java - 在 Java 中格式化 toString

Question

我已经尝试在我的 toString 中使用 String 格式化程序来使我的输出看起来不错，任何人都知道我该如何解决这个问题，我是 Java 格式化程序的新用户。

我正在使用这段代码从“.ser”格式的文件中读取数据代码看起来像这样，

    public void printList(List<Person> list, File filepath) throws IOException, ClassNotFoundException {
          Person person = new Person();
        if (filepath.isFile()) {
            ObjectInputStream inputStream = new ObjectInputStream(new FileInputStream(filepath));
            list = (List<Person>) inputStream.readObject(); 
            inputStream.close();
            ListIterator iterator = list.listIterator();
            int count = 0;
            while (iterator.hasNext()) {

                System.out.println((count +1)+ ". " +iterator.next());
                count++;
            }
            if(count > 0){
                System.out.println("\nThere are : " +  count +"  persons.\n");
            }else {
                System.out.println("The list is empty....!\nTo add to list enter 1:");
            }
        }
    }

我正在使用 ObjectInputStream 和 OutPutstream 来读取这些数据。我的代码工作正常，但我想修复的输出状态更好。

public class Person extends Addresses implements Serializable {

    private String firstName = "";
    private String lastName = "";
    private String age= "";
    private double length;
    private String postAddress = "";
    private String postNumber = "";
    private String postArea = "";


    public Person() {
        this("", "", "", 0.0, "", "", "");
    }

    public Person(String firstName, String lastName, string age, double length, String postAddress, String postNumber, String postArea) {
        super(postAddress, postNumber, postArea);
        this.firstName = firstName;
        this.lastName = lastName;
        this.signature = age;
        this.length = length;

    }

@Override
    public String toString() {

        String header = String.format(java.util.Locale.US, "\t%5s  \t%1s %1s \t%10.2f",
                this.age, this.firstName, this.lastName,  this.length);
        return header;
}

No  Age         Name                   Length[M]
1.  12          John Doe           1.80
2.  32          Mona lisa                1.69
3.  50          Benjamin Franklin         1.75
4.  45          Robert Oppenheimer            1.82

我想要的输出是

No  Age         Name                        Length[M]
1.  12          John Doe                    1.80
2.  32          Mona lisa                   1.69
3.  50          Benjamin Franklin           1.75
4.  45          Robert Oppenheimer          1.82

Answer 1

解决方案 1:您可以增加字符串限制，使所有生成的字符串都符合给定的限制，以便出现统一的制表符缩进。

解决方案 2:您可以像下面那样限制 char 限制，这样字符串总是在给定的限制内，这样就会有适当的缩进

 String result = String.format("|%.5s|", "Hello World");//  output -> |Hello|