Question

嗨，我有这个哈希图的数组列表

[{EndTime=09:00 AM, UserId=48, StartTime=08:00 AM, AppointmentId=79, Date=11/18/13},
{EndTime=09:00 AM, UserId=48, StartTime=08:00 AM, AppointmentId=80, Date=11/18/13},
{EndTime=09:00 AM, UserId=48, StartTime=08:00 AM, AppointmentId=81, Date=11/18/13},
{EndTime=09:00 AM, UserId=48, StartTime=08:00 AM, AppointmentId=82, Date=11/18/13},
{EndTime=09:00 AM, UserId=48, StartTime=08:00 AM, AppointmentId=83, Date=11/18/13},
{EndTime=09:00 AM, UserId=48, StartTime=08:00 AM, AppointmentId=85, Date=11/18/13}]

我想使用“AppoinmentID”从这里检查特定条目，我想获得不同的hashmap和其他所有其他的记录...我怎么能这样做？提前谢谢。

Answer 1

将这些值存储在hashmap中并不是一个好主意。你为什么不创建一个约会课。在这种情况下，删除约会对象将很容易。

public class Appointment
{
    private int     appointmentId;
    private int     userId;     // or private User user
    private Date    start;
    private Date    end;

    public Appointment(int id)
    {
        this.appointmentId = id;
    }

    // getters and setters

    @Override
    public boolean equals(Object obj)
    {
        if (this == obj)
        {
            return true;
        }
        if (obj == null)
        {
            return false;
        }
        if (this.getClass() != obj.getClass())
        {
            return false;
        }
        Appointment other = (Appointment) obj;
        if (this.appointmentId != other.appointmentId)
        {
            return false;
        }
        return true;
    }

}

现在，如果您要删除具有特定ID的特定项目：

List<Appointment> appointments = new ArrayList<Appointment>();
appointments.remove(new Appointment(theIdYouWantToDelete));

或更好的方式：

像这样存储它们

Map<Integer, Appointment> appointments = new HashMap<Integer, Appointment>();
appointments.put(appointment.getAppointmentId(), appointment);

并将其删除：

appointments.remove(theIdYouWantToDelete);

使用这种方法，您不需要equals方法。

为什么会这样：

如果要从List或Map中删除Object，Java会使用equals方法来识别它们。正如您所见，我只检查appointmentId。因此，如果2个对象的ID相同，则Java表示它们是同一个对象。如果你不重写equals，只检查==（内存中的同一个对象）的检查，但大多数情况并非如此。

Answer 2

你可以上课：

public class Apointment{
Stirng EndTime="09:00 AM";
int AppointmentId=79;
...
...
}

并且有一个带有apointmentId作为键的

的hashmap

HashMap<Integer,Apointment> map=new  HashMap<Integer,Apointment>();
Apointment ap=new Apointment(...);

map.put(ap.getAppointmentId(),ap);
..
..
..

如果你有apointmentID，你可以通过以下方式获得apointment对象：

Apointment ap=map.get(79);

Answer 3

1.创建一个类

public class Appointment
{
 public int     appointmentId;
 public int     userId;    
 public Date    startTime;
 public Date    endTime;
 public Appointment(int id,int aUserID,Date aStartTime,Date aEndTime)
 {
    this.appointmentId = id;
    this.userId = aUserID;
    this.startTime = aStartTime;
    thiis.endTime =  aEndTime;
 }
}

2。创建约会对象并在HashMap中存储

String dateFormat = "MMMM d, yyyy HH:mm"; //any date format
 DateFormat df = new SimpleDateFormat(dateFormat);       
 Date startDate = df.parse("January 2, 2010 13:00");
 Date endDate = df.parse("January 2, 2010 20:00");
 Appointment appointment1 = new Appointment(1,23,startDate,endDate);

 ...
 Map<Integer, Appointment> appointments = new HashMap<Integer, Appointment>();
 // add to hashmap making appointment id as key
 appointments.put(appointment1.appointmentId,appointment1);
 ......
 ...
 appointments.put(appointmentN.appointmentId,appointmentN);

3。删除约会对象

 appointments.remove(aAppointmentId)

4。获得约会对象

Appointment ap = appointments.get(aAppointmentId);
System.out.printLn("id "+ ap.appointmentId);
System.out.printLn("userId "+ ap.userId);
DateFormat df = new SimpleDateFormat(dateFormat);       
System.out.printLn("starttime "+  df.format(ap.startTime));
System.out.printLn("endtime "+  df.format(ap.endTime));

从ArrayList中删除元素<hashmap <string，string =“”>＆gt; </hashmap <string，>

3 个答案: