我试图创建一个查询 exists(X)
如果X存在于prolog数据库中,返回true。
Prolog数据库
store(best_smoothies, [alan,john,mary],
[ smoothie(berry, [orange, blueberry, strawberry], 2),
smoothie(tropical, [orange, banana, mango, guava], 3),
smoothie(blue, [banana, blueberry], 3) ]).
store(all_smoothies, [keith,mary],
[ smoothie(pinacolada, [orange, pineapple, coconut], 2),
smoothie(green, [orange, banana, kiwi], 5),
smoothie(purple, [orange, blueberry, strawberry], 2),
smoothie(smooth, [orange, banana, mango],1) ]).
store(smoothies_galore, [heath,john,michelle],
[ smoothie(combo1, [strawberry, orange, banana], 2),
smoothie(combo2, [banana, orange], 5),
smoothie(combo3, [orange, peach, banana], 2),
smoothie(combo4, [guava, mango, papaya, orange],1),
smoothie(combo5, [grapefruit, banana, pear],1) ]).
我的尝试
exists(X) :- store(_,_,S), isIn(X,S).
isIn(X, [smoothie(X,_,_)|[]]).
isIn(X, [smoothie(N,_,_)|T]) :- isIn(X,T).
令人惊讶的是,它只对 blue
, smooth
和 combo5
. 是不是我的递归有问题?(我认为没有,因为它能够读出 blue
, smooth
一直到 combo5
)
我得到的输出
?- exists(combo1).
false
?- exists(X).
X = blue;
X = smooth;
X = combo5;
false
感谢任何帮助。非常感谢!
编辑:修改后的尝试
exists(X) :- store(_,_,S), isIn(X,S).
isIn(X, [smoothie(X,_,_)]).
isIn(X, [smoothie(N,_,_)|T]) :- isIn(X,T).
EDIT: 修正
exists(X) :- store(_,_,S), isIn(X,S).
isIn(X, [smoothie(X,_,_)|_]).
isIn(X, [smoothie(N,_,_)|T]) :- isIn(X,T).
[smoothie(X,_,_) | []]
与 [smoothie(X,_,_)]
.
isIn( X, S )
相当于
isIn( X, S ) :-
last( S, smoothie(X,_,_) ).
这解释了你的观察。