lectures(monday, nolecture).
lectures(tuesday, vp).
lectures(tuesday, se).
lectures(tuesday, ddbms).
lectures(wednessday, ds).
lectures(wednessday, mpl).
lectures(thursday, vp).
lectures(thrusday, se).
lectures(friday, ds).
lectures(friday, mpl).
lectures(saturday, ai).
lectures(saturday, ddbms).
?- lectures(friday, X), write(X),nl.
问询仅作为ds的第一记录印发,我希望在特定日即星期五印发所有记录,结果应为ds和mpl。
各位
在问询结束时添加<代码>fail.:
| ?- lectures(friday, X), write(X), nl, fail.
ds
mpl
no
| ?-
另见。 这一非常相似的问题试图在提出新问题之前再用一点点。
b 过去,如果你想让它更加可行,你可以总结如下:
display_lectures(Day) :-
lectures(Day, X),
write(X),
nl,
fail
;
true.
页: 1 这里的一条是真正的禁令。 这使这一呼吁能够在更大程度上得到响应,因此我认为我已提到这一点。
And alternatively, here is another way to do it (if your implementation has some sort of forall/2) :
display_lectures(Day) :- forall(lectures(Day, X), (write(X), nl)).
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