1.user实体
package com.demo.dto;
public class User { private Integer id;
private String userName;
private String password;
private Integer age; private long c;
public User() { super(); // TODO Auto-generated constructor stub }
public User(Integer id, String userName, String password, Integer age) { super(); this.id = id; this.userName = userName; this.password = password; this.age = age; }
public Integer getId() { return id; }
public void setId(Integer id) { this.id = id; }
public String getUserName() { return userName; }
public void setUserName(String userName) { this.userName = userName == null ? null : userName.trim(); }
public String getPassword() { return password; }
public void setPassword(String password) { this.password = password == null ? null : password.trim(); }
public Integer getAge() { return age; }
public void setAge(Integer age) { this.age = age; }
public long getC() { return c; }
public void setC(long l) { this.c = l; }
@Override public String toString() { return "User [id=" id ", userName=" userName ", password=" password ", age=" age ", c=" c "]"; } public static void main(String[] args) { User u1=new User(1, "aa", "aap", 23); User u2=new User(2, "aa", "aap", 23); User u3=new User(3, "bb", "aap", 23); User u4=new User(4, "cc", "aap", 23); User u5=new User(5, "cc", "aap", 23); User u6=new User(6, "cc", "aap", 23); User u7=new User(7, "aa", "aap", 24); List<User> list=new ArrayList<User>(); list.add(u1);list.add(u2);list.add(u7);list.add(u3);list.add(u4);list.add(u5);list.add(u6); //原有list(根据第二个字段:userName和第四个字段:age 统计重复的记录数) //jdk8的方法统计个数: Map<String, Map<Integer, Long>> map = list.stream().collect(Collectors.groupingBy(User::getUserName,Collectors.groupingBy(User::getAge,Collectors.counting()))); //jdk8以下: Map<String,Map<Integer,Long>> map=new HashMap<String,Map<Integer,Long>>(); for (User user1 : list) { Map<Integer, Long> value=new HashMap<Integer,Long>(); long count=0; if(map.containsKey(user1.getUserName())){ continue; } for(int i=0;i<list.size();i ){ if(user1.getUserName().equals(list.get(i).getUserName())&&user1.getAge()==list.get(i).getAge()){ count =1; value.put(user1.getAge(),count); map.put(user1.getUserName(), value); }else if(user1.getUserName().equals(list.get(i).getUserName())&&user1.getAge()!=list.get(i).getAge()){ value.put(list.get(i).getAge(),Long.valueOf(1)); map.put(user1.getUserName(), value); } } } map.forEach((k, v) -> { System.out.println(k ">>>>" v); }); List<User> list2=new ArrayList<User>(); list.forEach(user ->{ map.forEach((k, v) -> { if(k==user.getUserName()){ Long remove = v.remove(user.getAge()); user.setC(null==remove?0:remove); } }); list2.add(user); }); //遍历最后想要的结果(User中c为统计后的个数,方便前台遍历集合时单元格合并行) list2.forEach(u ->{ System.out.println(u); }); } }
备注:运行结果如下
cc>>>>{23=3} bb>>>>{23=1} aa>>>>{23=2, 24=1} User [id=1, userName=aa, password=aap, age=23, c=2] User [id=2, userName=aa, password=aap, age=23, c=0] User [id=7, userName=aa, password=aap, age=24, c=1] User [id=3, userName=bb, password=aap, age=23, c=1] User [id=4, userName=cc, password=aap, age=23, c=3] User [id=5, userName=cc, password=aap, age=23, c=0] User [id=6, userName=cc, password=aap, age=23, c=0]
此处是为了实现如下效果: