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我有一个实用程序列表,这些实用程序来自:

abstract public class BaseUtil implements Callable{
    public String name;
    public StreamWrapper data;
    public void setData(StreamWrapper stream){ this.data = stream; }
    private static Class me = BaseUtil.class;
    private static Method[] availableUtilities = me.getDeclaredMethods();
    public static Method[] getUtilities(){ return availableUtilities; }
}

我希望在每个节点上都能够为其分配一个实用程序,例如:

Initialize(String utilName){
    activeUtility = utilName;
    gk = new GateKeeper(BaseUtil.getUtilities() <something>);
}

public class GateKeeper {
  GateKeeper(BaseUtil util) {
    this.util = util;
}

private BaseUtil util;

但我不确定如何从传入的字符串中获取特定的实用程序类。一个示例实用程序是:

public class WordLengthUtil extends BaseUtil {
    private String name = "WordLength";
    public Integer call() {
        System.out.println(this.data);
        return Integer.valueOf(this.data.length());
    }
}
4

1 回答 1

2

您可以使用反射:

String name = "WordLength";
String className = hashMap.get(name);

Callable plugin = (Callable) Class.forName(className).newInstance();

使用 HashMap 存储类名和字符串标识符之间的绑定

于 2015-12-04T06:31:09.267 回答