我正在为java类创建一个员工时钟。我的计划的这一部分用于报告个人的时间，并报告所有员工的时间。我的代码适用于个人，但我无法将其转换为所有员工的工作。我应该尝试循环遍历整个文件并检索它吗？控制语句中的信息导致我出现问题。另外，仅查看两周时间，使用日历和日期-14天是否是实现这一目标的好方法？
有关如何继续欣赏的任何反馈意见。
package PunchinPunchout;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.ArrayList;
import java.util.Calendar;
import java.util.Date;
import java.util.Scanner;
public class IDchecker {
private static BufferedReader br;
private static BufferedReader br1;
static int total;
static int total1;
public static void main(String args[]) throws IOException {
getsubject();
}
public static void getsubject() throws FileNotFoundException, IOException {
ArrayList<Integer> totalhours = new ArrayList<>();
br = new BufferedReader(new FileReader("timeclock1.txt"));
br1 = new BufferedReader(new FileReader("newemployee8.txt"));
String line = "";
String line1 = "";
Scanner sc = new Scanner(System.in);
System.out.print("Enter an employee ID number: ");
String idnumber = sc.next();//read the choice
sc.nextLine();// discard any other data entered on the line
while ((line1 = br1.readLine()) != null) {
if (line1.contains(idnumber)) {
System.out.println("Employee Name & ID ");
System.out.println(line1);
}
}
while ((line = br.readLine()) != null) {
if (line.contains(idnumber + " ") && line.contains("in")) {
System.out.println();
System.out.println(" Date Time ID Punched");
System.out.println(line);
String regexp = "[\\s:\\n]+"; // these are my delimiters
String[] tokens; // here i will save tokens
for (int i = 0; i < 1; i++) {
tokens = line.split(regexp);
total = Integer.parseInt(tokens[1]);
}
} else if (line.contains(idnumber + " ") && line.contains("out")) {
System.out.println(line);
String regexp = "[\\s:\\n]+";
String[] tokens;
for (int i = 0; i < 1; i++) {
tokens = line.split(regexp);
total1 = Integer.parseInt(tokens[1]);
System.out.print("Total hours for " + tokens[0] + " are: ");
}
int dailytotal = total1 - total;
System.out.println(dailytotal + " hours");
totalhours.add(dailytotal);
}
}
System.out.println();
int sum = totalhours.stream().mapToInt(Integer::intValue).sum();
System.out.println("The total hours for the last two weeks is " + sum + " hours.");
}
}
*来自timeclock1.txt的输出
05/05/2014 05:00:00 508 in
05/05/2014 09:00:00 508 out
05/05/2014 03:00:00 509 in
05/05/2014 09:00:00 509 out
05/05/2014 03:00:00 510 in
05/05/2014 08:00:00 510 out
05/05/2014 08:00:00 511 in
05/05/2014 10:00:00 511 out
*来自newemployee8.txt的输出
詹姆斯布什10bobby bush 11
john hunt 12
mick jag 13
jacob sanchez 14
答案 0 :(得分：1)
好的，这只是一个顶级的例子，但它强调了像Java这样的OO语言的强大功能......
根据您的要求，可以通过多种方式实现这一目标。我已经做了一些假设（比如in后跟同一个雇员的out，但是基本的要点已经证明了。
目的是将一些功能集中到可重用和可管理的块中，从而减少代码重复。简化了对数据的访问，因为它在内存中完成，速度更快......
首先，您需要创建员工和时钟数据的对象表示，这样可以更轻松地管理...
