Question

我仍然是编程的新手，如果有人可以帮我解决这个问题，我会很感激，基本上我有一个我要解组的电影文件，只有“Robert Benton”的人/导演作为system.out的输出

使用JAXBU的Java类

        package jaxbadv;

    import java.io.File;
    import java.util.Iterator;
    import java.util.List;

    import org.me.media.*;
    /**
     *
     * @author Ket
     */
    public class rbFilms {

    public static void main(String[] args) {
        // Create root XML node 'todaysShow' and get its main element 'movies_today'
        ShowingToday todaysShow = new ShowingToday();
        List<MovieType> movies_today =  todaysShow.getMovieCollection();
        // Create Movie instanses and add them to the 'movies_today' collection
        MovieType film;


        film = new MovieType();
        film.getTitle();
        film.getDirector();
        film.getYear();

        try {
            javax.xml.bind.JAXBContext jaxbCtx = javax.xml.bind.JAXBContext.newInstance(film.getClass().getPackage().getName());
            javax.xml.bind.Unmarshaller unmarshaller = jaxbCtx.createUnmarshaller();
            film = (MovieType) unmarshaller.unmarshal(new java.io.File("Now_Showing.txt")); //NOI18N

            //print out only movies produced after 1990
            MovieType nextMovie = new MovieType();
            Iterator itr = movies_today.iterator();
            while(itr.hasNext()) {
                nextMovie = (MovieType) itr.next();
                if(nextMovie.getDirector() == "Robert Benton") {
                    System.out.println(nextMovie.getTitle());
                }
            }


        } catch (javax.xml.bind.JAXBException ex) {
            // XXXTODO Handle exception
            java.util.logging.Logger.getLogger("global").log(java.util.logging.Level.SEVERE, null, ex); //NOI18N
        }




    }
}

XML文件

    <?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<Showing_Today xmlns="http://xml.netbeans.org/schema/Shows">
    <movie_collection>
        <Title>Red</Title>
        <Director>Robert Schwentke</Director>
        <Year>2010</Year>
    </movie_collection>
    <movie_collection>
        <Title>Kramer vs Kramer</Title>
        <Director>Robert Benton</Director>
        <Year>1979</Year>
    </movie_collection>
    <movie_collection>
        <Title>La Femme Nikita</Title>
        <Director>Luc Besson</Director>
        <Year>1997</Year>
    </movie_collection>
    <movie_collection>
        <Title>Feast of love</Title>
        <Director>Robert Benton</Director>
        <Year>2007</Year>
    </movie_collection>
</Showing_Today>

JAXB绑定生成的源 - ShowingToday

    package org.me.media;

import java.util.ArrayList;
import java.util.List;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlType;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "", propOrder = {
    "movieCollection"
})
@XmlRootElement(name = "Showing_Today")
public class ShowingToday {

    @XmlElement(name = "movie_collection")
    protected List<MovieType> movieCollection;

public List<MovieType> getMovieCollection() {
        if (movieCollection == null) {
            movieCollection = new ArrayList<MovieType>();
        }
        return this.movieCollection;
    }

}

Answer 1

您可能希望从一些更简单的练习开始，您在代码中存在一些基本的Java错误。例如，您有一些无效分配：

showingToday todaysShow = new ShowingToday(); // value isn't used
List<MovieType> movies_today =  todaysShow.getMovieCollection(); // value isn't used

有些地方正在初始化变量并对其进行无效get调用：

film = new MovieType(); // values is never used
film.getTitle();        // this and the other get calls are not needed
film.getDirector();
film.getYear();

您需要解决这些问题。

就您的特定JAXB问题而言，根据我的判断，您应该从XML流中反序列化ShowingToday实例，然后从中访问信息。代码与此类似：

try {
   final JAXBContext context = JAXBContext
         .newInstance(ShowingToday.class);
   final Unmarshaller unmarshaller = context.createUnmarshaller();
   final ShowingToday showingToday = unmarshaller.unmarshal(
         new StreamSource(new File("absolute path of file here")),
         ShowingToday.class).getValue();

} catch (final Exception e) {
   // Do something useful here
}

Java使用JAXB和选定的输出解组XML文档文件

1 个答案: