zouhong
============Java8之前的方式==========
Map<String, Integer> items = new HashMap<>();
items.put("A", 10);
items.put("B", 20);
items.put("C", 30);
items.put("D", 40);
items.put("E", 50);
items.put("F", 60);
for (Map.Entry<String, Integer> entry : items.entrySet()) {
    System.out.println("Item : " + entry.getKey() + " Count : " + entry.getValue());
}
============forEach + Lambda表达式==========
Map<String, Integer> items = new HashMap<>();
items.put("A", 10);
items.put("B", 20);
items.put("C", 30);
items.put("D", 40);
items.put("E", 50);
items.put("F", 60);
items.forEach((k,v)->System.out.println("Item : " + k + " Count : " + v));
items.forEach((k,v)->{
    System.out.println("Item : " + k + " Count : " + v);
    if("E".equals(k)){
        System.out.println("Hello E");
    }
});
 ———————————————— 

二遍历List: 
============Java8之前的方式==========

List<String> items = new ArrayList<>();
items.add("A");
items.add("B");
items.add("C");
items.add("D");
items.add("E");

for(String item : items){
    System.out.println(item);
}
============forEach + Lambda表达式==========
List<String> items = new ArrayList<>();
items.add("A");
items.add("B");
items.add("C");
items.add("D");
items.add("E");
//输出:A,B,C,D,E
items.forEach(item->System.out.println(item));
//输出 : C
items.forEach(item->{
    if("C".equals(item)){
        System.out.println(item);
    }
});

分类:

技术点:

相关文章:

  • 2021-12-08
  • 2021-10-06
  • 2021-05-30
  • 2021-10-01
  • 2021-11-02
  • 2021-10-10
  • 2022-01-15
  • 2021-05-19
猜你喜欢
  • 2021-08-08
  • 2021-10-12
  • 2020-03-30
  • 2021-05-23
  • 2020-04-24
  • 2020-01-13
  • 2021-08-03
相关资源
相似解决方案