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java - 按字母顺序排序数组列表时遇到问题

转载 作者:行者123 更新时间:2023-11-30 08:18:48 25 4
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我一直无法按字母顺序对输出进行排序(按名称)。我尝试了一些方法,它们已被注释掉。我尝试的最新方法是冒泡排序。

我的程序将所有输入写入二进制文件,然后在用户想要输出时读取它们。所以基本上当用户点击数字5时,它会运行sortEmployees()方法,然后按字母顺序显示记录,但目前我还没有找到一种方法来做到这一点。

我对数组列表没有信心,我没有经常使用它们......所以我几乎可以肯定我使用了错误的方法。我相信我没有使用最新的 JDK。我还在 stack + google 上查看了不同的问题,但找不到太多有帮助的。

目前的输出如下:

  1. 1,约翰逊,80000.00
  2. 2 等等...
  3. 3 等等...正如它们已在下面输入的那样。

我将在下面添加我的所有代码,以便您查看。

EmployeeFileDriver.java

    java.io.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import javax.swing.JOptionPane;


public class EmployeeFileDriver
{
private static ArrayList<Employee> list = new ArrayList<>();

public static void main(String args[]) throws IOException
{
final int QUIT_MAIN_MENU = 4;
String fileName = "Employee.txt";

readEmployeeFile(fileName);
//sample records
list.add(new Employee(1, "Johnson", 80000.00));
list.add(new Employee(2, "Singh", 75000.00));
list.add(new Employee(3, "Smith", 63000.00));
list.add(new Employee(4, "Jones", 52000.00));

int choice = mainMenu();

while(choice != QUIT_MAIN_MENU)
{
switch (choice)
{
case 1:
list.add(inputEmployee());
break;
case 2:
searchID();
break;
case 3:
listAllEmployees();
break;
case 5:
sortEmployees(list[]);
break;
} //end switch (choice)

choice = mainMenu(); // recall mainMenu()
} // end while

saveEmployeeFile(fileName); //save objects to file before closing
System.out.println("Records saved to file");
}

public static int mainMenu()
{
String heading, menu, userSelection;

heading = "<h3><u>EMPLOYEE SYSTEM</u></h3>";

menu = "<HTML><center>" + heading + "<hr><h2><u>Main Menu</u></h2><br /><table>" +
"<tr><td>1. Enter Employee Details</td></tr>" +
"<tr><td>2. Search Employee ID</td></tr>" +
"<tr><td>3. List all Employees</td></tr>" +
"<tr><td>4. Save and Exit</td></tr>" +
"<tr><td>5. Sort alphabetically</td></tr>" +
"</table><br /><hr></center></HTML>" +
"\nEnter Selection: ";

//get user selection for main menu
userSelection = JOptionPane.showInputDialog(null , menu,"Input",
JOptionPane.PLAIN_MESSAGE);

int option = Integer.parseInt(userSelection);

//return option to main method
return option;
}

public static void readEmployeeFile(String fileName)
{
try
{
//open InputFile
FileInputStream fiStream = new FileInputStream(fileName);
ObjectInputStream inStream = new ObjectInputStream(fiStream);

list = (ArrayList)inStream.readObject(); //read whole ArrayList object
//rather than individual Employee objects

System.out.println("\tEMPLOYEE LIST (from File)");
System.out.println("ID \t Name \t Salary\n");
for(Employee e: list)
{
System.out.printf("%-10d %-16s $%8.2f\n", e.getID(),
e.getName(), e.getSalary());
}
inStream.close();
}
catch (FileNotFoundException e)
{
System.err.println("Could not open file\n" + e.toString());
}
catch (ClassNotFoundException e)
{
System.err.println("Class not found \n" + e.toString());
}
catch (IOException e)
{
System.err.println("Employee file: \n" + e.toString());
}
}

