Java Apache公共多值映射迭代
我有一个多重映射假设公司，它以字符串作为键，另一个多重映射假设员工作为值。Employee Multimap将字符串作为键，另一个Multimap作为值。我的问题是如何检索和迭代存储在multimap中的multimap？我正在使用apachecommonmultimap 示例：compMap有2家公司。CompA和CompB 每个公司有10名员工。（员工可以出现多次，因此使用multimap） Coll包含compA的employee multimap，但是如何从employee multimap检索特定员工（如果他出现4次） 代码：Java Apache公共多值映射迭代,java,apache-commons,multimap,multivalue,Java,Apache Commons,Multimap,Multivalue,我有一个多重映射假设公司，它以字符串作为键，另一个多重映射假设员工作为值。Employee Multimap将字符串作为键，另一个Multimap作为值。我的问题是如何检索和迭代存储在multimap中的multimap？我正在使用apachecommonmultimap 示例：compMap有2家公司。CompA和CompB 每个公司有10名员工。（员工可以出现多次，因此使用multimap） Coll包含compA的employee multimap，但是如何从employee multim
我希望下面的场景是您所期待的。
单位
---系
---班级
---雇员
-身份证
-名字
包com.java.examples；
导入java.util.HashMap；
导入java.util.Iterator；
导入java.util.Map；
导入java.util.Map.Entry；
班级员工{
}
阶级公司{
}
班级部{
}
公共类Test1{
公共静态void main（字符串[]args）{
Map inputEmployeeMap=新HashMap（）；
inputEmployeeMap.put（“1”，newEmployee（））；
inputEmployeeMap.put（“2”，newEmployee（））；
inputEmployeeMap.put（“3”，newEmployee（））；
Map inputClassList=新建HashMap（）；
inputClassList.put（“class1”，inputEmployeeMap）；
inputClassList.put（“class2”，inputEmployeeMap）；
Map inputDeptList=new HashMap（）；
inputDeptList.put（“ece”，inputClassList）；
inputDeptList.put（“csc”，inputClassList）；
Map inputCompanyList=新的HashMap（）；
inputCompanyList.put（“ABCCompany”，inputDeptList）；
//解析图
解析多值映射（inputCompanyList）；
}
私有静态多值映射(
映射输入公司列表）{
迭代器it=inputCompanyList
.entrySet（）.iterator（）；
while（it.hasNext（））{
Map.Entry companyList=（Map.Entry）it.next（）；
System.out.println（“公司名称=“+companyList.getKey（））；//公司
//名字
Map deptList=（Map）companyList.getValue（）；//部门列表
//部门列表的迭代器。
迭代器companyListIterator=deptList.entrySet（）.Iterator（）；
while（companyListIterator.hasNext（））{
Map.Entry部门列表=（Map.Entry）公司注册人
.next（）；
System.out.print（“Dept Name=“+departmentList.getKey（））；//department
//类图
Map classList=（Map）departmentList.getValue（）；
迭代器departmentListIterator=classList.entrySet（）
.iterator（）；
Map.Entry empNames=（Map.Entry）departmentListIterator.next（）；
//类名
System.out.print（“类名=“+empNames.getKey（））；
//员工地图
Map empNamesMap=（Map）empNames.getValue（）；
迭代器eNamesIt=empNamesMap.entrySet（）.Iterator（）；
while（eNamesIt.hasNext（））{
Map.Entry name=（Map.Entry）eNamesIt.next（）；
//打印emp Id
System.out.print（“id”+name.getKey（））；
System.out.print（“name=“+name.getValue（））；
}
System.out.println（“\n”）；
}
System.out.println（“\n”）；
}
}
}
希望能有帮助。
public class AbcTest {
public static void main(String[] args) {
Address address1 = new Address("street1");
Address address2 = new Address("street2");
Employee employee1 = new Employee("employee1");
employee1.getAddresses().add(address1);
employee1.getAddresses().add(address2);
Employee employee2 = new Employee("employee2");
employee2.getAddresses().add(address2);
Company companyA = new Company("compA");
companyA.getEmployees().add(employee1);
companyA.getEmployees().add(employee1); // you can add employee to the list as many times as you want, if you really need this?
