java - 返回空指针异常但预期有 MovieListException？

Question

这个问题已经有答案了:

What is a NullPointerException, and how do I fix it? (12 个回答)

已关闭 5 年前。

所以我正在运行 JUnit 测试，它获取电影的评级(Hashmap: HashMap<String, Rating> movieList) )但是如果没有对电影的评分抛出 MovieListException。

JUNIT 预期与实际 /image/DnFE1.jpg

了解抛出异常，这样我就不知道自己做错了什么？

• 在评级类中，我检查对象值是否为 null，然后抛出 MovieListException
• 在 MovieList类此行出现 null 错误:Integer value = movieList.get(string).getRating();即使我正在捕获并返回 e 对象(我认为它需要返回？基于 JUNIT TEST？)

ISSUE: I am getting a null pointer instead of my MovieListException

测试类

    /* Test 5: Can't get a rating for an unrated movie
     */
    @Test(expected = MovieListException.class)
    public void nonexistentRating() throws MovieListException {
        String result = movies.getRating("The Ghost in the Invisible Bikini");

    }

MovieList 类

public class MovieList {
private HashMap<String, Rating> movieList = new HashMap<String, Rating>();

public void addMovie(String string) {
    // TODO Auto-generated method stub
    movieList.put(string, new Rating(RatingType.NONE, null));
}



public String getRating(String string) throws MovieListException {
    // TODO Auto-generated method stub
    String ratingString = "";

    try {
        Integer value = movieList.get(string).getRating();

        if(value != null && value > 0) {
            for(int i = 0; i < value; i++) {
                ratingString += "*";
            } 
            return ratingString;
        }                   
    } catch(MovieListException e) {
        System.out.print(e.getMessage());
        return e.getMessage();
    }


    return "No rating";
}

public void setRating(String string, Rating rating) {
    // TODO Auto-generated method stub
    movieList.replace(string,  rating);
}

package movieListQuestion;

评级等级

public class Rating {


private Integer ratingValue;
private RatingType typeValue;

public Rating(RatingType type, Integer rating) {
    this.ratingValue = rating;
    this.typeValue = type;
}   

public Integer getRating() throws MovieListException {
    if(this.ratingValue == null) {
        throw new MovieListException("No rating");
    }
    return this.ratingValue;
}

}

异常类

    package movieListQuestion;

/* A trivial exception class for the Movie List program.
 */
@SuppressWarnings("serial")
public class MovieListException extends Exception {

    public MovieListException() {
        super();
    }

    public MovieListException(String message) {
        super(message);
    }

}

更新了 MovieList 类中的 getRating 方法

public String getRating(String string) throws MovieListException {
        // TODO Auto-generated method stub
        String ratingString = "";

            Integer value = movieList.get(string).getRating();

            if (!movieList.containsKey(string)) {
               throw new MovieListException("Movie not found");
            }

            if(value != null && value > 0) {
                for(int i = 0; i < value; i++) {
                    ratingString += "*";
                } 
                return ratingString;
            }                   


        return "No rating";
    }

Answer 1

一个NullPointerException在movieList.get(string).getRating()很明显，您搜索的电影不在 movieList 中

您可能会决定抛出相同的 MovieListException当找不到电影时。你可以这样做

if (!movieList.containsKey(string)) {
   throw new MovieListException("Movie not found");
}
//Rest of the code

//Or using Optional
Rating rating = Optional.ofNullable(movieList.get(string))
        .orElseThrow(() -> new MovieListException("Movie not found"));
Integer value = rating.getRating();

此外，您不应该捕获 MovieListException在getRating方法