而不是使用SQL的,我只想用PL/SQL,因爲似乎沒有必要引入不必要的上下文切換:
declare
v_a varchar2(10);
v_b varchar2(10);
v_c varchar2(10);
v_string varchar2(33);
procedure split_string (p_string in varchar2,
p_a out varchar2,
p_b out varchar2,
p_c out varchar2)
is
begin
if regexp_count(p_string, '/') = 2 then
p_a := regexp_substr(p_string, '[^/]+', 1, 1);
p_b := regexp_substr(p_string, '[^/]+', 1, 2);
p_c := regexp_substr(p_string, '[^/]+', 1, 3);
elsif regexp_count(p_string, '/') = 1 then
p_a := regexp_substr(p_string, '[^/]+', 1, 1);
p_c := regexp_substr(p_string, '[^/]+', 1, 2);
end if;
end;
begin
v_string := 'x/y/z';
split_string(v_string, v_a, v_b, v_c);
dbms_output.put_line('v_string = "'||v_string||'", v_a = "'||v_a||'", v_b = "'||v_b||'", v_c = "'||v_c||'"');
v_string := 'x/y';
split_string(v_string, v_a, v_b, v_c);
dbms_output.put_line('v_string = "'||v_string||'", v_a = "'||v_a||'", v_b = "'||v_b||'", v_c = "'||v_c||'"');
v_string := 'x/y/z/1';
split_string(v_string, v_a, v_b, v_c);
dbms_output.put_line('v_string = "'||v_string||'", v_a = "'||v_a||'", v_b = "'||v_b||'", v_c = "'||v_c||'"');
v_string := 'x';
split_string(v_string, v_a, v_b, v_c);
dbms_output.put_line('v_string = "'||v_string||'", v_a = "'||v_a||'", v_b = "'||v_b||'", v_c = "'||v_c||'"');
end;
/
v_string = "x/y/z", v_a = "x", v_b = "y", v_c = "z"
v_string = "x/y", v_a = "x", v_b = "", v_c = "y"
v_string = "x/y/z/1", v_a = "", v_b = "", v_c = ""
v_string = "x", v_a = "", v_b = "", v_c = ""
如果你絕對必須使用SQL,沒有必要通過使用連接 - 你可以分開的結果分爲3列匹配3個變量你想輸入結果到:
with strings as (select 'x/y/z' str from dual union all
select 'x/y' str from dual union all
select 'x/y/z/1' str from dual union all
select 'x' str from dual)
select str,
case when regexp_count(str, '/') in (1, 2) then regexp_substr(str, '[^/]+', 1, 1) end v_a,
case when regexp_count(str, '/') = 2 then regexp_substr(str, '[^/]+', 1, 2) end v_b,
case when regexp_count(str, '/') = 2 then regexp_substr(str, '[^/]+', 1, 3)
when regexp_count(str, '/') = 1 then regexp_substr(str, '[^/]+', 1, 2)
end v_c
from strings;
STR V_A V_B V_C
------- --------------------- --------------------- ---------------------
x/y/z x y z
x/y x y
x/y/z/1
x
的問題是不太清楚。爲了訪問查詢的元素,你應該讀取光標,它可以通過幾種方法完成。例如,您可以在'FOR'循環中迭代元素並將值分配給變量。 –