companyA.getEmployees().add(employee2);
// now to get the employee with give name simly iterate over list of employees
Iterator<Employee> employeeIt = companyA.getEmployees().iterator();
Employee wantedEmployee = null;
while (employeeIt.hasNext() && (wantedEmployee == null)) {
Employee next = employeeIt.next();
if (next.getName().equals("employee1")) {
wantedEmployee = next;
}
}
System.out.println(wantedEmployee);
}
}
class Company {
final String name;
final List<Employee> employees;
Company(String name) {
this.name = name;
this.employees = new ArrayList<>();
}
public String getName() {
return name;
}
public List<Employee> getEmployees() {
return employees;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if ((o == null) || (getClass() != o.getClass())) {
return false;
}
Company company = (Company) o;
if ((employees != null) ? (!employees.equals(company.employees)) : (company.employees != null)) {
return false;
}
if ((name != null) ? (!name.equals(company.name)) : (company.name != null)) {
return false;
}
return true;
}
@Override
public int hashCode() {
int result = (name != null) ? name.hashCode() : 0;
result = (31 * result) + ((employees != null) ? employees.hashCode() : 0);
return result;
}
}
class Employee {
final String name;
final Set<Address> addresses;
Employee(String name) {
this.name = name;
this.addresses = new HashSet<>();
}
public String getName() {
return name;
}
public Set<Address> getAddresses() {
return addresses;
}
@Override
public String toString() {
return "Employee{" +
"name='" + name + '\'' +
'}';
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if ((o == null) || (getClass() != o.getClass())) {
return false;
}
Employee employee = (Employee) o;
if ((addresses != null) ? (!addresses.equals(employee.addresses)) : (employee.addresses != null)) {
return false;
}
if ((name != null) ? (!name.equals(employee.name)) : (employee.name != null)) {
return false;
}
return true;
}
@Override
public int hashCode() {
int result = (name != null) ? name.hashCode() : 0;
result = (31 * result) + ((addresses != null) ? addresses.hashCode() : 0);
return result;
}
}
class Address {
final String street;
Address(String street) {
this.street = street;
}
public String getStreet() {
return street;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if ((o == null) || (getClass() != o.getClass())) {
return false;
}
Address address = (Address) o;
if ((street != null) ? (!street.equals(address.street)) : (address.street != null)) {
return false;
}
return true;
}
@Override
public int hashCode() {
return (street != null) ? street.hashCode() : 0;
}
}
公共类AbcTest{
公共静态void main（字符串[]args）{
地址1=新地址（“街道1”）；
地址2=新地址（“街道2”）；
雇员雇员1=新雇员（“雇员1”）；
employee1.getAddresses（）.add（address1）；
employee1.getAddresses（）.add（address2）；
雇员雇员2=新雇员（“雇员2”）；
employee2.getAddresses（）.add（address2）；
公司A=新公司（“compA”）；
companyA.getEmployees（）.add（employee1）；
companyA.getEmployees（）.add（employee1）；//如果确实需要，您可以将员工添加到列表中任意次数？
companyA.getEmployees（）.add（employee2）；
//现在，要获取名为的员工，请简单地迭代员工列表
迭代器employeeIt=companyA.getEmployees（）.Iterator（）；
员工wantedEmployee=null；
while（employeeIt.hasNext（）&&（wantedEmployee==null））{
Employee next=employeeIt.next（）；
if（next.getName（）.equals（“employee1”））{
wantedEmployee=next；
}
}
System.out.println（wantedEmployee）；
}
}
阶级公司{
最后的字符串名；
最终员工名单；
公司（字符串名称）{
this.name=名称；
this.employees=new ArrayList（）；
}
公共字符串getName（）{
返回名称；
}
公开名单{
返回员工；
}
@凌驾
公共布尔等于（对象o）{
if（this==o）{
返回真值
I hope the below you are expecting the below scenario.
Company<Map>
--- Dept<Map>
---Classes<Map>
--- Employee<Map>
-id
-name
package com.java.examples;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
class Employee {
}
class Company {
}
class Dept {
}
public class Test1 {
public static void main(String[] args) {
Map<String, Employee> inputEmployeeMap = new HashMap<String, Employee>();
inputEmployeeMap.put("1", new Employee());
inputEmployeeMap.put("2", new Employee());
inputEmployeeMap.put("3", new Employee());
Map<String, Map<String, Employee>> inputClassList = new HashMap<String, Map<String, Employee>>();
inputClassList.put("class1", inputEmployeeMap);
inputClassList.put("class2", inputEmployeeMap);
Map<String, Map<String, Map<String, Employee>>> inputDeptList = new HashMap<String, Map<String, Map<String, Employee>>>();
inputDeptList.put("ece", inputClassList);
inputDeptList.put("csc", inputClassList);
Map<String, Map<String, Map<String, Map<String, Employee>>>> inputCompanyList = new HashMap<String, Map<String, Map<String, Map<String, Employee>>>>();
inputCompanyList.put("ABCCompany", inputDeptList);
// parseMap
parseMultiValueMap(inputCompanyList);
}
private static void parseMultiValueMap(
Map<String, Map<String, Map<String, Map<String, Employee>>>> inputCompanyList) {
Iterator<Entry<String, Map<String, Map<String, Map<String, Employee>>>>> it = inputCompanyList
.entrySet().iterator();
while (it.hasNext()) {
Map.Entry companyList = (Map.Entry) it.next();
System.out.println("Company Name = " + companyList.getKey()); // company
// name
Map deptList = (Map) companyList.getValue(); // department list
// iterator for department List.
