找不到类型的属性...自定义 Spring Data 存储库
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No property found for type... custom Spring Data repository
提问by Héctor
I'm trying to implement a custom Spring repository. I have the interface:
我正在尝试实现自定义 Spring 存储库。我有接口：
public interface FilterRepositoryCustom {
List<User> filterBy(String role);
}
the implementation:
实施：
public class FilterRepositoryImpl implements FilterRepositoryCustom {
...
}
and the "main" repository, extending my custom repository:
和“主”存储库，扩展我的自定义存储库：
public interface UserRepository extends JpaRepository<User, String>, FilterRepositoryCustom {
...
}
I'm using Spring Boot and, according to the docs:
我正在使用 Spring Boot，并且根据文档：
By default, Spring Boot will enable JPA repository support and look in the package (and its subpackages) where @SpringBootApplication is located.
默认情况下，Spring Boot 将启用 JPA 存储库支持并查看 @SpringBootApplication 所在的包（及其子包）。
When I run my application, I get this error:
当我运行我的应用程序时，我收到此错误：
org.springframework.data.mapping.PropertyReferenceException: No property filterBy found for type User!
org.springframework.data.mapping.PropertyReferenceException：未找到用户类型的属性 filterBy！
回答by Aman Tuladhar
The problem here is that you are creating FilterRepositoryImplbut you are using it in UserRepository. You need to create UserRepositoryImplto make this work.
这里的问题是您正在创建FilterRepositoryImpl但您在UserRepository. 您需要创建UserRepositoryImpl才能完成这项工作。
Basically
基本上
public interface UserRepositoryCustom {
List<User> filterBy(String role);
}
public class UserRepositoryImpl implements UserRepositoryCustom {
...
}
public interface UserRepository extends JpaRepository<User, String>, UserRepositoryCustom {
...
}
Spring Data 2.x update
This answer was written for Spring 1.x. As Matt Forsythepointed out, the naming expectations changed with Spring Data 2.0. The implementation changed from the-final-repository-interface-name-with-an-additional-Impl-suffixto the-custom-interface-name-with-an-additional-Impl-suffix.
Spring Data 2.x 更新
这个答案是为 Spring 1.x 编写的。正如Matt Forsythe指出的那样，命名期望随着 Spring Data 2.0 发生了变化。实现从 更改the-final-repository-interface-name-with-an-additional-Impl-suffix为the-custom-interface-name-with-an-additional-Impl-suffix。
So in this case, the name of the implementation would be: UserRepositoryCustomImpl.
所以在这种情况下，实现的名称将是：UserRepositoryCustomImpl。
回答by user64141
Another way this error can happen if the impl class for FilterRepositoryCustom isn't picked up in your spring configuration:
如果您的 spring 配置中未选择 FilterRepositoryCustom 的 impl 类，则可能发生此错误的另一种方式：
@EnableJpaRepositories(basePackageClasses = {RepoPackageMarker.class, FilterRepositoryCustomImpl.class})
回答by Vivek Vhatkar
Is it a must that the customMethod()in the CustomRepository can only have parameters defined that are either
是否必须customMethod()在 CustomRepository 中定义的参数要么是
1.Entity class name - customMethod(User user),
1.实体类名 - customMethod(User user),
2.Entity class attributes - customMethod(String firstName), here firstName is an attribute of User Entity class.
2.Entity类属性- customMethod(String firstName)，这里firstName是User Entity类的一个属性。
Can I not have something like customMethod(CustomCriteria criteria), the criteria class contain the various attributes that are used to construct a dynamic query.
我不能有像 customMethod(CustomCriteria criteria) 这样的东西吗，标准类包含用于构造动态查询的各种属性。
e.g. getStatusByCriteria(CustomCriteria criteria), CustomCriteriais a simple pojo annotated with @Component so that spring identifies it.
eggetStatusByCriteria(CustomCriteria criteria), CustomCriteria是一个用 @Component 注释的简单 pojo，以便 spring 识别它。
When I tried this I get an error:
当我尝试这个时，我收到一个错误：
org.springframework.data.mapping.PropertyReferenceException: No property criteria found for type UserRepository!
org.springframework.data.mapping.PropertyReferenceException：未找到 UserRepository 类型的属性标准！
回答by Bruno 82
I had the same problem in a project of mine. I solved the problem by adding a line in my pom.xml
我在我的一个项目中遇到了同样的问题。我通过在pom.xml 中添加一行解决了这个问题
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<configuration>
<includes>
<include>com/my/package/entities/*.java</include>
<include>com/my/package/repositories/*.java</include>
<include>com/my/package/repositories/impl/*.java</include> <!-- add this -->
</includes>
</configuration>
</plugin>
回答by Java Man
Old way:
旧方式：
Entity aThing = repository.findOne(1L);
New way:
新方法：
Optional<Entity> aThing = repository.findById(1L);
