关于Java：从用户输入设置枚举

Setting an enum from user input

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选择Java备份并尝试熟悉EnUM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。我已经创建了所有的东西，但我挂断了设置一个联系类型(家庭，朋友，业务等…)我在一个单独的Java类中设置了一个EnUM类。

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public class ContactType
{
public enum contactType
{
Family,
Church,
Friend,
BusninessColleague,
ServicePerson,
Customer,
Other
}
}

我的联系人类看起来像：

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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 ="[email protected]";
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之外，其他所有功能都可以工作，所以为了保持简短，我刚刚包含了这段代码片段)：

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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c1.setContactType(strContactType);中做了一些错误的事情，因为我得到了"类型contacts中的方法set contacttype(contacttype)不适用于参数(string)"，因为我完全不知道如何修复它，将contacttype设置为用户输入的任何内容。