Iterator companyListIterator = deptList.entrySet().iterator();
while (companyListIterator.hasNext()) {
Map.Entry departmentList = (Map.Entry) companyListIterator
.next();
System.out.print(" Dept Name = " + departmentList.getKey()); // department
// class map
Map classList = (Map) departmentList.getValue();
Iterator departmentListIterator = classList.entrySet()
.iterator();
Map.Entry empNames = (Map.Entry) departmentListIterator.next();
// class name
System.out.print(" class name = " + empNames.getKey());
// employee map
Map empNamesMap = (Map) empNames.getValue();
Iterator eNamesIt = empNamesMap.entrySet().iterator();
while (eNamesIt.hasNext()) {
Map.Entry name = (Map.Entry) eNamesIt.next();
// prints the emp Id
System.out.print(" id " + name.getKey());
System.out.print(" name = " + name.getValue());
}
System.out.println("\n");
}
System.out.println("\n");
}
}
}
Hope it helps.
public class AbcTest {
public static void main(String[] args) {
Address address1 = new Address("street1");
Address address2 = new Address("street2");
Employee employee1 = new Employee("employee1");
employee1.getAddresses().add(address1);
employee1.getAddresses().add(address2);
Employee employee2 = new Employee("employee2");
employee2.getAddresses().add(address2);
Company companyA = new Company("compA");
companyA.getEmployees().add(employee1);
companyA.getEmployees().add(employee1); // you can add employee to the list as many times as you want, if you really need this?
companyA.getEmployees().add(employee2);
// now to get the employee with give name simly iterate over list of employees
Iterator<Employee> employeeIt = companyA.getEmployees().iterator();
Employee wantedEmployee = null;
while (employeeIt.hasNext() && (wantedEmployee == null)) {
Employee next = employeeIt.next();
if (next.getName().equals("employee1")) {
wantedEmployee = next;
}
}
System.out.println(wantedEmployee);
}
}
class Company {
final String name;
final List<Employee> employees;
Company(String name) {
this.name = name;
this.employees = new ArrayList<>();
}
public String getName() {
return name;
}
public List<Employee> getEmployees() {
return employees;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if ((o == null) || (getClass() != o.getClass())) {
return false;
}
Company company = (Company) o;
if ((employees != null) ? (!employees.equals(company.employees)) : (company.employees != null)) {
return false;
}
if ((name != null) ? (!name.equals(company.name)) : (company.name != null)) {
return false;
}
return true;
}
@Override
public int hashCode() {
int result = (name != null) ? name.hashCode() : 0;
result = (31 * result) + ((employees != null) ? employees.hashCode() : 0);
return result;
}
}
class Employee {
final String name;
final Set<Address> addresses;
Employee(String name) {
this.name = name;
this.addresses = new HashSet<>();
}
public String getName() {
return name;
}
public Set<Address> getAddresses() {
return addresses;
}
@Override
public String toString() {
return "Employee{" +
"name='" + name + '\'' +
'}';
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if ((o == null) || (getClass() != o.getClass())) {
return false;
}
Employee employee = (Employee) o;
if ((addresses != null) ? (!addresses.equals(employee.addresses)) : (employee.addresses != null)) {
return false;
}
if ((name != null) ? (!name.equals(employee.name)) : (employee.name != null)) {
return false;
}
return true;
}
@Override
public int hashCode() {
int result = (name != null) ? name.hashCode() : 0;
result = (31 * result) + ((addresses != null) ? addresses.hashCode() : 0);
return result;
}
}
class Address {
final String street;
Address(String street) {
this.street = street;
}
public String getStreet() {
return street;
}
@Override
public boolean equals(Object o) {
if (this == o) {
return true;
}
if ((o == null) || (getClass() != o.getClass())) {
return false;
}
Address address = (Address) o;
if ((street != null) ? (!street.equals(address.street)) : (address.street != null)) {
return false;
}
return true;
}
@Override
public int hashCode() {
return (street != null) ? street.hashCode() : 0;
}
}