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();


}

}

 

posted on 2018-12-26 13:22  龙从一  阅读(754)  评论(0编辑  收藏  举报