Print every word in the new line for a space in a string

Hi all

I would print every word in the new line, if there is a space in the string.

Is there a better way using regexp?

declare

number of v_count;

v_text varchar2 (1000): = 'Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - Production | " ';

v_single_text varchar2 (1000);

number of v_space_pos;

Start

Select regexp_count (v_text,' ') in double v_count;

While v_count > 0

loop

v_space_pos: = instr (v_text,' ');

v_single_text: = substr(v_text,1,v_space_pos-1);

dbms_output.put_line (v_single_text);

v_text: = substr (v_text, v_space_pos + 1);

v_count: = v_count-1;

end loop;

end;

Thank you

Rambeau

HI - try as

with t as (select 'Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - Production' |) "" double str)

Select regexp_substr (str,'[^] +', 1, level)

t

connect by level<=regexp_count(str,'[^>

Tags: Database

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