MC *_*ror 5

好吧,如果您的所有案例都具有这种确切的形式,那么您可以简单地执行以下操作:

String surname;
switch (name) {
    case "name1":
        surname = "surname1";
        break;
    case "name2":
        surname = "surname2";
        break;
    default:
        surname = "";
        break;
}
String fullname = String.format("%s %s", name, surname);

或者使用 aMap来查找值:

Map<String, String> mapping = new HashMap<>();
mapping.put("name1", "surname1");
mapping.put("name2", "surname2");
mapping.put("name3", "surname3");

String surname = mapping.get(name); // Assuming name is always a valid map entry
String fullname = String.format("%s %s", name, surname);

Bri*_*etz 5

其他地方概述的Map方法可能是 Java 8 最好的方法,但请注意,从 Java 14 开始,您可以使用 switch 表达式:

String surname = switch (name) { 
    case "name1" -> "surname1";
    case "name2" -> "surname2";
    ...
};
String fullName = String.format("%s %s", name, surname);
 

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