How to Get Rightmost 10 Places of a String in Oracle

Finding last index of a string in Oracle

Use -1 as the start position:

INSTR('JD-EQ-0001', '-', -1)

How to read string from right PLSQL

substr('DV-2011-01-000004', length('DV-2011-01-000004')-6 + 1 )

Extract number from string with Oracle function

You'd use REGEXP_REPLACE in order to remove all non-digit characters from a string:

select regexp_replace(column_name, '[^0-9]', '')
from mytable;

or

select regexp_replace(column_name, '[^[:digit:]]', '')
from mytable;

Of course you can write a function extract_number. It seems a bit like overkill though, to write a funtion that consists of only one function call itself.

create function extract_number(in_number varchar2) return varchar2 is
begin
  return regexp_replace(in_number, '[^[:digit:]]', '');
end; 

Remove last character from string in sql plus

A closing parenthesis is in the wrong place. It should be:

SUBSTR(ooo.CO_NAME, 1, LENGTH(ooo.CO_NAME) - 1)

Java: Get last element after split

Save the array in a local variable and use the array's length field to find its length. Subtract one to account for it being 0-based:

String[] bits = one.split("-");
String lastOne = bits[bits.length-1];

Caveat emptor: if the original string is composed of only the separator, for example "-" or "---", bits.length will be 0 and this will throw an ArrayIndexOutOfBoundsException. Example: https://onlinegdb.com/r1M-TJkZ8

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