我已經創建了一個新類「ServiceRequest」,如下所示。我沒有在這裏添加getter和setter來節省空間。創建一個Java類的新對象返回一個空對象
package testListenerPackage;
import java.util.Date;
public class ServiceRequest {
public static final ServiceRequest REQUEST_ARRIVAL = new ServiceRequest("Request_Arrival");
public static final ServiceRequest REQUEST_COMPLETION = new ServiceRequest("Request_Completion");
public static final ServiceRequest REQUEST_UNDER_PROCESS = new ServiceRequest("Request_Under_Process");
private String serviceRequest;
private String requestName;
private int requestID;
private long arrivalTime;
private long startServiceTime;
private long endServiceTime;
private long totalServiceTime;
private String requestStatus;
public enum RequestStatus{
NEW, COMPLETED
}
public ServiceRequest()
{
}
public ServiceRequest(String serviceRequest) {
serviceRequest = serviceRequest;
}
}
當我嘗試在像這樣的其他類中創建此對象的實例時,它將返回一個空對象。
public ServiceRequest generateServiceRequest()
{
ServiceRequest serviceRequest = new ServiceRequest("Ali baba");
serviceRequest.setRequestID(1);
serviceRequest.setRequestName("Read");
serviceRequest.setRequestStatus(ServiceRequest.REQUEST_ARRIVAL.toString());
serviceRequest.setArrivalTime(System.currentTimeMillis());
return serviceRequest;
}
有人可以告訴我該怎麼辦?
你是什麼意思「返回空對象」?你是否得到一個NullPointerException?哪條線? – 2011-12-19 18:45:19
另外你的構造函數中的'serviceRequest = serviceRequest'行沒有完成任何事情。您需要將其更改爲'this.serviceRequest = serviceRequest;' – 2011-12-19 18:45:44
看起來應該起作用。是否有其他可能導致問題的代碼?你是否將其剝離到_just_發佈的代碼? – cdeszaq 2011-12-19 18:46:22