java如何使用HashMap作为LinkedHashSet

Question

当我运行这个程序时：-

import java.util.HashMap;
import java.util.LinkedHashSet;
import java.util.Map;

public class try1 {

    public static void main(String args[]) {

        Map<String, Object> props = new HashMap<String, Object>();
        props.putIfAbsent("videos", new LinkedHashSet<String>());
        System.out.println(LinkedHashSet<String>(props.get("videos")).add("video_url");

    }
}

我有3个错误：-

try1.java:14: error: cannot find symbol
        System.out.println(LinkedHashSet<String>(props.get("videos")).add("yoyo"));//.add("video_url");
                           ^
  symbol:   variable LinkedHashSet
  location: class try1
try1.java:14: error: cannot find symbol
        System.out.println(LinkedHashSet<String>(props.get("videos")).add("yoyo"));//.add("video_url");
                                         ^
  symbol:   variable String
  location: class try1
try1.java:14: error: cannot find symbol
        System.out.println(LinkedHashSet<String>(props.get("videos")).add("yoyo"));//.add("video_url");
                                                                     ^
      symbol:   method add(String)
      location: class Object
    3 errors

我试图使用HashMap值作为LinkedHashSet，但得到了这些错误。我该怎么办

Answer 1

我猜你想用java。util。LinkedList作为Map的值

Map<String, List<String>> props = new HashMap<>();
props.putIfAbsent("videos", new LinkedList<String>());
props.get("videos").add("video_url");
System.out.println(props);

但是如果你想使用一个与put-order相同的顺序进行迭代的映射

Map<String, String> props = new LinkedHashMap<>();
props.putIfAbsent("videos", "video_url");
System.out.println(props);

Answer 2

正确的代码如下所示：

public static void main(String[] args) {
        Map<String, Object> props = new HashMap<String, Object>();
        props.putIfAbsent("videos", new LinkedHashSet<String>());
        System.out.println(((LinkedHashSet)(props.get("videos"))).add("video_url"));
    }

Answer 3

语法中有一个错误，首先需要将对象转换为LinkedHashSet，然后向其中添加一个字符串

   System.out.println( ((LinkedHashSet<String>)(props.get("videos")) ).add("video_url") );