Jess模式匹配＆＃39;不是＆＃39;规则的条件

Question

我在jess上玩。我想制定不使用＆＃39;〜＆＃39;。

的规则

这是我的代码。

(clear)
(reset)

(deftemplate person
    (slot name)
    (slot eyes)
    (slot hair))

(defrule person-without-brown-hair
    (person (name ?name) (hair ?hair&:(not brown)))
=>
    (printout t ?name " does not have brown hair." crlf))

(assert (person (name "Jeon") (eyes green) (hair brown)))
(assert (person (name "Park") (eyes blue) (hair red)))

(run)

我的预期是多变的？头发不是棕色的。

Answer 1

请注意，函数(not ...)否定了它的参数。 TRUE为true，FALSE为false，任何其他符号为true。

要测试广告位值是否等于某个值，请使用(eq ...)或(neq ...)或(= ...)或(<> ...)。

(defrule person-without-brown-hair
    (person (name ?name) (hair ?hair&:(neq ?hair brown)))
=>
    (printout t ?name " does not have brown hair." crlf))

结果：

Park does not have brown hair.
1

直接匹配当然更可取。