我想通过JPL查询在java中使用Prolog,我阅读了文档(http://www.swi-prolog.org/packages/jpl/java_api/getting_started.html) 序言谓词是:
child_of(joe, ralf).
child_of(mary, joe).
child_of(steve, joe).
child_of(steve, ralf).
descendent_of(X, Y) :-
child_of(X, Y).
descendent_of(X, Y) :-
child_of(Z, Y),
descendent_of(X, Z).
我的代码看起来像这样
Variable X = new Variable();
Query q4 =
new Query(
"descendent_of",
new Term[] {X,new Atom("joe")}
);
java.util.Hashtable solution;
while ( q4.hasMoreSolutions() ){
solution = q4.nextSolution();
System.out.println( "X = " + solution.get(X));
}
根据我的prolog谓词,我的java代码应该检索'mary'和'steve',但我明白了:
X = null
X = null
我做错了什么?提前谢谢
编辑:这是我的整个测试
Query q1 =
new Query(
"consult",
new Term[] {new Atom("C:\\Users\\cardozo\\Documents\\fer\\info2\\lore\\test.pl")}
);
return q1;
System.out.println( "consult " + (q.query() ? "succeeded" : "failed"));
Query q2 =
new Query(
"child_of",
new Term[] {new Atom("joe"),new Atom("X")}
);
Boolean resp= q2.query();
System.out.println("child_of(joe,X) is " + resp.toString()
);
Query q3 =
new Query(
"descendent_of",
new Term[] {new Atom("steve"),new Atom("ralf")}
);
System.out.println(
"descendent_of(joe,ralf) is " +
( q3.query() ? "provable" : "not provable" )
);
Variable X = new Variable();
Query q4 =
new Query(
"descendent_of",
new Term[] {X,new Atom("joe")}
);
java.util.Hashtable solution;
q4.query();
while ( q4.hasMoreSolutions() ){
solution = q4.nextSolution();
System.out.println( "X = " + solution.get("X"));
}
这就是我在java控制台中得到的结果
run:
% C:\Users\cardozo\Documents\fer\info2\lore\test.pl compiled 0.00 sec, 8 clauses
consult succeeded
child_of(joe,X) is false
descendent_of(joe,ralf) is provable
X = null
X = null
BUILD SUCCESSFUL (total time: 0 seconds)
答案 0 :(得分:3)
我找到了解决方案,我必须像这样使用类化合物(包含在jpl中)
Query q4 = new Query(new Compound("descendent_of", new Term[] { new Variable("X"), new Atom("joe")}));
while ( q4.hasMoreSolutions() ){
solution = q4.nextSolution();
System.out.println( "X = " + solution.get("X"));
}
我得到了解决方案
X = mary
X = steve
答案 1 :(得分:2)
我会尝试按名称变量:
solution.get("X")
修改强>
使用像
这样的文字查询查询q4 =新查询(“descendent_of(X,joe)”)