【发布时间】:2015-04-26 14:17:12
【问题描述】:
选择 java 备份并尝试熟悉枚举类型。我正在尝试创建一个通讯簿,用户可以在其中创建联系人。我已经创建了所有内容,但是我一直在设置联系人类型(家庭、朋友、商务等)我已经在一个单独的 java 类中设置了一个枚举类。
public class ContactType
{
public enum contactType
{
Family,
Church,
Friend,
BusninessColleague,
ServicePerson,
Customer,
Other
}
}
我的联系人类看起来像:
public class Contacts
{
private contactType contact;
private String name;
private String streetAddress;
private String city;
private String state;
private String zipCode;
private String phone;
private String email;
private String photo;
public Contacts ( )
{
contact = null;
name = "XXX XXX";
streetAddress = "XXX";
state = "XX";
zipCode = "00000";
phone = "XXX-XXXX";
email = "XXX@XXX.COM";
photo = "XXX.jpg";
}
public Contacts (ContactType contactType, String name, String streetAddress, String city,
String state, String zipCode, String phone, String email, String photo)
{
this.contact = contactType;
this.name = name;
this.streetAddress = streetAddress;
this.city = city;
this.state = state;
this.zipCode = zipCode;
this.phone = phone;
this.email = email;
this.photo = photo;
}
public ContactType getContactType ( )
{
return contact;
}
public void setContactType (ContactType input)
{
this.contact = input;
}
//rest of code
最后是我的驱动程序,它包括一个菜单(除了设置联系人类型外,其他所有功能都有效,所以为了简短起见,我只包含了该 sn-p):
switch (iSelection)
{
case 1:
c1 = new Contacts(); //creates a new contact
break;
case 2:
strContactType = JOptionPane.showInputDialog ("Please enter contact type (Family, Church, BusinessColleague, ServicePerson, Customer, or Other)");
contactType.valueOf(strContactType);
JOptionPane.showMessageDialog (null, strContactType);
c1.setContactType (strContactType);
break;
我知道我在这里做错了c1.setContactType(strContactType);,因为我完全无知,我不知道“联系人类型中的方法 setContactType(ContactType) 不适用于参数 (String)”如何修复它以将 contactType 设置为用户输入的任何内容。
【问题讨论】:
-
我已经删除了您的javascript 标签,因为您的问题与使用这种语言无关。请理解这是两种完全不同的编程语言,就像 ham 和 hamburger 密切相关,如果你错误地标记了你的问题,你将无法让合适的专家来审查它,这可能会损害你获得答案的机会体面的帮助。