员工示例
public class Employee {
private final int id;
private final String name;
public Employee(String text) {
String[] parts = text.split(" ");
id = Integer.parseInt(parts[2]);
name = parts[0] + " " + parts[1];
}
public String getName() {
return name;
}
public int getId() {
return id;
}
}
TimeClockEntry示例
public class TimeClockEntry {
private Date inTime;
private Date outTime;
private int employeeID;
public TimeClockEntry(String text) throws ParseException {
String parts[] = text.split(" ");
employeeID = Integer.parseInt(parts[2]);
setClockTimeFrom(text);
}
public void setClockTimeFrom(String text) throws ParseException {
String parts[] = text.split(" ");
if ("in".equalsIgnoreCase(parts[3])) {
inTime = CLOCK_DATE_TIME_FORMAT.parse(parts[0] + " " + parts[1]);
} else if ("out".equalsIgnoreCase(parts[3])) {
outTime = CLOCK_DATE_TIME_FORMAT.parse(parts[0] + " " + parts[1]);
}
}
public int getEmployeeID() {
return employeeID;
}
public Date getInTime() {
return inTime;
}
public Date getOutTime() {
return outTime;
}
}
现在，我们需要某种“管理器”来管理这两个类的细节，这些管理器应该提供允许用来检索它们管理的信息的访问方法。这些经理还将负责从文件中加载数据......
EmployeeManager示例
public class EmployeeManager {
private Map<Integer, Employee> employees;
public EmployeeManager() throws IOException {
employees = new HashMap<>(25);
try (BufferedReader br = new BufferedReader(new FileReader(new File("NewEmployee8.txt")))) {
String text = null;
while ((text = br.readLine()) != null) {
Employee emp = new Employee(text);
employees.put(emp.getId(), emp);
}
}
}
public List<Employee> getEmployees() {
return Collections.unmodifiableList(new ArrayList<Employee>(employees.values()));
}
public Employee getEmployee(int id) {
return employees.get(id);
}
}
TimeClockManager示例
public class TimeClockManager {
private Map<Integer, List<TimeClockEntry>> timeClockEntries;
public TimeClockManager() throws IOException, ParseException {
timeClockEntries = new HashMap<>(25);
try (BufferedReader br = new BufferedReader(new FileReader(new File("TimeClock1.txt")))) {
String text = null;
TimeClockEntry entry = null;
int line = 0;
while ((text = br.readLine()) != null) {
if (line % 2 == 0) {
entry = new TimeClockEntry(text);
} else {
entry.setClockTimeFrom(text);
List<TimeClockEntry> empEntries = timeClockEntries.get(entry.getEmployeeID());
if (empEntries == null) {
empEntries = new ArrayList<>(25);
timeClockEntries.put(entry.getEmployeeID(), empEntries);
}
empEntries.add(entry);
}
line++;
}
}
}
public List<TimeClockEntry> getByEmployee(Employee emp) {
List<TimeClockEntry> list = timeClockEntries.get(emp.getId());
list = list == null ? new ArrayList<>() : list;
return Collections.unmodifiableList(list);
}
}
现在，在内部，这些经理通过使用Map来管理数据，以便更容易找到数据，具体而言，这最关键的是员工的身份
现在，一旦我们拥有了这些，我们可以随时向您索取信息......
public Report() {
try {
EmployeeManager empManager = new EmployeeManager();
TimeClockManager timeClockManager = new TimeClockManager();
for (Employee emp : empManager.getEmployees()) {
System.out.println("[" + emp.getId() + "] " + emp.getName());
for (TimeClockEntry tce : timeClockManager.getByEmployee(emp)) {
System.out.println(" "
+ CLOCK_DATE_TIME_FORMAT.format(tce.getInTime())
+ " to "
+ CLOCK_DATE_TIME_FORMAT.format(tce.getOutTime()));
}
}
} catch (IOException | ParseException exp) {
exp.printStackTrace();
}
}
另一种方法是将两个经理合并为一个班级。基本的想法是加载员工和时钟数据，时钟数据将成为Employee的属性，您可以直接访问它。
这是一个稍微优雅的解决方案，因为您将所有数据包含在单个构造中，但可能无法满足您的需求
完全可运行的示例
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.text.SimpleDateFormat;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Date;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import oracle.jrockit.jfr.parser.ParseException;
public class Report {
