List相关操作集合（不断更新...）

1. List去重合并

    // 两个不同的list 根据id去重合并
    List<T> list = new ArrayList<>();
    List<T> list1 = new ArrayList<>();
      list.forEach(o -> {
         list1.removeIf(next ->next.getId().equals(o.getId()));
    });
    list.addAll(list1)

2.List取某属性的所有值/List转数组

public static void main(String[] args) {
        List<Book> list = new ArrayList<>();

        list.add(new Book("1", "sql基础大全", 200));
        list.add(new Book("2", "Java基础", 500));
        log.info(list);
        list.forEach(book -> {
            book.toString();
            log.info(book.toString());
        });
        List<String> nameList = list.stream().map(Book -> Book.getName()+"_"+Book.getPrice()).collect(Collectors.toList());
        String[] strings = nameList.toArray(new String[nameList.size()]);
        for (String string : strings) {
            log.info("begin show string msg -------------");
            log.info(string);
        }
        log.info(nameList);
        log.info(strings);
    }

    @Data
    public static class Book{
        private String id;
        private String name;
        private int price;

        public Book(String s, String s1, int i) {
            setId(s);
            setName(s1);
            setPrice(i);
        }
    }

3.判断List中是否存在某一个值

List<Book> list = new ArrayList<>();

list.add(new Book("1", "sql基础大全", 200));
list.add(new Book("2", "Java基础", 500));

 if (list.stream().filter(w -> String.valueOf(w.getId()).equals("1")).findAny().isPresent()){
            System.out.println("存在");
        }

4.判断string数组中存在某个值

   String[] array = {"1","2","3","4"};
   boolean flag = Arrays.asList(array).contains("5");
   System.out.println("flag:" + flag);//--- flag:true