//save employee objects to file on exit
public static void saveEmployeeFile(String fileName)
{
try
{
//open OutputFile
FileOutputStream foStream = new FileOutputStream(fileName);
//FileWriter fw = new FileWriter(fileName);
ObjectOutputStream outStream = new ObjectOutputStream(foStream);

outStream.writeObject(list); //save whole ArrayList object to file

outStream.close();
}
catch (FileNotFoundException e)
{
System.out.println("Could not open file\n" + e.toString());
}
catch(IOException e)
{
System.out.println("I/O error\n" + e.toString());
}
}

public static Employee inputEmployee()
{
int id;
String name;
double salary;
Employee emp;

id = Integer.parseInt(
JOptionPane.showInputDialog(null , "Enter employee ID"));
name = JOptionPane.showInputDialog(null , "Enter employee name");
salary = Double.parseDouble(
JOptionPane.showInputDialog(null , "Enter employee salary"));
emp = new Employee(id, name, salary);

return emp;
}

public static void searchID()
{
int id;
boolean found = false;

id = Integer.parseInt(
JOptionPane.showInputDialog(null, "Enter employee ID to search"));

for(Employee e: list)
{
if(id == e.getID())
{
JOptionPane.showMessageDialog(null, e.toString());
found = true;
}
}
if(found == false)
JOptionPane.showMessageDialog(null, "Employee ID not found");
}

public static void listAllEmployees()
{
String output = "";

for(Employee e: list)
{
output += e.toString();
}

JOptionPane.showMessageDialog(null, output);
}

// public void sortEmployees(){
// //Collections.sort(list, (e1, e2) -> e1.getname().compareTo(e2.id()));
// //[list sortUsingSelector:@selector(caseInsensitiveCompare:)];
// Collections.sort(list, new Comparator<Employee>()
// {
// @Override
// public int compare(String name)
// {
// return name.compareToIgnoreCase("ABCDE");
// }
// });
// }
public static void sortEmployees()
{

int j;
boolean flag = true; // will determine when the sort is finished
String temp;

while ( flag )
{
flag = false;
for ( j = 0; j < list.length - 1; j++ )
{
if ( list [ j ].compareToIgnoreCase( list [ j+1 ] ) > 0 )
{ // ascending sort
temp = list [ j ];
list [ j ] = list [ j+1]; // swapping
list [ j+1] = temp;
flag = true;
}
}
}
String output = "";
for(String e: list)
{
output += e.toString();
}

JOptionPane.showMessageDialog(null, output);
}
}

员工.java

    public class Employee 
{

private int id;
private String name;
private double salary;


public Employee (int id, String name, double salary)
{
this.id = id;
this.name = name;
this.salary = salary;
}

public void setName(String name)
{
this.name = name;
}

public String getName()
{
return this.name;
}

public double getSalary()
{
return this.salary;
}

public int getID()
{
return this.id;
}

public String toString()
{
return "\nID_NO:\t"+id+"\nNAME:\t"+name+"\nSALARY\t"+salary+"\n";
}

}

最佳答案

这应该可以完成工作。

static List<Employee> list = new ArrayList<>();

static void sortEmployees() {
list.sort(new Comparator<Employee>() {

@Override
public int compare(Employee e1, Employee e2) {
return e1.getName().compareTo(e2.getName());
}});
}

注意list.sort(...)自 Java 8 起可用。对于 Java 7 及使用前 Collections.sort(list, new Comparator<Employee>() {...});

<小时/>

对于java8流,如果你想保持原来的顺序:

List<Employee> sortedList = list.stream()
.sorted((e1, e2) -> e1.getName().compareTo(e2.getName()))
.collect(Collectors.toList());
<小时/>

另一个变体是 Employee实现Comparable界面并在 Employee 上使用自然排序对象(如您定义的那样 - 这是名称的字母顺序):

public class Employee implements Comparable {

@Override
public int compareTo(Object o) {
Employee e = (Employee)o;
return name.compareTo(e.getName());
}

// ...
}

然后你就可以这样做(Java 8):

static void sortEmployees() {
list.sort(null); // passing a null Comparator => use natural order
}

或者在 Java 7 及之前版本中:

static void sortEmployees() {
Collections.sort(list);
}

关于java - 按字母顺序排序数组列表时遇到问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29274666/

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