public static void main(String[] args) {
new Report();
}
public Report() {
try {
EmployeeManager empManager = new EmployeeManager();
TimeClockManager timeClockManager = new TimeClockManager();
for (Employee emp : empManager.getEmployees()) {
System.out.println("[" + emp.getId() + "] " + emp.getName());
for (TimeClockEntry tce : timeClockManager.getByEmployee(emp)) {
System.out.println(" "
+ CLOCK_DATE_TIME_FORMAT.format(tce.getInTime())
+ " to "
+ CLOCK_DATE_TIME_FORMAT.format(tce.getOutTime()));
}
}
} catch (IOException | ParseException exp) {
exp.printStackTrace();
}
}
public class EmployeeManager {
private Map<Integer, Employee> employees;
public EmployeeManager() throws IOException {
employees = new HashMap<>(25);
try (BufferedReader br = new BufferedReader(new FileReader(new File("NewEmployee8.txt")))) {
String text = null;
while ((text = br.readLine()) != null) {
if (!text.trim().isEmpty()) {
Employee emp = new Employee(text);
employees.put(emp.getId(), emp);
}
}
}
}
public List<Employee> getEmployees() {
return Collections.unmodifiableList(new ArrayList<Employee>(employees.values()));
}
public Employee getEmployee(int id) {
return employees.get(id);
}
}
public class TimeClockManager {
private Map<Integer, List<TimeClockEntry>> timeClockEntries;
public TimeClockManager() throws IOException, ParseException {
timeClockEntries = new HashMap<>(25);
try (BufferedReader br = new BufferedReader(new FileReader(new File("TimeClock1.txt")))) {
String text = null;
TimeClockEntry entry = null;
int line = 0;
while ((text = br.readLine()) != null) {
if (!text.trim().isEmpty()) {
if (line % 2 == 0) {
entry = new TimeClockEntry(text);
} else {
entry.setClockTimeFrom(text);
List<TimeClockEntry> empEntries = timeClockEntries.get(entry.getEmployeeID());
if (empEntries == null) {
empEntries = new ArrayList<>(25);
timeClockEntries.put(entry.getEmployeeID(), empEntries);
}
empEntries.add(entry);
}
line++;
}
}
}
}
public List<TimeClockEntry> getByEmployee(Employee emp) {
List<TimeClockEntry> list = timeClockEntries.get(emp.getId());
list = list == null ? new ArrayList<>() : list;
return Collections.unmodifiableList(list);
}
}
public class Employee {
private final int id;
private final String name;
public Employee(String text) {
System.out.println("[" + text + "]");
for (char c : text.toCharArray()) {
System.out.print((int) c + ",");
}
System.out.println("");
String[] parts = text.split("\\s+");
id = Integer.parseInt(parts[2]);
name = parts[0] + " " + parts[1];
}
public String getName() {
return name;
}
public int getId() {
return id;
}
}
public static final SimpleDateFormat CLOCK_DATE_TIME_FORMAT = new SimpleDateFormat("dd/MM/yyyy HH:mm:ss");
public static final SimpleDateFormat CLOCK_DATE_FORMAT = new SimpleDateFormat("dd/MM/yyyy");
public class TimeClockEntry {
private Date inTime;
private Date outTime;
private int employeeID;
public TimeClockEntry(String text) throws ParseException {
System.out.println("[" + text + "]");
for (char c : text.toCharArray()) {
System.out.print((int) c + ",");
}
System.out.println("");
String parts[] = text.split("\\s+");
employeeID = Integer.parseInt(parts[2]);
setClockTimeFrom(text);
}
public void setClockTimeFrom(String text) throws ParseException {
String parts[] = text.split("\\s+");
if ("in".equalsIgnoreCase(parts[3])) {
inTime = CLOCK_DATE_TIME_FORMAT.parse(parts[0] + " " + parts[1]);
} else if ("out".equalsIgnoreCase(parts[3])) {
outTime = CLOCK_DATE_TIME_FORMAT.parse(parts[0] + " " + parts[1]);
}
}
public int getEmployeeID() {
return employeeID;
}
public Date getInTime() {
return inTime;
}
public Date getOutTime() {
return outTime;
}
}
}