Question

我有一个pojo课程：

class Employee{
    private int empCode; 
    private String name;
//getters and setters and Constructor
}

我有另一个类，它使用参数构造函数实例化所有Employee Object。并且所有Employee pojo对象都添加在Arraylist中。

class EmployeeData{
    Employee employee = new Employee(111,"XXX");
    Employee employee1 = new Employee(222, "YYY");
    Employee employee2 = new Employee(444, "BBB");
    List<Employee> listData  = new ArrayList<Employee>();
    listData.add(employee);
    listData.add(employee1);
    listData.add(employee2);

    public List<Employee> getList(List<Employee> list){
        return list;
    }
}

最后我有一个Main类，它有另一个列表和一个合并方法。

class EmployeeTest{
    Employee employee = new Employee(111,"NNN");
    Employee employee1 = new Employee(222, "YYY");
    Employee employee2 = new Employee(333, "KKK");
    List<Employee> list  = new ArrayList<Employee>();
    list.add(employee);
    list.add(employee1);
    list.add(employee2);  
    public List<Employee> merge(List<Employee> list){
    //code goes here
    return listData;
}

我需要比较列表中的id，如果id匹配但名称不同，我将用列表的值覆盖listData的值。
如果id不匹配，并且在listData中不存在333。所以我们将添加id和值，最后
如果listData中的某些ID与列表不匹配，我们会删除它，例如我们将从listData中删除444，最后我们将返回listData。

我试图做这样的事情，但我没有得到正确的输出。请帮忙。

public List<Employee> merge(List<Employee> list){
    if(list==null){
        return null;
    }
    List<Employee> list1 = employeeDao.getList();
    for(int i =0;i<list1.size();i++){
        for(int j=0; j<list1.size();j++){
            if(list.get(i).getEmpCode().equals(list2.get(j).getEmpCode())){
                list1.set(j, list.get(i));
                j++;
            }else{
                j++;
            }
        }
    }
    return list;
}

我无法考虑最后两个逻辑。请帮忙

Answer 1

检查您是否得到正确答案。这不是好方法，但仍然。

import java.util.ArrayList;
import java.util.List;
import java.util.Map;


class Employee {
    private int empCode;
    private String name;
    public Employee(int empCode, String name) {
        super();
        this.empCode = empCode;
        this.name = name;
    }
    public int getEmpCode() {
        return empCode;
    }
    public void setEmpCode(int empCode) {
        this.empCode = empCode;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    @Override
    public String toString() {
        return "Employee [empCode=" + empCode + ", name=" + name + "]";
    }

}



public class Practice {

    public static void main(String[] args) {

        Employee employee = new Employee(111,"XXX");
        Employee employee1 = new Employee(222, "YYY");
        Employee employee2 = new Employee(444, "BBB");


        Employee employee3 = new Employee(111,"NNN");
        Employee employee4 = new Employee(222, "YYY");
        Employee employee5 = new Employee(333, "KKK");

        List<Employee> employeeDataList = new ArrayList<Employee>();
        employeeDataList.add(employee);
        employeeDataList.add(employee1);
        employeeDataList.add(employee2);


        List<Employee> employeeTestList = new ArrayList<Employee>();
        employeeTestList.add(employee3);
        employeeTestList.add(employee4);
        employeeTestList.add(employee5);

        merge(employeeDataList, employeeTestList);

    }

    public static List<Employee> merge(List<Employee> employeeDataList, List<Employee> employeeTestList) {
        List<Employee> list = new ArrayList<Employee>();

        for(Employee emp : employeeTestList) {
            int lengthOfEmployeeDataList = employeeDataList.size();
            int count = 0;
            for(Employee empData : employeeDataList){
                if(emp.getEmpCode() == empData.getEmpCode()) {
                    //EMPLOYEE CODE MATCHES
                    if(!emp.getName().equals(empData.getName())) {
                        //WHEN NAME ARE NOT EQUAL
                        empData.setName(emp.getName()); //NAME IN TEST LIST IS ASSIGNED TO DATALIST
                        list.add(empData);
                        break;
                    }
                    else {
                        list.add(emp);
                        break;
                    }
                }
                else if(!(emp.getEmpCode() == empData.getEmpCode()) && count < lengthOfEmployeeDataList){
                    //ID DOESNT MATCHES
                    count++;
                }
                if(count == lengthOfEmployeeDataList){
                    //ALL THE EMPLOYEE Objects has been Checked and None Matching
                    list.add(emp);
                }

            }
        }
        for(Employee emp : employeeDataList) {
            if(!list.contains(emp)) {
                employeeDataList.remove(emp);
            }
        }
        System.out.println(list);
        return list;
    }


}

Answer 2

如果List不是唯一的解决方案，我建议使用Class HashMap并将id用作键。然后你可以使用HashMap api，这对这些动作更有帮助，例如的putIfAbsent（） https://docs.oracle.com/javase/8/docs/api/java/util/HashMap.html

比较给定2个字符串的pojo值

2 个答案: