我想打印这种格式的人，每个人都在一个栏目中，每个人都用空格或标签分隔，即
约翰史密斯 男性男 01 02如何修改toString（）以打印出漂亮的格式？目前，每个字段和每个人都打印在一行中。
public class Test {
public static void main(String[] args) {
ArrayList<Person> persons = new ArrayList<Person>();
persons.add(new Person("John", "Male", "01"));
persons.add(new Person("Smith", "Male", "02"));
persons.add(new Person("Jack", "Male", "03"));
persons.add(new Person("Mary", "Female", "04"));
persons.add(new Person("Alice", "Feamle", "05"));
for(Person p : persons){
System.out.print(p + " ");
}
}
static class Person{
private String name;
private String sex;
private String ID;
public Person(String n, String s, String id){
name = n;
sex = s;
ID = id;
}
public String toString(){
StringBuilder sb = new StringBuilder();
sb.append(ID + "\n");
sb.append(name + "\n");
sb.append(sex + "\n");
return sb.toString();
}
}
}
我真正想要的是这种格式。我有一个优先级队列，为了调试，我需要以这种格式打印出优先级队列的内容。
优先级队列跟踪。
Here are the contents of our priority queue (sorted by priority) just before dequeueing each node when using the Manhattan priority function on puzzle04.txt.
Step 0: priority = 4
moves = 0
manhattan = 4
3
0 1 3
4 2 5
7 8 6
Step 1: priority = 4 priority = 6
moves = 1 moves = 1
manhattan = 3 manhattan = 5
3 3
1 0 3 4 1 3
4 2 5 0 2 5
7 8 6 7 8 6
Step 2: priority = 4 priority = 6 priority = 6
moves = 2 moves = 1 moves = 2
manhattan = 2 manhattan = 5 manhattan = 4
3 3 3
1 2 3 4 1 3 1 3 0
4 0 5 0 2 5 4 2 5
7 8 6 7 8 6 7 8 6
Step 3: priority = 4 priority = 6 priority = 6 priority = 6 priority = 6
moves = 3 moves = 3 moves = 2 moves = 3 moves = 1
manhattan = 1 manhattan = 3 manhattan = 4 manhattan = 3 manhattan = 5
3 3 3 3 3
1 2 3 1 2 3 1 3 0 1 2 3 4 1 3
4 5 0 4 8 5 4 2 5 0 4 5 0 2 5
7 8 6 7 0 6 7 8 6 7 8 6 7 8 6
Step 4: priority = 4 priority = 6 priority = 6 priority = 6 priority = 6 priority = 6
moves = 4 moves = 3 moves = 4 moves = 2 moves = 3 moves = 1
manhattan = 0 manhattan = 3 manhattan = 2 manhattan = 4 manhattan = 3 manhattan = 5
3 3 3 3 3 3
1 2 3 1 2 3 1 2 0 1 3 0 1 2 3 4 1 3
4 5 6 0 4 5 4 5 3 4 2 5 4 8 5 0 2 5
7 8 0 7 8 6 7 8 6 7 8 6 7 0 6 7 8 6
答案 0 :(得分：1)
查看String.Format方法。我认为这正是您正在寻找的格式化您返回的字符串。这是文档的String.Format
答案 1 :(得分：1)
我建议您为get类使用Person方法，以便在遍历每个人时，您可以将此人的每个属性添加到字符串中，然后稍后显示。像这样：
import java.util.ArrayList;
public class PersonTest {
public static void main(String[] args) {
ArrayList<Person> persons = new ArrayList<Person>();
persons.add(new Person("John", "Male", "01"));
persons.add(new Person("Smith", "Male", "02"));
persons.add(new Person("Jack", "Male", "03"));
persons.add(new Person("Mary", "Female", "04"));
persons.add(new Person("Alice", "Feamle", "05"));
// Strings to display at the end of the iterations
StringBuilder names = new StringBuilder(), sexes = new StringBuilder(), ids = new StringBuilder();
for(Person p : persons) {
// tab after each property
names.append(p.getName() + "\t");
sexes.append(p.getSex() + "\t");
ids.append(p.getId() + "\t");
}
System.out.println(names);
System.out.println(sexes);
System.out.println(ids);
}
static class Person {
private String name;
private String sex;
private String ID;
public Person(String n, String s, String id) {
name = n;
sex = s;
ID = id;
}
// getter methods to get the properties of this person
public String getName() {
return name;
}
public String getSex() {
return sex;
}
public String getId() {
return ID;
}
}
}
输出是这样的：
John Smith Jack Mary Alice
Male Male Male Female Feamle
01 02 03 04 05
答案 2 :(得分：0)
您无法通过覆盖toString()方法来执行此操作。你必须编写一些字符串util方法，它将占用List个人并构建列表的字符串表示并返回它。完成填充Person个对象后，请调用此util方法。
要构建字符串，您可以使用StringBuilder对象。
Util方法看起来像这样
StringBuilder sb = new StringBuilder();
for(Person p : persons){
sb.append(p.getName()+"\t");
}
sb.append("\n")
for(Person p : persons){
sb.append(p.getSex() + "\t");
}
sb.append("\n")
for(Person p : persons){
sb.append(p.getID() + "\t");
}
答案 3 :(得分：0)
这样做的基本方法是：
public class Test {
public static void main(String[] args) {
ArrayList<Person> persons = new ArrayList<Person>();
persons.add(new Person("John", "Male", "01"));
persons.add(new Person("Smith", "Male", "02"));
persons.add(new Person("Jack", "Male", "03"));
persons.add(new Person("Mary", "Female", "04"));
persons.add(new Person("Alice", "Feamle", "05"));
StringBuilder sb = new StringBuilder();
for(Person p : persons){
sb.append(p.getName() + " ");
}
sb.append("\n");
for(Person p : persons){
sb.append(p.getSex() + " ");
}
sb.append("\n");
for(Person p : persons){
sb.append(p.getID() + " ");
}
sb.append("\n");
System.out.print(sb.toString());
}
static class Person{
private String name;
private String sex;
private String ID;
public Person(String n, String s, String id){
name = n;
sex = s;
ID = id;
}
public String getName() {
return name;
}
public String getSex() {
return sex;
}
public String getID() {
return ID;
}
}
答案 4 :(得分：0)
我认为会是：
for(int i = 0; i < Persons.size()){
System.out.print(Persons.get(i).name + " ");
}
答案 5 :(得分：0)
很快就会有一些人受到限制：
public class Test {
public static void main(String[] args) {
ArrayList<Person> persons = new ArrayList<Person>();
persons.add(new Person("John", "Male", "01"));
persons.add(new Person("Smith", "Male", "02"));
persons.add(new Person("Jack", "Male", "03"));
persons.add(new Person("Mary", "Female", "04"));
persons.add(new Person("Alice", "Feamle", "05"));
StringBuilder name = new StringBuilder();
StringBuilder sex = new StringBuilder();
StringBuilder ID = new StringBuilder();
String delimeter = "";
for(Person p : persons){
name.append(delimeter).append(p.name);
sex.append(delimeter).append(p.sex);
ID.append(delimeter).append(p.ID);
delimeter = "\t";
}
System.out.println(name);
System.out.println(sex);
System.out.println(ID);
}
输出John Smith Jack Mary Alice
Male Male Male Female Feamle
01 02 03 04 05
如果名称很长并且打破了对齐，可能会引起一些额外的担忧。