sqlfiddle - 如何在 sql.js 中编写 datediff() - “sql.js 应该具有相同的输出”？

Question

我有一个使用 sql.js 的在线代码评估表我有这个在 sqlfiddle com 中运行良好的 mysql 代码：

select  
buy.buy_id,
datediff(date_step_end ,date_step_beg)   as Количество_дней,

if(days_delivery < datediff(date_step_end ,date_step_beg),
datediff(date_step_end ,date_step_beg)-days_delivery,0)    as Опоздание

from city inner join client on city.city_id = client.city_id 
inner join buy on buy.client_id = client.client_id 
inner join buy_step on buy_step.buy_id = buy.buy_id 
inner join step on step.step_id = buy_step.step_id 
where buy_step.step_id = 3 and date_step_end is not null

它无法正确评估上面的代码：

sql.js should have the same output

SyntaxError: Parse error on line 5:
...p_end ,date_step_beg)  

     
from 
-----------------------^
Expecting 'COMMA', 'IN', 'LIKE', 'ARROW', 'DOT', 
 'CARET', 'EQ', 'SLASH', 'TILDA', 'GLOB', 'NOT_LIKE', 
 'BARBAR', 'MINUS',  'DOUBLECOLON', got 'RPAR'

got 'RPAR' 表示右括号。我认为这是无法识别的 datediff() 函数。

如何使它在 sql.js 中工作？