对于数据库去重相当简单,只需要在查询语句上添加DISTINCT关键字就可以搞定,但是针对Nosql数据库而言DISTINCT未必支持,这时候我们需要自己手动实现。这里有一个数据不重复的最简单的原理:字符串的hashcode比较,当然记录的唯一指标不是ID,而是数据本身,我们去掉ID再来比较是否重复就可以了。
Java 模拟代码:
- package com.boonya.mongo.client;
-
- import java.util.ArrayList;
- import java.util.HashMap;
- import java.util.List;
- import java.util.Map;
-
- public class LoveTest {
-
- public static void main(String[] args) {
- List<Person> list=new ArrayList<Person>();
-
- Person p=new Person("1", "boonya", 20);
- Person p2=new Person("https://files.jxasp.com/image/2", "boonya", 20);
- Person p3=new Person("3", "boonya", 20);
- Person p4=new Person("4", "boonya", 20);
-
- list.add(p);
- list.add(p2);
- list.add(p3);
- list.add(p4);
-
- Map<String,Object> map=new HashMap<String, Object>();
- for (Person person : list) {
- if(!map.containsKey(""+person.toString().hashCode())){
- map.put(person.toString().hashCode()+"", person);
- }
- }
-
- for (Object person : map.values()) {
- Person uniquePerson=(Person) person;
- System.out.println(uniquePerson.getId()+" "+uniquePerson);
- }
- }
-
-
- }
- class Person{
- private String id;
- private String name;
- private int age;
- public String getId() {
- return id;
- }
- public void setId(String id) {
- this.id = id;
- }
- public String getName() {
- return name;
- }
- public void setName(String name) {
- this.name = name;
- }
- public int getAge() {
- return age;
- }
- public void setAge(int age) {
- this.age = age;
- }
- public Person(String id, String name, int age) {
- super();
- this.id = id;
- this.name = name;
- this.age = age;
- }
-
- @Override
- public String toString() {
- return "Person [" + (name != null ? "name=" + name + ", " : "")
- + "age=" + age + "]";
- }
-
-
- }
测试结果 :
1 Person [name=boonya, age=20]
hash碰撞的情况本文不考虑。
Java8 去重:
List<String> unique = list.stream().distinct().collect(Collectors.toList());