如何在单个Java方法中返回int和string属性？

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（）

中引用此方法，以将所有值作为字符串的数组列表返回），因为所有属性都具有不同的数据类型，所以执行此操作时遇到问题（

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());

公共字符串getBaseballPlayer（int i）{
ArrayList bArray=新的ArrayList（）；
add（gethomely（））；
add（getState（））；
添加（getHighSchool（））；
add（getPosition（））；

但是，它只适用于字符串方法，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(){
}

公共类BaseballPlayer扩展播放器实现表{
私有整数；
专用字符串pos；
公共棒球手（a、b、c、d、，
字符串e、字符串f、字符串g、高度h）{
超级（a、ft、in、c、d、e、f、ht）；
num=a；
pos=b；
}
公共棒球手（）{}
//返回特定属性的值。输入参数开始
第一个属性为0，第二个属性为1，依此类推
在…上
//您可以在getBaseballPlayer（）中使用getBaseballPlayer（inti）和for
循环获取每个getBaseballPlayer（int i）。
公共字符串getBaseballPlayer（int i）{
ArrayList bArray=新的ArrayList（）；
add（gethomely（））；
add（getState（））；
添加（getHighSchool（））；
add（getPosition（））；
返回（营房）；
}
//以字符串的ArrayList形式返回所有属性的值。
公共阵列列表GetBaseballPlayer（）{
}

---

我只是在寻找返回每个属性值的最简单方法，然后使用该方法在另一个方法中将每个值作为字符串的arraylist返回。

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

好的，如果您的需求无法更改，并且如果您希望棒球对象中的所有内容都在一个字符串中，那么您可以将所有参数连接到一个分隔符，如“：”

例如：

---

在调用端，可以使用“split（）”从该字符串中获取各个值字符串的方法。

如果你想做得完美，你想做的是什么。它在简单JSON的基础上，可以将任意类、数据结构和其他类型编码成JSON，这样你就可以从JSON表示中轻松地重建这些对象。考虑到它的实际功能有多强大，使用起来很容易

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就更容易了。在您的情况下，这似乎已经足够好了。JSON的好处在于它是一个标准。您不必选择编码方案，而且您已经有了用任何您能想到的语言编写的库，可以阅读您的字符串已转换为数据

下面是一个大致遵循代码所做操作的示例：

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);
    }

}

在这里，每个球员都被单独转换成JSON。再多几行代码，您就可以将棒球运动员对象数组转换成单个字符串

---

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。只需再进行一点设置，就可以描述如何将每个对象转换为JSON并返回。IMO最简单的选择是将它们包装到一个对象中，然后返回！我没有否决，但您不应该使用

I

所以mewhere？Hello@GBlodgett。这个问题只是给出了所需内容的粗略概念。它甚至不会编译。因此，我只想画出相同的图片，以便提问者能够轻松理解。我真的不知道为什么有人必须对问题和答案投反对票。我已经清楚地提到，这不是一个好的实践将所有内容都转换为字符串。但如果他的要求根本无法更改，则别无选择。我们应该提供一种实现方法。：）许多人会认为，如果问题只是给出了所需内容的粗略概念（即要求不明确）被提问的问题应该被关闭，而不是被回答。请参阅讨论的主题谢谢你强调@GBlodgett.我是新来的。因此，在回答之前，我可能需要了解交易。无论如何，有人提到，如果对坏问题有一个好的答案，就不应该被否决。接下来的答案不应该被否决d、 不管怎么说，既然不是你，你就无能为力了。谢谢你的帮助。没问题。我不会让否决票打扰你。如果你继续做出有益的贡献，你会得到更多的选票和声誉。

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);
    }

}

[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)]