该项目是关于将半自然语言转换为SQL表。代码:
label(S) --> label_h(C), {atom_codes(A, C), string_to_atom(S, A)}, !.
label_h([C|D]) --> letter(C), letters_or_digits(D), !.
letters_or_digits([C|D]) --> letter_or_digit(C), letters_or_digits(D), !.
letters_or_digits([C]) --> letter_or_digit(C), !.
letters_or_digits([]) --> "", !.
letter(C) --> [C], {"a"=<C, C=<"z"}, !.
letter(C) --> [C], {"A"=<C, C=<"Z"}, !.
letter_or_digit(C) --> [C], {"a"=<C, C=<"z"}, !.
letter_or_digit(C) --> [C], {"A"=<C, C=<"Z"}, !.
letter_or_digit(C) --> [C], {"0"=<C, C=<"9"}, !.
table("student").
sbvr2sql --> label(Name), " is an integer.", {assert(fields(Name, "INT"))}.
sbvr2sql --> label(Name), " is a string.", {assert(fields(Name, "VARCHAR(64)"))}.
sbvr2sql(Table, Property) --> label(Table), " has ", label(Property), ".".
这是如何正常工作的:
?- sbvr2sql("age is an integer.", []).
true
?- sbvr2sql("firstName is a string.", []).
true.
?- sbvr2sql(T, P, "student has firstName.", []).
T = "student",
P = "firstName".
?- fields(F, T).
F = "age",
T = [73, 78, 84] n
F = "firstName",
T = [86, 65, 82, 67, 72, 65, 82, 40, 54|...].
?- sbvr2sql(T, P, "student has firstName.", []), fields(P, _).
T = "student",
P = "firstName".
但这在这里不起作用:
?- table(T).
T = [115, 116, 117, 100, 101, 110, 116]. % "student"
?- sbvr2sql(T, P, "student has firstName.", []), table(T).
false.
显然，它无法识别
table("student")为true。如上所述，它会将“学生”识别为标签。是什么赋予了？ 最佳答案
我无法重现该错误，但我怀疑它可能在您的label/3规则中。当我使用此规则的以下定义时:
label([C|S]) -->
[C], {[Sp|_] = " ", C \= Sp, [Dot|_] = ".", C \= Dot}, !,
label(S).
label([],X,X).
我得到正确的结果:
?- sbvr2sql(TS, PS, "student has firstName.", []), table(TS),
atom_codes(P,PS), atom_codes(T,TS).
TS = [115, 116, 117, 100, 101, 110, 116],
PS = [102, 105, 114, 115, 116, 78, 97, 109, 101],
P = firstName,
T = student.
通常，我建议在进行DCG操作之前将字符串标记化为原子列表。这样，由于Prolog的字符串输出笨拙，调试起来要容易得多。