Question

我有一个EmployeeDetails对象的arrayList和另一个EmployeeSalary对象的arrayList。我想创建一个同时具有EmployeeDetails和EmployeeSalary属性的新ArrayList.My EmployeeDetails类具有属性＆＃34; id＆＃34;和＆＃34;名称＆＃34;。我的EmployeeSalary类有属性＆＃34; id＆＃34;和＃34;薪水＆＃34;。我想要一个带有属性＆＃34; id＆＃34;，＆＃34; name＆＃34;的arraylist和＆＃34;薪水＆＃34;。

EmployeeDetails Class

public class EmployeeDetails {

    private String id;
    private String name;
    public String getId() {
        return id;
    }
    public void setId(String id) {
        this.id = id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
}

EmployeeSalary类

public class EmployeeSalary {

    private String id;
    private String sal;
    public String getId() {
        return id;
    }
    public void setId(String id) {
        this.id = id;
    }
    public String getSal() {
        return sal;
    }
    public void setSal(String sal) {
        this.sal = sal;
    }
}

Answer 1

EmployeeSalary应该扩展Employeedetails

public class EmployeeSalary extends EmployeeDetails{

private String sal;

...

之后，为了合并两个列表，应该调用arraylist1.addAll（arraylist2）。

Answer 2

首先，设置一张地图，以便从ID中访问工资：

Map<String, String> salaryMappings = new HashMap();
for(EmployeeSalary salary : salaries) {
    salaryMappings.put(salary.getId(), salary.getSal());
}

（您的工资ArrayList名为salaries）。

接下来，你需要一个类来存储信息（注意我不是为了保持帖子的简短而不是编写访问者，尽管你应该添加它们）：

class EmployeeInformation {
    String id;
    String name;
    String sal;

    public EmployeeInformation(String id, String name, String sal) {
        this.id = id;
        this.name = name;
        this.sal = sal;
    }
}

最后，将值复制到新的ArrayList中：

List<EmployeeInformation> infos = new ArrayList<EmployeeInformation>();
for(EmployeeDetails detail : details) {
    infos.add(new EmployeeInformation( details.getId(), details.getName(), salaryMappings.get(details.getId()) ));
}

Answer 3

无需使用HashMap或Map。只需使用java 8 lambda :)

创建EmployeeInformation.java类。

public class EmployeeInformation {

    String id;
    String name;
    String sal;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

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

    public String getSal() {
        return sal;
    }

    public void setSal(String sal) {
        this.sal = sal;
    }

    public EmployeeInformation(String id, String name, String sal) {
        this.id = id;
        this.name = name;
        this.sal = sal;
    }
}

找到完整的解决方案。

List<EmployeeDetails> employeeDetailsList = new ArrayList<>();

        EmployeeDetails employeeDetails1 = new EmployeeDetails();
        employeeDetails1.setId("A");
        employeeDetails1.setName("EMP1");
        employeeDetailsList.add(employeeDetails1);


        EmployeeDetails employeeDetails2 = new EmployeeDetails();
        employeeDetails2.setId("B");
        employeeDetails2.setName("EMP2");
        employeeDetailsList.add(employeeDetails2);

        List<EmployeeSalary> employeeSalariesList = new ArrayList<>();

        EmployeeSalary employeeSalary1 = new EmployeeSalary();
        employeeSalary1.setId("A");
        employeeSalary1.setSal("SAL1");
        employeeSalariesList.add(employeeSalary1);

        EmployeeSalary employeeSalary2 = new EmployeeSalary();
        employeeSalary2.setId("B");
        employeeSalary2.setSal("SAL2");
        employeeSalariesList.add(employeeSalary2);



        List<EmployeeInformation> employeeInformationList = new ArrayList<>();


        employeeDetailsList.forEach(employeeDetails -> {
            String _id = employeeDetails.getId();
            String _name = employeeDetails.getName();
            employeeSalariesList.stream().filter(employeeSalary -> employeeSalary.getId().equalsIgnoreCase(_id)).forEach(employeeSalary -> {
                EmployeeInformation employeeInformation = new EmployeeInformation(_id, _name, employeeSalary.getSal());
                employeeInformationList.add(employeeInformation);

            });

        });

        for (EmployeeInformation employeeInformation : employeeInformationList) {
            System.out.println(employeeInformation.getId() + "-" + employeeInformation.getName() + "-" + employeeInformation.getSal());
        }

输出 - ＆gt;

A-EMP1-SAL1
B-EMP2-SAL2

Answer 4

您可以将这两个集合转换为id - ＆gt; value的地图，然后进行辅助查找。

以下构建Map<String, String>的集合，其中包含结果中所需的键：

List<EmployeeDetails> details = ...;
List<EmployeeSalary> salaries = ...;

Map<String, EmployeeDetails> detailsMap = details.stream()
        .collect(Collectors.groupingBy(EmployeeDetails::getId, 
            Collectors.reducing(null, (e1, e2) -> e1)));
Map<String, EmployeeSalary> salariesMap = salaries.stream()
          .collect(Collectors.groupingBy(EmployeeSalary::getId, 
            Collectors.reducing(null, (e1, e2) -> e1)));

List<Map<String, String>> summaries = detailsMap.entrySet().stream()
        .map(entry -> {
               Map<String, String> m = new HashMap<>();

    m.put("id", entry.getKey());
    m.put("name", entry.getValue().getName());
    m.put("salary", salariesMap.get(entry.getKey()).getSal());

    return m;
}).collect(Collectors.toList());

您可能会选择为摘要构建单独的类，包含id，name和salary字段，而不是使用一个Map。但这是一个如何去做的想法。

Answer 5

我认为，最好的解决方案是创建另一个类（例如EmployeeData）并在那里收集所有必需的数据。如果您不希望（或不能）修改EmployeeDetails和EmployeeSalry的定义。

class EmployeeData {

    private final String id;
    private String name;
    private String sal;

    public EmployeeData(String id) {
        this.id = id;
    }
}

private static List<EmployeeData> merge(List<EmployeeDetails> details, List<EmployeeSalary> salaries) {
    Map<String, EmployeeData> data = new HashMap<>();
    String id;

    for (EmployeeDetails detail : details) {
        id = detail.getId();
        if (!data.containsKey(id))
            data.put(id, new EmployeeData(id));
        data.get(id).setName(detail.getName());
    }

    for (EmployeeSalary salary : salaries) {
        id = salary.getId();
        if (!data.containsKey(id))
            data.put(id, new EmployeeData(id));
        data.get(id).setSal(salary.getSal());
    }

    return new ArrayList<>(data.values());
}

合并具有Employee的不同属性的两个数组列表，以创建组合两个arraylist的属性的新数组列表

5 个答案: