例如,以列表遍历字符串<>和List <名单<字符串>>使用相同的功能,我可以写这样的事:如何使用相同的递归函数迭代Map <String,String>和Map <String,Map <String,String >>?
import java.util.*;
public class Test{
public static void print(Object obj) {
if(obj instanceof List){
List list=(List)obj;
System.out.print("[");
for(Object obj2 : list){
print(obj2);
}
System.out.print("]");
}else{
System.out.print(obj+",");
}
}
public static void main(String[] args){
String l0="a";
System.out.println(l0);
List<String> l1=Arrays.asList("a","b");
print(l1);
System.out.println("");
List<List<String> > l2=Arrays.asList(Arrays.asList("a","b"),Arrays.asList("c","d"));
print(l2);
}
}
输出:
a
[a b ]
[[a b ][c d ]]
现在我想遍历地图< String,String>和Map < String,Map < String,String >>同样,我试过了:
import java.util.*;
public class Test{
public static void print(Object obj) {
if(obj instanceof Map){
System.out.print("{");
Map map=(Map)obj;
for(Map.Entry<Object,Object> entry : map.entrySet()){
print(entry.getKey());
System.out.print(":");
print(entry.getValue());
System.out.print(",");
}
System.out.print("}");
}else{
System.out.print(obj);
}
}
public static void main(String[] args){
String m0="a";
print(m0);
System.out.println("");
Map<String,String> m1=new HashMap<String,String>();
m1.put("surname","Tom");
m1.put("lastname","Bob");
print(m1);
System.out.println("");
Map<String,HashMap<String,String>> m2=new HashMap<String,HashMap<String,String>>();
HashMap<String,String> mm1=new HashMap<String,String>();
mm1.put("surname","Tom");
mm1.put("lastname","Bob");
mm1.put("nickname","Penguin");
m2.put("owner",mm1);
HashMap<String,String> mm2=new HashMap<String,String>();
mm2.put("name","Lucky");
mm2.put("type","cat");
m2.put("pet",mm2);
print(m2);
}
}
其预期的输出是一样的东西:
a
{surname:Tom,lastname:Bob,}
{owner:{surname:Tom,nickname:Penguin,lastname:Bob,},pet:{name:Lucky,type:cat,},}
,但它不能编译:
Test.java:20: error: incompatible types: Object cannot be converted to Entry<Object,Object>
for(Map.Entry<Object,Object> entry : map.entrySet()){
的原因是什么?是否有可能修复它?如果不是,我如何递归地迭代嵌套的地图,就像在显示的开始递归迭代List一样?
为什么定义一个'Object',而不是打印了'Map'参数()? – davidxxx
你可以试试:Map