英文:
SpringBoot JPA Repository - get selected fields shows error "Column not found"
问题
###我想要获取选定的列，最好是动态的。###
目前，我正在指定静态列 - id、title、description。
Category.java
@Entity(name="Category")
@Table(name="categories")
public class Category {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
private String slug;
private String title;
private String description;
private String preview_image;
private int isFeatured;
/* Getters, setters, etc */
}
CategoryRepository.java
public interface CategoryRepository extends JpaRepository<Category, Long> {
@Query(
value = "SELECT id, title, description FROM categories",
nativeQuery = true
)
List<Category> findAll();
}
这会导致错误：
未找到列 'is_featured'。异常类：class
org.springframework.dao.InvalidDataAccessResourceUsageException
异常 [无法执行查询；SQL [SELECT id, title, description
FROM categories]; 嵌套异常是
org.hibernate.exception.SQLGrammarException: 无法执行查询]
有没有办法解决这个错误？我正在考虑使用另一个模型而不是Category，但稍后需要使字段动态化。如果我不知道要返回哪些字段，就很难创建模型类。
另外，有没有办法使这个 @Query 代码动态化，以便根据参数返回列？
感谢您的任何帮助。提前感谢！ 
英文:
###I want to fetch selected columns, preferably dynamically.###
For now, I'm specifying static columns - id, title, description.
Category.java
@Entity(name="Category")
@Table(name="categories")
public class Category {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
private String slug;
private String title;
private String description;
private String preview_image;
private int isFeatured;
/* Getters, setters etc */
}
CategoryRepository.java
public interface CategoryRepository extends JpaRepository<Category, Long> {
@Query(
value = "SELECT id, title, description FROM categories",
nativeQuery = true
)
List<Category> findAll();
}
This gives the error:
> Column 'is_featured' not found. Exception class: class
> org.springframework.dao.InvalidDataAccessResourceUsageException
> Exception [could not execute query; SQL [SELECT id, title, description
> FROM categories]; nested exception is
> org.hibernate.exception.SQLGrammarException: could not execute query]
Any idea how this error can be resolved? I was thinking of using another model instead of Category but I need to make the fields dynamic later on. It's hard to make model class if I'm unaware of which fields to return.
Also, is there a way I can make this @Query code dynamic such that it returns columns mentioned in parameter?
Any help would be appreciated. Thanks in advance!
答案1
得分: 0
因为Spring和Hibernate的命名策略导致的。默认情况下，它会将camelCase转换为SNAKE_CASE。所以在你的情况下，isFeatured会变为is_featured。
如果你不想改变命名策略，只需在你的属性上添加@Column("isFeatured")。这将覆盖该属性的默认行为。
你可以在这里找到更多信息：https://stackoverflow.com/q/25283198/3493036
关于动态查询，你应该在Spring-data-jpa的文档中查找有关功能，以及关于querydsl的内容。
英文:
Its because of the naming strategy of Spring and Hibernate. It will convert camelCase to SNAKE_CASE by default. So in your case its isFeatured -> is_featured.
If you dont want to change the naming strategy just add @Column("isFeatured") on your property. This will override the default behavior of this property.
Here you can find more https://stackoverflow.com/q/25283198/3493036
Regarding dynamic query. You should look up the features in the documentation of Spring-data-jpa and about querydsl
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