关于Map取值对应到的键问题

Question

String[] codeA="黯 鏖 鳌 鏊 霭 螯 聱 鹌 锿 锕";
String[] codeB="鬓 鞴 魃 髌 髀 鳔 鳊 鲅 鐾 踣";
String[] codeC="黪 黜 魑 鲳 雠 龊 龀 躔 蹴 蹰";
String[] codeD="黩 黛 髑 骶 鞑 靼 鲽 鲷 貂 踱";Map<String, String[]> map = new HashMap<String, String[]>();
map.put("a", codeA);
map.put("b", codeB);
map.put("c", codeC);
map.put("d", codeD);
因为我把codeA,B,C,D都存放到map里去了，也就意味着codeA或者B他们缩对应的字的键为a,b
现在我想要把这些字的全部键位都输出出来，请问该怎么做呢？？？

Answer 1

不知道LZ是不是要遍历所有的key，是的话看看下的代码 public static void main(String[] args){
String[] codeA={"黯","鏖","鳌 ","鏊","霭","螯","聱","鹌","锿","锕"};
String[] codeB={"鬓","鞴","魃","髌","髀","鳔","鳊","鲅","鐾 ","踣"};
String[] codeC={"黪","黜","魑","鲳","雠 ","龊"," 龀"," 躔"," 蹴"," 蹰"};
String[] codeD={"黩","黛","髑 ","骶 ","鞑 ","靼"," 鲽 ","鲷 ","貂 ","踱"}; Map<String, String[]> map = new HashMap<String, String[]>();
map.put("a", codeA);
map.put("b", codeB);
map.put("c", codeC);
map.put("d", codeD);
Set<String> keyset = map.keySet();
Iterator<String> iterator = keyset.iterator();
while(iterator.hasNext()){
System.out.println(iterator.next());
}
}

Answer 2

我的意思是要求把codeA  codeB  codeC   codeD里的每个值所对应的键一起输出出来噢，是这个意思。

Answer 3

这么说吧，就是把codeA codeB codeC codeD的值所对应的键全部都放到一个集合里去，然后输出出来。

Answer 4

试试看  要什么  public static void main(String[] args){
        String[] codeA={"黯","鏖","鳌 ","鏊","霭","螯","聱","鹌","锿","锕"};
        String[] codeB={"鬓","鞴","魃","髌","髀","鳔","鳊","鲅","鐾 ","踣"};
        String[] codeC={"黪","黜","魑","鲳","雠 ","龊"," 龀"," 躔"," 蹴"," 蹰"};
        String[] codeD={"黩","黛","髑 ","骶 ","鞑 ","靼"," 鲽 ","鲷 ","貂 ","踱"};        Map<String, String[]> map = new HashMap<String, String[]>();
        map.put("a", codeA);
        map.put("b", codeB);
        map.put("c", codeC);
        map.put("d", codeD);
               for(String key : map.keySet() ){
            System.out.println(key + "   " + map.get(key));
        }
    }

Answer 5

 Set<String> keyset = map.keySet();
        Iterator<String> iterator = keyset.iterator();
        while(iterator.hasNext()){
            System.out.println(iterator.next());
        }
这个不就是把所有的键都输出吗，输出a b c d不是这样吗？

Answer 6

public static void main(String[] args){
        String[] codeA={"黯","鏖","鳌 ","鏊","霭","螯","聱","鹌","锿","锕"};
        String[] codeB={"鬓","鞴","魃","髌","髀","鳔","鳊","鲅","鐾 ","踣"};
        String[] codeC={"黪","黜","魑","鲳","雠 ","龊"," 龀"," 躔"," 蹴"," 蹰"};
        String[] codeD={"黩","黛","髑 ","骶 ","鞑 ","靼"," 鲽 ","鲷 ","貂 ","踱"};        Map<String, String[]> map = new HashMap<String, String[]>();
        map.put("a", codeA);
        map.put("b", codeB);
        map.put("c", codeC);
        map.put("d", codeD);
               for(String key : map.keySet() ){
            System.out.println(key + "   " + map.get(key));
        }
    }貌似这个不错。

Answer 7

public static void main(String[] args) {
String[] codeA={"黯","鏖","鳌 ","鏊","霭","螯","聱","鹌","锿","锕"};
        String[] codeB={"鬓","鞴","魃","髌","髀","鳔","鳊","鲅","鐾 ","踣"};
        String[] codeC={"黪","黜","魑","鲳","雠 ","龊"," 龀"," 躔"," 蹴"," 蹰"};
        String[] codeD={"黩","黛","髑 ","骶 ","鞑 ","靼"," 鲽 ","鲷 ","貂 ","踱"};        Map<String, String[]> map = new HashMap<String, String[]>();
        map.put("a", codeA);
        map.put("b", codeB);
        map.put("c", codeC);
        map.put("d", codeD);
for (Entry<String, String[]> e : map.entrySet()) {
for (String v : e.getValue()) {
System.out.println(e.getKey()+":"+v);
}
}
}