Find the LINE FEED and replace it.

Question

Hi
I am using ORACLE 11g R2. I am having some text data in a variable as below :-
Variable = V_TEXT
V_TEXT :=
'/u01/oracle/data/file1.txt
/u01/oracle/data/file2.txt
/u01/oracle/data/file3.txt'I want to place an '|' (pipe symbol) at the end of each line. So the expected output is as below :-
'/u01/oracle/data/file1.txt|
/u01/oracle/data/file2.txt|
/u01/oracle/data/file3.txt|' I tried with replace() function and replaced return character i.e. CHR(10) with '|' but it gives me an wrong output and also at the end of last line we dont have an return character.
after searching a lot on internet I found that i can do this with the help of regular expression.
As far as i have studied : -
The line feed character is the last character in the line or we can say it is end of line. This character is noted in oracle as 'x0a'
I want to replace this character in every line and place an '|' symbol.
can anyone guide me how to do it ??
Thanks in advance.

Answer 1

Aha... Now that is a different expected output as in your original post ;-)
One way of doing it could be something like this:
declare
  v_text varchar2(32767) := '/u01/oracle/data/file1.txt
/u01/oracle/data/file2.txt
/u01/oracle/data/file3.txt';
begin
  v_text := regexp_replace(
               regexp_replace(v_text,chr(10)||'\s*'||chr(10),chr(10))
             ,'$','|',1,0,'m'
  dbms_output.put_line(v_text);
end;
/Inner regexp replaces "linefeed followed by zero or more whitespace followed by linefeed" with a single linefeed. So this removes the blank lines.
Then the outer regexp takes care of the | characters.
(It might be possible to wrap it up in a single regexp, though I can't think of a way just off hand ;-) )