返回 bigint 时，函数抱怨 [不允许从函数返回结果集]

Question

我忘记了放置分隔符指令，一些分号，并且使用了 tsl 语法，例如 [select variable = field]，这在 mysql 中无效。
使用 tsl 语法时的 Mysql 错误是 [不允许从函数返回结果集] 并且没有多大帮助。
@AndreKR 把这一切都指向我，谢谢。
我使用 mysqlworkbench 5.2.30 CE。 功函数变为：

delimiter //
CREATE FUNCTION nextval (seq_name varchar(100))  
  RETURNS bigint(20)  
    READS SQL DATA  
  NOT DETERMINISTIC  
    BEGIN  
     DECLARE workval bigint(20);  
     SELECT count(1) into workval  
        FROM tip_sequence  
        WHERE sequencename = seq_name;  
     IF workval <> 1 THEN  
        DELETE  
            FROM tip_sequence  
            WHERE sequencename = seq_name;  
        INSERT  
            INTO tip_sequence (sequencename, sequenceval, sequencestep)  
            VALUES (seq_name, 1, 1);  
     END IF;
     SELECT sequenceval into workval  
        FROM tip_sequence  
        WHERE sequencename = seq_name;  
     UPDATE tip_sequence  
        SET sequenceval = sequenceval + sequencestep  
        WHERE sequencename = seq_name;  
     RETURN workval;
    END//
delimiter ;

Answer 1

我认为SELECT workval = count(1) FROM ... 语法无效。我想你的意思是：SELECT count(1) INTO workval FROM ...。