用oracle定义一个函数,怎么定义?
答案:1 悬赏:40
解决时间 2021-01-18 18:04
- 提问者网友:不爱我么
- 2021-01-17 18:07
用oracle定义一个函数,怎么定义?
最佳答案
- 二级知识专家网友:詩光轨車
- 2021-01-17 18:49
create or replace function test(p1 in number, p2 in varchar2)
return varchar2 is
v_out varchar2(200);
begin
select flag_name
into v_out
from PU_META_PLAT.MD_META_DIM_CODE
where dim_table_id = p1
and flag_code in
(SELECt trim(substr(',' || p2 || ',',
instr(',' || p2 || ',', ',', 1, LEVEL) + 1,
instr(',' || p2 || ',', ',', 1, LEVEL + 1) -
instr(',' || p2 || ',', ',', 1, LEVEL) - 1))
FROM dual
CONNECT BY LEVEL <=
(length(',' || p2 || ',') -
length(replace(',' || p2 || ',', ',', ''))) / 1 - 1));
return v_out;
exception
when others then
return null;
end;
return varchar2 is
v_out varchar2(200);
begin
select flag_name
into v_out
from PU_META_PLAT.MD_META_DIM_CODE
where dim_table_id = p1
and flag_code in
(SELECt trim(substr(',' || p2 || ',',
instr(',' || p2 || ',', ',', 1, LEVEL) + 1,
instr(',' || p2 || ',', ',', 1, LEVEL + 1) -
instr(',' || p2 || ',', ',', 1, LEVEL) - 1))
FROM dual
CONNECT BY LEVEL <=
(length(',' || p2 || ',') -
length(replace(',' || p2 || ',', ',', ''))) / 1 - 1));
return v_out;
exception
when others then
return null;
end;
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