我有一个ArrayList成员,每个成员都有自己的信息,包括姓名,年龄,出生日期等,这些信息在打印到终端时用toString方法调用.
这是代码：
@Override
public String toString() {
return "|ID: " + id + "| Name: " + name + "| Age: " +
calculateAge(age,LocalDate.now()) + "| Tlf.: " + tlfNo + "| Active: " +
activeMember + "| Next payment: " + nextPayment + "|";
}
这是输出：
|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|
如何获得不同的信息,使它们彼此对齐？
您可以为此创建一个方法,因为toString根本无法使用：
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(来自番石榴,但这即使没有外部库也很容易编写).
在我的情况下,Pojo看起来像这样：
static class Pojo {
private final int id;
private final String name;
private final int age;
private final boolean active;
// constructor,getters
}
结果看起来像：
Pojo one = new Pojo("Casper",49,true,1);
Pojo two = new Pojo("Allan",100,10_000);
System.out.println(printPojos(one,two));
ID: 1 | Name: Casper| Age: 49 | Active: true
ID: 10000| Name: Allan | Age: 100| Active: true
