• java中实现对list的模糊查询


    比如我有下面这样一个List,里面存放的是多个Employee对象。然后我想对这个List进行按照Employee对象的名字进行模糊查询。有什么好的解决方案么?
    比如我输入的查询条件为“wang”,那么应该返回只包含employee1的List列表。

    List list = new ArrayList();
    Employee employee1 = new Employee();
    employee1.setName("wangqiang");
    employee1.setAge(30);
    list.add(employee1);
    Employee employee2 = new Employee();
    employee2.setName("lisi");
    list.add(employee2);
    employee2.setAge(25);

    方式一:

    public List search(String name,List list){
       List results = new ArrayList();
       Pattern pattern = Pattern.compile(name);
       for(int i=0; i < list.size(); i++){
          Matcher matcher = pattern.matcher(((Employee)list.get(i)).getName());
          if(matcher.matches()){
             results.add(list.get(i));
          }
       }
       return results;
    }
    

    上面那个是大小写敏感的,如果要求大小写不敏感,改成:
    Pattern pattern = Pattern.compile(name,Pattern.CASE_INSENSITIVE);

    并且上面那个是精确查询,如果要模糊匹配,matcher.find()即可以进行模糊匹配。

    public List search(String name,List list){
       List results = new ArrayList();
       Pattern pattern = Pattern.compile(name);
       for(int i=0; i < list.size(); i++){
          Matcher matcher = pattern.matcher(((Employee)list.get(i)).getName());
          if(matcher.find()){
             results.add(list.get(i));
          }
       }
       return results;
    }
    

     方式二:

    public class ListLike {
    
    //定义员工类
    public class Employee {
    private String name;
    private int age;
    public int getAge() {
    return age;
    }
    public void setAge(int age) {
    this.age = age;
    }
    public String getName() {
    return name;
    }
    public void setName(String name) {
    this.name = name;
    }
    }
    
    public List list=new ArrayList();
    
    //增加员工
    public List addList(String name,int age){
    Employee employee1 = new Employee();
    employee1.setName(name);
    employee1.setAge(age);
    list.add(employee1);
    return list;
    }
    
    //显示所有员工
    public void ShowList(){
    for(int i=0;i<list.size();i++){
    System.out.println(((Employee)(list.get(i))).getName()+" "+((Employee)(list.get(i))).getAge());
    }
    }
    
    //模糊查询
    public List likeString(String likename){
    for(int i=0;i<list.size();i++){
    if(((Employee)(list.get(i))).getName().indexOf(likename)<=-1)
    list.remove(i);
    }
    return list;
    
    }
    
    public static void main(String arg[]){
    ListLike ll=new ListLike();
    ll.addList("wuxiao",13);
    ll.addList("wangwang",11);
    ll.addList("wanghua",12);
    ll.addList("xiaowang",13);
    ll.addList("xiaoxiao",13);
    
    ll.likeString("wang");
    ll.ShowList();
    
    
    }
    
    }
    
  • 相关阅读:
    【LeetCode & 剑指offer刷题】数组题18:Plus One
    SQL Server 2005 的动态管理视图DMV和函数DMF
    数据库SQL优化大总结之 百万级数据库优化方案
    误删SQL Server日志文件后怎样附加数据库
    教你建立SQL数据库的表分区
    Sql Server 阻塞的常见原因和解决办法
    SQL索引优化方法
    详解如何定义SQL Server外关键字约束
    写出高性能SQL语句的十三条法则
    SQL SERVER内部函数大全
  • 原文地址:https://www.cnblogs.com/1925yiyi/p/8078890.html
Copyright © 2020-2023  润新知