Question

所以我有一个电影评级列表，如果新用户对电影评分，它会被添加到列表中。但是，如果现有用户已对其进行评级，则再次对其进行评级将导致旧评级更改为新评级。到目前为止，这是我的代码：

rateFilm :: String -> String -> Int -> [Film] -> [Rating]
rateFilm _ _ _ []                   = []
rateFilm user title number (x:xs)   = if title == name then do
                                        if user == existingUser then do
                                            removeItem (user,number) ratings
                                            ratings ++ [(user,number)]
                                        else do 
                                            ratings ++ [(user,number)]
                                else  do
                                    rateFilm user title number xs
                                where   (name, director, year, ratings) = x
                                        (existingUser,score)            = getUser user ratings

removeItem :: Rating -> [Rating] -> [Rating]
removeItem _ [] = []
removeItem x (y:ys) | x == y    = removeItem x ys
                | otherwise = y : removeItem x ys

这部电影的信息是：

[("Blade Runner","Ridley Scott",1982,[("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4)]),
 ("The Fly","David Cronenberg",1986,[("Megan",4),("Fred",7),("Chris",5),("Ian",0),("Amy",5)]),
 ("Psycho","Alfred Hitchcock",1960,[("Bill",4),("Jo",4),("Garry",8),("Kevin",7),("Olga",8),("Liz",10),("Ian",9)]),
 ("Body Of Lies","Ridley Scott",2008,[("Sam",3),("Neal",7),("Kevin",2),("Chris",5),("Olga",6)])]

当新用户添加评级时，它很好，输出将如下：

rateFilm "Alex" "Blade Runner" 8 testDatabase1
[("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Alex",8)]

但是，当现有用户执行此操作时，会发生以下情况：

rateFilm "Amy" "Blade Runner" 8 testDatabase1
[("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Amy",8),
("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Amy",8),
("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Amy",8),
("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Amy",8),
("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Amy",8),
("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Amy",8),
("Amy",6),("Bill",9),("Ian",7),("Kevin",9),("Emma",4),("Sam",5),("Megan",4),("Amy",8)]

关于为什么和/或如何修复它的任何想法？非常感谢。

Answer 1

你真的，真的不想在列表monad上运行，除非你知道那里发生了什么。列表monad用于非确定性计算。请参阅wikibook section或applicative section of the typeclassopedia。
您想从评分中删除(existingUser, score)而不是(user, number)。

总而言之，我们获得了rateFilm的以下内容：

rateFilm user title number (x:xs)   =
    if title == name then
        if user == existingUser then
            removeItem (user,score) ratings ++ [(user,number)]
        else
            ratings ++ [(user,number)]
    else
        rateFilm user title number xs
    where   (name, director, year, ratings)  = x
            (existingUser, score)            = getUser user ratings

从List中删除元素并将其替换为新值

1 个答案: