如何在mysql中使用datediff函数计算罚款

Question

create function fine1(date1 date,date2 date) returns int 
-> begin 
-> declare fine int; 
-> if datediff(date1,date2)>10 then 
-> set fine=(datediff-10)*5; 
-> return fine; 
-> end; 
-> // 

我写了这段代码，但它给了我一个错误。

Answer 1

你似乎缺少END IF

create function fine1(date1 date,date2 date) returns int 
-> begin 
-> declare fine int; 
-> if datediff(date1,date2)>10 then 
-> set fine=(datediff(date1,date2)-10)*5; <-- was not correct. 
-> end if; <-- was missing 
-> return fine; 
-> end; 

而且，好像你正在使用的计算，你已经错过了传递参数datediff功能。

set fine=(datediff-10)*5; 

应该

set fine=(datediff(date1,date2)-10)*5;