3

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）prolog谓词是这些：

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));
        }

根据我的序言谓词，我的 java 代码应该检索“mary”和“steve”，但我得到了这个：

X = null
X = null

我做错了什么？提前致谢

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"));
}

这就是我在我的 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)
4

2 回答 2

3

我找到了解决方案，我必须像这样使用 Compound 类（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
于 2012-07-15T21:33:13.743 回答
2

我会尝试按名称获取变量：

solution.get("X")

编辑

使用文字查询，例如

查询 q4 = new Query("descendent_of(X,joe)")

于 2012-07-15T20:00:19.453 回答