Android List(集合)中的对象以某一个字段排序案例

时间:2021-05-21

在Android开发中,有时我们需要对一个对象的集合按照某一个字段进行排序,

Bean

public class Student { private int studentId; private String studentName; private int age; public Student(int studentId , String studentName, int age){ this.studentId=studentId; this.studentName=studentName; this.age=age; } public int getStudentId() { return studentId; } public void setStudentId(int studentId) { this.studentId = studentId; } public String getStudentName() { return studentName; } public void setStudentName(String studentName) { this.studentName = studentName; } public int getAge() { return age; } public void setAge(int age) { this.age = age; }}

实现排序

实现排序比较类 Comparator ,里面实现排序规则。

public class test { /** * @param args */ public static void main(String[] args) { Student stu1 = new Student (1,"zhangsan",28); Student stu2 = new Student (2,"zhagnsan",19); Student stu3 = new Student (3,"wangwu",19); Student stu4 = new Student (4,"wangwu",19); Student stu5 = new Student (5,"zhaoliu",18); ArrayList<Student> list = new ArrayList<Student>(); list.add(stu1); list.add(stu2); list.add(stu3); list.add(stu4); list.add(stu5); //排序规则,这里是以年龄先排序,如果年龄相同 Comparator<Student> comparator = new Comparator<Student>() { public int compare(Student s1, Student s2) { // 先排年龄 if (s1.getAge() != s2.getAge()) { return s1.getAge() - s2.getAge(); } else if (!s1.getStudentName().equals(s2.getStudentName())) { // 年龄相同则按姓名排序 return s1.getStudentName().compareTo(s2.getStudentName()); } else { // 姓名也相同则按学号排序 return s1.getStudentId() - s2.getStudentId(); } } }; //这里就会自动根据规则进行排序 Collections.sort(list,comparator); for(int i=0;i<list.size();i++){ Student stu=list.get(i); System.out.println("年龄:"+stu.getAge()+" 姓名:"+stu.getStudentName()+" 学号:"+stu.getStudentId()); } }}

排序结果

结果:

年龄:18 姓名:zhaoliu 学号:5

年龄:19 姓名:wangwu 学号:3

年龄:19 姓名:wangwu 学号:4

年龄:19 姓名:zhagnsan 学号:2

年龄:28 姓名:zhangsan 学号:1

也可以想下面这样写:

Collections.sort(list, new Comparator<GoodsBean>() { @Override public int compare(GoodsBean bean1, GoodsBean bean2) { if (Integer.valueOf(bean1.getScore()).compareTo(Integer.valueOf(bean2.getScore())) == 0) { return Integer.valueOf(bean1.getRecommend_num()).compareTo(Integer.valueOf(bean2.getRecommend_num())); } else { return Integer.valueOf(bean1.getScore()).compareTo(Integer.valueOf(bean2.getScore())); } } });

这样就可以对一个集合中的数据各种排序了。

补充知识:java利用映射表名称反射创建实体类并赋属性值

1.hibernate中首先进行初始化,将对应的表名和类名以键值对的方式存放到map中

  private Map<String, String> mappings;//全局变量  /**  * 根据数据库表名获取实体类  */  public void initMappings() {    if (mappings == null) {    mappings = new HashMap<String, String>();    SessionFactory factory = this.getSessionFactory();    Map metaMap = factory.getAllClassMetadata();    for (String key : (Set<String>) metaMap.keySet()) {      AbstractEntityPersister classMetadata = (AbstractEntityPersister) metaMap.get(key);      String tableName = classMetadata.getTableName().toLowerCase();      int index = tableName.indexOf(".");      if (index >= 0) {        tableName = tableName.substring(index + 1);      }      String className = classMetadata.getEntityMetamodel().getName();      mappings.put(tableName, className);      }    }  }

2.调用方法,传入表名得到对应的实体类名

  public String getEntityNameByTableName(String tableName) {    initMappings();    return mappings.get(tableName);  }

3.根据实体类名创建实体类

  /**  *listobj:要赋的属性值集合,顺序要和实体类属性顺序一致  */  public Object getByReflect(String tableName, List listobj)throws Exception {     Class<?> model = Class.forName(tableName);     Object object = new Object();      if (model != null) {        Field[] field = model.getDeclaredFields();       String[] modelName = new String[field.length];        String[] modelType = new String[field.length];       object = model.newInstance();       Method m = null;        for (int i = 1; i <field.length ; i++) {          String name = field[i].getName();          Object value = null;         name = name.substring(0, 1).toUpperCase() + name.substring(1);          String type = field[i].getGenericType().toString();          if (type.equals("class java.lang.String")) {            m = model.getDeclaredMethod("set" + name, String.class);            if(listobj.get(i - 1) instanceof Double){              Double d=(Double) listobj.get(i-1);              value=String.valueOf(d);           }else{             value =(String)listobj.get(i - 1);            }          }          if (type.equals("class java.lang.Integer")) {            m = model.getDeclaredMethod("set" + name, Integer.class);           Double d = (Double) listobj.get(i - 1);           value = Integer.valueOf(d.intValue());          }          if (type.equals("class java.lang.Short")) {            m = model.getDeclaredMethod("set" + name, Short.class);            value = (Short) listobj.get(i - 1);          }          if (type.equals("class java.lang.Float")) {            m = model.getDeclaredMethod("set" + name, Float.class);           value = (Float) listobj.get(i - 1);          }          if (type.equals("class java.lang.Double")) {            m = model.getDeclaredMethod("set" + name, Double.class);            value = (Double) listobj.get(i - 1);          }          if (type.equals("class java.lang.Boolean")) {            m = model.getDeclaredMethod("set" + name, Boolean.class);            value = (Boolean) listobj.get(i - 1);          }          if (m != null) {            m.invoke(object, value);       }      }    }     return object;   }

以上这篇Android List(集合)中的对象以某一个字段排序案例就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持。

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