我正在尝试创建一个 prolog 程序,它接收查询以作为字符串运行(通过 json),然后打印结果(成功或失败)。
:- use_module(library(http/json)).
happy(alice).
happy(albert).
with_albert(alice).
does_alice_dance :- happy(alice),with_albert(alice),
format('When alice is happy and with albert, she dances ~n').
with_alice(albert).
does_albert_dance :- happy(albert),with_alice(albert),
format('When albert is happy and with alice, he dances ~n').
fever(martin).
low_appetite(martin).
sick(X):-fever(X),low_appetite(X).
main(json(Request)) :-
nl,
write(Request),
nl,
member(facts=Facts, Request),
format('Facts : ~w ~n',[Facts]),
atomic_list_concat(Facts, ', ', Atom),
format('Atom : ~w ~n',[Atom]),
atom_to_term(Atom,Term,Bindings),
format('Term : ~w ~n',Term),
write(Bindings).
执行此查询后:
主要(json([事实=['病(马丁)','does_alice_dance','does_albert_dance']]))。
我有:
[facts=[sick(martin), does_alice_dance, does_albert_dance]]
Facts : [sick(martin),does_alice_dance,does_albert_dance]
Atom : sick(martin), does_alice_dance, does_albert_dance
Term : sick(martin),does_alice_dance,does_albert_dance
[]
true
我想做的是评估 Term。我尝试使用 is/2 和调用谓词使其工作,但它似乎不起作用。
使用
通话(期限)
(我在主体的尾部添加),我有这个错误:
Sandbox restriction!
Could not derive which predicate may be called from
call(C)
main(json([facts=['sick(martin)',does_alice_dance,does_albert_dance]]))
使用
结果是期限
(结果是我添加用于存储结果的变量),我有这个错误:
Arithmetic: `does_albert_dance/0' is not a function
请问有没有解决方案来评估序言中的字符串表达式?