_E_1_2_3_4_5 ' and I am trying to extract the data ' A_B_C_D_E '. I am trying to remove the _1_2_3_4_5 & the 1_ portion from the string. Which is essentially the numeric portion in the string. any special characters after the last alphabet must also be removed. In this example the _ after the character E must also not be present.

and the Query I am trying is as below

SELECT 
REGEXP_SUBSTR('1_A_B_C_D_E_1_2_3_4_5','[^0-9]+',1,1) 
from dual

The Data I get from the above query is as below: -

_A_B_C_D_E_

I am trying to figure a way to remove the underscore towards the end. Any other way to approach this?

Answer 1

Assuming the "letters" come first and then the "digits", you could do something like this:

select regexp_substr('A_B_C_D_E_1_2_3_4_5','.*[A-Z]') from dual;

This will pull all the characters from the beginning of the string, up to the last upper-case letter in the string ( .* is greedy, it will extend as far as possible while still allowing for one more upper-case letter to complete the match).

Answer 2

I have the string '1_A_B_C_D_E_1_2_3_4_5' and I am trying to extract the data 'A_B_C_D_E'

Use REGEXP_REPLACE :

SQL> SELECT trim(BOTH '_' FROM
  2         (REGEXP_SUBSTR('1_A_B_C_D_E_1_2_3_4_5','[0-9]+', ''))) str
  3  FROM dual;

STR
---------
A_B_C_D_E

How it works:

1. REGEXP_REPLACE will replace all numeric occurrences '[0-9]+' from the string. Alternatively, you could also use POSIX character class '[^[:digit:]]+'
2. TRIM BOTH '_' will remove any leading and lagging _ from the string.

Also using REGEXP_SUBSTR :

SELECT trim(BOTH '_' FROM 
       (REGEXP_SUBSTR('1_A_B_C_D_E_1_2_3_4_5','[^0-9]+'))) str 
FROM dual;
STR
---------
A_B_C_D_E