如何在单个 Java 方法中返回 int 和 string 属性？

Question

我想从BaseballPlayer类返回所有属性的值。 需要执行此操作的方法必须是公共字符串getBaseballPlayer(int i)方法（因为我需要在getBaseballPlayers()引用此方法以将所有值作为字符串getBaseballPlayers()返回）我在执行此操作时遇到了麻烦，因为所有属性具有不同的数据类型（ int 、 String 、 Height ）。

我试过这样做：

public String getBaseballPlayer(int i){

ArrayList <String> bArray = new ArrayList <String>();  
    bArray.add(getHometown());
    bArray.add(getState());                       
    bArray.add(getHighSchool());
    bArray.add(getPosition());

但是，它仅适用于字符串方法，并不一定返回实际值，而是返回每个字符串属性的 get 方法。

public class BaseballPlayer extends Player implements Table {    

  private int num;
  private String pos;

public BaseballPlayer( int a, String b, String c, int d, 
String e, String f, String g, Height h){


        super(a,ft,in,c,d,e,f,ht);
        num = a;
        pos = b; 

}

public BaseballPlayer(){}


//Returns the value of a specific attribute. The input parameter start 
  with 0 for the first attribute, then 1 for the second attribute and so 
  on.
//you can use getBaseballPlayer(int i) in getBaseballPlayers( ) with a for 
  loop getting each getBaseballPlayer(int i).


public String getBaseballPlayer(int i){

ArrayList <String> bArray = new ArrayList <String>();  
    bArray.add(getHometown());
    bArray.add(getState());                       
    bArray.add(getHighSchool());
    bArray.add(getPosition());
    return (bArray);    
}

//Returns the value of all attributes as an ArrayList of Strings.
public ArrayList <String> getBaseballPlayers(){
}

我只是在寻找返回每个属性值的最简单方法，然后使用该方法将每个值作为另一个方法中的字符串数组列表返回。

Answer 1

将整个对象作为一个字符串返回不是一个好习惯。 除非，否则，您被迫这样做，不要尝试这样做。

好吧，如果您的要求无法更改，并且您希望 Baseball 对象中的所有内容都在一个字符串中，您可以使用“:”等分隔符连接所有参数。

例如：

public String getBaseballPlayer(int i){
    return getHometown() + ":" + getState() + ":" +getHighSchool() + ":" + getPosition();
}

在调用端，您可以使用 String 的“split()”方法从此字符串中获取各个值。

Answer 2

你想做的，如果你想做得完美，就是Gson的存在。 它建立在简单的 JSON 之上，可以将任意类、数据结构和其他类型编码为 JSON，这样您就可以轻松地从 JSON 表示中重建这些对象。 考虑到它的实际功能有多强大，它很容易使用。

如果您不需要对 JSON 本身无法处理的类型进行编码，那么使用常规 JSON 会更容易。 在您的情况下，这似乎已经足够了。 JSON 的伟大之处在于它是一个标准。 您不必选择编码方案，并且您已经拥有用您能想到的任何语言编写的库，可以读取您的字符串化数据。

这是一个大致遵循您的代码正在执行的操作的示例：

import org.codehaus.jackson.map.ObjectMapper;

import java.io.IOException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.List;

public class BaseballPlayer {

    private String name;
    private String hometown;
    private String state;
    private int age;
    private double height;
    private String position;

    static ObjectMapper mapper = new ObjectMapper();

    public BaseballPlayer( String name, String hometown, String state, int age, double height, String position) {
        this.name = name;
        this.hometown = hometown;
        this.state = state;
        this.age = age;
        this.height = height;
        this.position = position;
    }

    public void setName(String name) {
        this.name = name;
    }

    public void setHometown(String hometown) {
        this.hometown = hometown;
    }

    public void setState(String state) {
        this.state = state;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public void setHeight(float height) {
        this.height = height;
    }

    public void setPosition(String position) {
        this.position = position;
    }

    public String toString() {
        return String.format("Name: %s from %s, %s (height: %.1f)", name, hometown, state, height);
    }

    public BaseballPlayer(){}

    // Turn a BaseballPlayer object into a String
    public String getAsJSON() {
        Map<String, Object> info = new HashMap<>();
        info.put("name", name);
        info.put("hometown", hometown);
        info.put("state", state);
        info.put("age", age);
        info.put("height", height);
        info.put("position", position);

        try {
            return mapper.writeValueAsString(info);
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

    // Here's the method you ask for.  I don't know what 'i' is supposed
    // to do, since we're in the class for a single baseball player.  You
    // could create a class that contains a list of baseball players, but
    // why not just use a List by itself, as I've done.
    public String getBaseballPlayer(int i) {
        return getAsJSON();
    }

    // Turn a list of BaseballPlayer objects into a list of Strings
    public static List<String> playersToStrings(List<BaseballPlayer> players) {
        List<String> r = new ArrayList<>();
        for (BaseballPlayer player : players) {
            r.add(player.getAsJSON());
        }
        return r;
    }

    // Turn a list of Strings into a list of BaseballPlayer objects
    public static List<BaseballPlayer> stringsToPlayers(List<String> playerStrings) {
        List<BaseballPlayer> r = new ArrayList<>();
        for (String playerString : playerStrings) {
            try {
                BaseballPlayer player = mapper.readValue(playerString, BaseballPlayer.class);
                r.add(player);
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return r;
    }

    public static void main(String... args) {

        // Create a list of BaseballPlayer objects and print them
        List<BaseballPlayer> players = new ArrayList<>();
        players.add(new BaseballPlayer("Joe", "Boston", "MA", 25, 6.1, "First Base"));
        players.add(new BaseballPlayer("Sam", "San Francisco", "CA", 28, 5.8, "Pitcher"));
        players.add(new BaseballPlayer("Kelly", "Chicago", "IL", 32, 6.4, "Catcher"));
        System.out.println(players);

        // Convert the list to a list of Strings and print the list
        List<String> playerStrings = playersToStrings(players);
        System.out.println(playerStrings);

        // Convert the Strings back into BaseballPlayer objects and print them
        players = stringsToPlayers(playerStrings);
        System.out.println(players);
    }

}

这是结果输出：

[Name: Joe from Boston, MA (height: 6.1), Name: Sam from San Francisco, CA (height: 5.8), Name: Kelly from Chicago, IL (height: 6.4)]
[{"hometown":"Boston","name":"Joe","state":"MA","position":"First Base","age":25,"height":6.1}, {"hometown":"San Francisco","name":"Sam","state":"CA","position":"Pitcher","age":28,"height":5.8}, {"hometown":"Chicago","name":"Kelly","state":"IL","position":"Catcher","age":32,"height":6.4}]
[Name: Joe from Boston, MA (height: 6.1), Name: Sam from San Francisco, CA (height: 5.8), Name: Kelly from Chicago, IL (height: 6.4)]

在这里，每个玩家都被单独转换为 JSON。 再多几行代码，您就可以将 Baseball Player 对象数组转换为单个 String。

如果这个仅 JSON 的解决方案对您来说还不够好，请查看 Gson。 它可以保留所有 Java 类型。 只需要更多的设置来描述您的每个对象应该如何转换为 JSON 并返回。