如何使 mysql 从 select 结果中得到 return 的函数？

Question

我想在 mySql 8 上创建这个函数。它会创建一个像 00001、00002

这样的序列号

CREATE FUNCTION dbOne.create_sequence_number(lastNumber CHAR(255), numberLength INT, lastValue CHAR(255) ) RETURNS char(255)
BEGIN
    DECLARE select_var CHAR(255);
    SET select_var = (SELECT 
        CASE WHEN lastNumber = lastValue 
        THEN
        LPAD( '1', numberLength, '0' ) 
        ELSE 
        LPAD(CAST(( CAST(COALESCE ( lastNumber, '0' ) AS INT) + 1 ) AS VARCHAR, numberLength, '0' ) INTO select_var);
    RETURN select_var;
END

我不知道这个查询有什么问题，但我总是遇到这个错误。

SQL Error [1064] [42000]: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INT) + 1 ) AS VARCHAR, numberLength, '0' ) INTO select_var);
    RETURN select_var' at line 9
  You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INT) + 1 ) AS VARCHAR, numberLength, '0' ) INTO select_var);
    RETURN select_var' at line 9
  You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INT) + 1 ) AS VARCHAR, numberLength, '0' ) INTO select_var);
    RETURN select_var' at line 9

我也试过这个查询。

CREATE FUNCTION erhav2_db.create_sequence_number(lastNumber CHAR(255), numberLength INT, lastValue CHAR(255) ) RETURNS char(255)
BEGIN
    DECLARE select_var CHAR(255);
    SELECT 
        (CASE WHEN lastNumber = lastValue 
        THEN
        lpad( '1', numberLength, '0' ) 
        ELSE 
        lpad(CAST(( CAST(COALESCE ( lastNumber, '0' ) AS INT) + 1 ) AS VARCHAR, numberLength, '0' ))) INTO select_var;
    RETURN select_var;
END

但仍然给我同样的错误。我的函数查询可能出了什么问题？

Answer 1

CREATE FUNCTION dbOne.create_sequence_number(
    lastNumber /* CHAR(255) */ UNSIGNED, 
    numberLength INT, 
    lastValue CHAR(255) 
) 
RETURNS CHAR(255)
RETURN LPAD(CASE WHEN lastNumber = lastValue
                 THEN 1
                 ELSE COALESCE(lastNumber, 0) + 1
                 END,
            numberLength, 
            '0');

多个数据类型转换过多 - MySQL 将根据操作上下文隐式更改数据类型。

所有操作都可以在单个语句中执行，这使得声明变量和 BEGIN-END（以及分隔符重新分配）都变得不必要。

代码需要 lastNumber 才能转换为数字数据类型。否则，您和我的代码都将在严格 SQL 模式下失败。所以我建议将 lastNumber CHAR(255) 输入参数数据类型的数据类型更改为 UNSIGNED / INT - 这将允许在函数调用阶段检测值不正确，而不是在函数代码中。

如何使 mysql 从 select 结果中得到 return 的函数？

How to make mysql function that return from the select result?

mysql

sql

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