英文:
How to remove a field from a List of Object using java 8, below is the example of what I want to achieve
问题
我有一个名为Category的实体类（示例如下）
@Entity
class Category {
private int id;
private String name;
private String address;
private Date createDate;
private Date modifiedDate;
}
我正在使用JPA，当我执行
findAll(); // JPA repository
我也会获取createdDate和modifiedDate，但我不想获取它们。
那么有没有一种方法可以通过移除createDate和modifiedDate字段来筛选获取的列表？
英文:
I have an Entity class as Category (below is sample)
@Entity
class Category {
private int id;
private String name;
private String address;
private Date createDate;
private Date modifiedDate;
}
I am using JPA, and when I am doing
findAll(); // JPA repository
I am getting createdDate and modifiedDate as well, But I don't want to fetch them.
So Is there a way to filter the fetched list by removing the createDate and modifiedDate fields. ?
答案1
得分: 1
@Entity
class Category {
private int id;
private String name;
private String address;
private Date createDate;
private Date modifiedDate;
Category(){
}
// This constructor will be used to fetch only desired columns
Category(String id, String name, String address){
this.id = id;
this.name = name;
this.address = address;
}
}
在你的 repository 中，你需要定义 @Query 如下：
@Query("SELECT new com.test.Category(cat.id, cat.name, cat.address) FROM Category cat")
List<Category> findAll();
注意： 你需要将 com.test.Category 替换为你实际的 Category 类所在的包路径。
英文:
@Entity
class Category {
private int id;
private String name;
private String address;
private Date createDate;
private Date modifiedDate;
Category(){
}
// This constructor will be used to fetch only desired columns
Category(String id, String name, String address){
this.id = id;
this.name = name;
this.address = address;
}
}
In your repository you need to define the @Query as
@Query("SELECT new com.test.Category(cat.id, cat.name, cat.address) FROM Category cat")
List<Category> findAll();
Note: You need to replace com.test.Category to your actual package for Category
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