/***@paramm1*@paramm2*@paramstring 根据某个key来合并*@return*/publicstaticList<Map<String,Object>> merge(List<Map<String,Object>> m1,List<Map<String,Object>>m2,String string){m1.addAll(m2);Set<String> set =newHashSet<>();returnm1.stream().collect(Collectors.groupingBy(o ->{//暂存...
* @Description: 合并两个list<map>,并将userId相同的其它属性合并 * @Title: mergeList * @param: @return * @return: List<Map<String,Object>> * @throws */publicstaticList<Map<String,Object>>mergeList(){//构建List集合1List<Map<String,Object>>list1=newArrayList<>();Map<String,Object>data=...
map1.put("num1", 1); Map<String, Object> map2 = new HashMap<>(); map2.put("date", "2021-11-25"); map2.put("num1", 2); Map<String, Object> map3 = new HashMap<>(); map3.put("date", "2021-11-26"); map3.put("num1", 3); Map<String, Object> map4 = new Has...
合并出来的List就是b,而不是一个新的List . 如果创建新的可以: ArrayList <String> c=(ArrayList <String> )a.clone; c.addAll(b); 实例: List1: List<Map<String, Object>> changedEntity = new ArrayList<Map<String, Object>>(); List2: List<Map<String,Object>> datalist = new ArrayList<Map...
map<String,Object> map1 = new HashMap<>(); map1.put("a_id",1); map1.put("in_num",10); map<String,Object> map2 = new HashMap<>(); map1.put("a_id",3); map1.put("in_num",10); map<String,Object> map3 = new HashMap<>(); map1.put("a_id",4); map1.put("in...
很简单，换个思路，先把两个list转成json数组合并，再转成list.json和list互转网上好多工具类
将结果收集到map中 先定义如下Person对象 class Person{ public String name; public int age; Person(String name, int age){ this.name = name; this.age = age; } @Override public String toString(){ return String.format("Person{name='%s', age=%d}", name, age); ...
给你一个参考的代码：HashMap listMap1 = new HashMap();for (int i = 0; i < 3; i++) { MyClass class1=new MyClass();class1.setClassID(i+1);class1.setClassName("班级"+(i+1));ArrayList list = new ArrayList();int count=(int) (Math.random() * 7)+1;for (int ...
您的INVALID_KEY和DUPLICATE_KEY从未被使用过，所以请删除它们，并将enum还原为boolean。
Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists ...