SQL中计算字符串总ASCII值的最优方法是什么？

嘿，我完全懂你手动逐个字符写ASCII('X')累加的痛苦——这活儿不仅繁琐还容易出错！下面针对不同主流SQL数据库，给你提供不用手动拆分字符的简便解法：

Oracle 解法（适配你的DUAL表场景） #

利用CONNECT BY生成字符位置序列，自动遍历每个字符计算ASCII值并求和：

SELECT SUM(ASCII(SUBSTR('DESIGNATION OF EMPLOYEE', LEVEL, 1))) AS "TOTAL ASCII"
FROM DUAL
CONNECT BY LEVEL <= LENGTH('DESIGNATION OF EMPLOYEE');

原理：LEVEL会生成从1到字符串总长度的连续数字，SUBSTR按位置提取单个字符，最后用SUM累加所有字符的ASCII值。

SQL Server 解法 #

用递归CTE生成字符位置，再逐个计算求和：

WITH CharPositions AS (
    SELECT 1 AS Pos, 'DESIGNATION OF EMPLOYEE' AS Str
    UNION ALL
    SELECT Pos + 1, Str
    FROM CharPositions
    WHERE Pos < LEN(Str)
)
SELECT SUM(ASCII(SUBSTRING(Str, Pos, 1))) AS "TOTAL ASCII"
FROM CharPositions;

MySQL 8.0+ 解法 #

同样用递归CTE实现自动遍历：

WITH RECURSIVE CharPositions AS (
    SELECT 1 AS Pos, 'DESIGNATION OF EMPLOYEE' AS Str
    UNION ALL
    SELECT Pos + 1, Str
    FROM CharPositions
    WHERE Pos < LENGTH(Str)
)
SELECT SUM(ASCII(SUBSTRING(Str, Pos, 1))) AS `TOTAL ASCII`
FROM CharPositions;

PostgreSQL 解法 #

利用字符串拆分函数简化操作：

SELECT SUM(ASCII(ch)) AS "TOTAL ASCII"
FROM UNNEST(STRING_TO_ARRAY('DESIGNATION OF EMPLOYEE', NULL)) AS ch;

这些方法的好处是通用适配任意长度的字符串，不用因为字符串内容变化而修改代码，彻底摆脱手动拆分的麻烦～

内容的提问来源于stack exchange，提问作者AMAN KAG