Question

我有一个返回大数据的端点，我想删除其中的一部分。

例如：

A级

public class A{

private String id;
private Date createOn;
private String processed;
}

B类

public class B extends MongoDBObject{
private String id;
private Date createOn;
private String processed;
}

控制器

@RestController
@RequestMapping("/v1/read")
public class ReadController{

@Autowired
private StatementBundleService bundleService;

@CrossOrigin
@GetMapping(value = "/statementBundles")
public List<A> listStatements() {
   List<A> result = new ArrayList<A>();

   List<B> bundles = bundleService.getAll();

   for(B bundle: bundles) {
       result.add(new A(bundle));
   }

   return result;
}

我试图找出最好的方法是返回A和A类都没有“处理过”属性的B列表。

我应该只使用for each循环还是iterator？还应该将属性设置为null还是其他方法？

Answer 1

我怀疑是否可以在不迭代的情况下更改属性。虽然您可以尝试java8来进行快速简单的输出。看看里面。

public class Java8 {
public static void main(String[] args) {
    List<Student> myList = new ArrayList<Student>();
    myList.add(new Student(1, "John", "John is a good Student"));
    myList.add(new Student(1, "Paul", "Paul is a good Player"));
    myList.add(new Student(1, "Tom", "Paul is a good Teacher"));

    System.out.println(myList);//old list
    myList = myList.stream().peek(obj -> obj.setBiography(null)).collect(Collectors.toList());
    System.out.println(myList);//new list
}

/*Output*/
//[Student [id=1, Name=John, biography=John is a good Student], Student [id=1, Name=Paul, biography=Paul is a good Player], Student [id=1, Name=Tom, biography=Paul is a good Teacher]]
//[Student [id=1, Name=John, biography=null], Student [id=1, Name=Paul, biography=null], Student [id=1, Name=Tom, biography=null]]

}

按原样的学生班级

public class Student{
private int id;
private String Name;
private String biography;

public Student(int id, String name, String biography) {
    super();
    this.id = id;
    Name = name;
    this.biography = biography;
}
public int getId() {
    return id;
}       
public String getBiography() {
    return biography;
}
public void setBiography(String biography) {
    this.biography = biography;
}
@Override
public String toString() {
    return "Student [id=" + id + ", Name=" + Name + ", biography=" + biography + "]";
}       
}

如何删除ArrayList中对象的属性

1 个答案: