小编Luu*_*y19的帖子

用PL/SQL中的字符串替换每个字母的ASCII码

是一种更好,更短的方式来执行此代码:

/*Replace all letters by their respective ASCII code - 55*/
as_iban := REPLACE(as_iban, 'A', '10');
as_iban := REPLACE(as_iban, 'B', '11');
as_iban := REPLACE(as_iban, 'C', '12');
as_iban := REPLACE(as_iban, 'D', '13');
as_iban := REPLACE(as_iban, 'E', '14');
as_iban := REPLACE(as_iban, 'F', '15');
as_iban := REPLACE(as_iban, 'G', '16');
as_iban := REPLACE(as_iban, 'H', '17');
as_iban := REPLACE(as_iban, 'I', '18');
as_iban := REPLACE(as_iban, 'J', '19');
as_iban := REPLACE(as_iban, 'K', '20');
as_iban := REPLACE(as_iban, 'L', '21');
as_iban := REPLACE(as_iban, 'M', '22');
as_iban := REPLACE(as_iban, 'N', …
Run Code Online (Sandbox Code Playgroud)

sql ascii plsql

2
推荐指数
1
解决办法
8927
查看次数

标签 统计

ascii ×1

plsql ×1

sql ×1