如何使用toString方法輕松，整齊地打印多個對象，這些對象也是對齊的？

Question

我有一個成員的ArrayList，每個都有自己的信息，包括姓名，年齡，出生日期等，當打印到終端時用toString方法調用。

這是代碼：

@Override
    public String toString() {
        return "|ID: " + id + "| Name: " + name + "| Age: " + 
              calculateAge(age, LocalDate.now()) + "| Tlf.: " + tlfNo + "| Active: " + 
              activeMember + "| Next payment: " + nextPayment + "|";
    }

這是輸出：

|ID: 12| Name: Casper| Age: 49| Tlf.: 12345678| Active: true| Next payment: 2018-02-12|
|ID: 13| Name: Allan| Age: 69| Tlf.: 12345678| Active: true| Next payment: 2018-01-12|
|ID: 16| Name: Christina| Age: 100| Tlf.: 12345678| Active: false| Next payment: 2018-02-04|
|ID: 19| Name: RICK| Age: 49| Tlf.: 12345678| Active: false| Next payment: 2018-04-14|

如何讓不同的信息彼此對齊？

Answer 1

你可以這樣做：

 public String toString() {
        return String.format("|id: %10d |Name: %30s| Age: %02d | Tlf.: %d| Active: %s| Next Payment: %s |", id, getNameTrimmed() , age , tlfNo ,    activeMember, nextPayment );
 }

 private  String getNameTrimmed(){
    if(name.length()>30) {
        return name.substring(0,27)+"...";
    }
    return name;
}

這樣，id將有10個字符（如果你有更長的ID，它仍然會破壞格式）名稱將有30，所以你需要添加一個方法給你30個字符名稱

Answer 2

你可以為它創建一個方法，因為toString簡單，不起作用：

static String printPojos(Pojo... pojos) {
    int maxName = Arrays.stream(pojos)
                        .map(Pojo::getName)
                        .mapToInt(String::length)
                        .max()
                        .orElse(0);

    int maxId = Arrays.stream(pojos)
                      .mapToInt(x -> x.getName().length())
                      .max()
                      .orElse(0);

    StringJoiner result = new StringJoiner("\n");
    for (Pojo pojo : pojos) {
        StringJoiner sj = new StringJoiner("|");
        sj.add("ID: " + Strings.padEnd(Integer.toString(pojo.getId()), maxId, ' '));
        sj.add(" Name: " + Strings.padEnd(pojo.getName(), maxName, ' '));
        sj.add(" Age: " + Strings.padEnd(Integer.toString(pojo.getAge()), 3, ' '));
        sj.add(" Active: " + Strings.padEnd(Boolean.toString(pojo.isActive()), 5, ' '));

        result.merge(sj);
    }

    return result.toString();
}

我正在使用Strings::padEnd （來自guava ，但即使沒有外部庫，這也很容易編寫）。

在我的情況下Pojo看起來像這樣：

static class Pojo {

    private final int id;
    private final String name;
    private final int age;
    private final boolean active;

    // constructor, getters

}

結果如下：

    Pojo one = new Pojo("Casper", 49, true, 1);
    Pojo two = new Pojo("Allan", 100, true, 10_000);

    System.out.println(printPojos(one, two));



ID: 1    | Name: Casper| Age: 49 | Active: true 
ID: 10000| Name: Allan | Age: 100| Active: true 

Answer 3

我有一個簡單的問題實現，不需要外部庫。 如果你想使用外部庫，那么Apache-commons或Guava中有足夠的方法可用。 它可能會更聰明地完成，但這將成功：

class Member {

  String id;
  String name;
  int age;
  LocalDate date;
  int tlfNo;
  boolean activeMember;
  LocalDate nextPayment;

  public Member(String id, String name, int age, int tlfNo, boolean activeMember, LocalDate nextPayment) {
    this.id = id;
    this.name = name;
    this.age = age;
    this.tlfNo = tlfNo;
    this.activeMember = activeMember;
    this.nextPayment = nextPayment;
  }

  @Override
  public String toString() {
    return String.format("|ID: %s| Name: %s| Age: %s| Tlf.: %s| Active: %s| Next payment: %s|", 
      doFormat(id, "id", 10), doFormat(name, "name", 10),             
      doFormat(String.valueOf(age), "age", 10), 
      doFormat(String.valueOf(tlfNo), "tlfNo", 10), 
      doFormat(String.valueOf(activeMember), "active", 10),             
      doFormat(nextPayment.toString(), "nextPayment", 20));
  }

  String doFormat(String value, String fieldName, int maxLenght) {
    if (value.length() > maxLenght) {
      throw new IllegalArgumentException(String.format("cannot format string longer than %s for %s, value = %s", maxLenght, fieldName, value));
    }
    return padLeftSpaces(value, maxLenght);
  }

  public String padtLefSpaces(String inputString, int length) {
    if (inputString.length() >= length) {
      return inputString;
    }
    StringBuilder sb = new StringBuilder();
    while (sb.length() < length - inputString.length()) {
      sb.append(' ');
    }
    sb.append(inputString);

    return sb.toString();
  }
}

我們的想法是用空格填充所有字段，只替換占用的空間。 pad left方法是從bealdung借來的