对输入数据判重，使用Map，将对象作为Key，使用map的containsKey方法来着重是否是重复记录。正常的处理流程：(1)将输入的数据封装成对象(2)判重，将没有重复的数据存储到map(3)对没有重复的数据，进行校验(4)将部分输入的数据，转换成数据库存储时的格式
问题：因为对已经存储在map中的对象中的字段进行了修改，即使后面的输入的对象与前一个重复，判重就会失效。反思：这个逻辑的流程是:(1)得到输入–》(2)判重–》(3)将部分字段的值更改成数据库存储的值–》(4)持久化第三部，更改了对象的值，因为会导致判重的预期失效，见下例子
import java.util.HashMap;import java.util.Map;/*2015-9-20*/public class MapDemo { public static void main(String[] args) { Map<Person, String> info = new HashMap<Person, String>(); Person person = new Person(“xiaoming”, 1); info.put(person, person.getName()); print(info, person); person.setSalary(2); System.out.println(String.format(“Get a pay raise:%d to %d”, 1, 2)); print(info, person); person = new Person(“xiaoming”, 1); print(info, person); } private static void print(Map<Person, String> info, Person person) { System.out.println(String.format(“get ‘%s’ from ‘%s’.Result: ‘%s'”, person, info, info.get(person))); }}class Person { private String name; private int salary; public Person(String name, int salary) { super(); this.name = name; this.salary = salary; } public String getName() { return name; } public void setName(String name) { this.name = name; } public int getSalary() { return salary; } public void setSalary(int salary) { this.salary = salary; } @Override public String toString() { return “Person [name=” + name + “, salary=” + salary + “]”; }}
Output:
get ‘Person [name=xiaoming, salary=1]’ from ‘{Person [name=xiaoming, salary=1]=xiaoming}’.Result: ‘xiaoming’Get a pay raise:1 to 2get ‘Person [name=xiaoming, salary=2]’ from ‘{Person [name=xiaoming, salary=2]=xiaoming}’.Result: ‘xiaoming’get ‘Person [name=xiaoming, salary=1]’ from ‘{Person [name=xiaoming, salary=2]=xiaoming}’.Result: ‘null’
如果雨后还是雨，如果忧伤过后还是忧伤，
