Spring Data JPA:找不到类型的属性(字段名称包含查询方法谓词关键字)

Question

所以我一直在努力解决这个错误。所以我得到了一个相当标准的错误，我理解它告诉我的是什么，我只是不明白它是如何得到这个错误的。这是错误

Caused by: org.springframework.data.mapping.PropertyReferenceException: No property stored found for type StorageItemEntity! Did you mean 'storedIn'?
at org.springframework.data.mapping.PropertyPath.<init>(PropertyPath.java:90) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.mapping.PropertyPath.create(PropertyPath.java:437) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.mapping.PropertyPath.create(PropertyPath.java:413) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.mapping.PropertyPath.lambda$from$0(PropertyPath.java:366) ~[spring-data-commons-2.4.8.jar:2.4.8]
at java.base/java.util.concurrent.ConcurrentMap.computeIfAbsent(ConcurrentMap.java:330) ~[na:na]
at org.springframework.data.mapping.PropertyPath.from(PropertyPath.java:348) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.mapping.PropertyPath.from(PropertyPath.java:331) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.repository.query.parser.Part.<init>(Part.java:81) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.repository.query.parser.PartTree$OrPart.lambda$new$0(PartTree.java:249) ~[spring-data-commons-2.4.8.jar:2.4.8]
at java.base/java.util.stream.ReferencePipeline$3$1.accept(ReferencePipeline.java:195) ~[na:na]
at java.base/java.util.stream.ReferencePipeline$2$1.accept(ReferencePipeline.java:177) ~[na:na]
at java.base/java.util.Spliterators$ArraySpliterator.forEachRemaining(Spliterators.java:948) ~[na:na]
at java.base/java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:484) ~[na:na]
at java.base/java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:474) ~[na:na]
at java.base/java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:913) ~[na:na]
at java.base/java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234) ~[na:na]
at java.base/java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:578) ~[na:na]
at org.springframework.data.repository.query.parser.PartTree$OrPart.<init>(PartTree.java:250) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.repository.query.parser.PartTree$Predicate.lambda$new$0(PartTree.java:383) ~[spring-data-commons-2.4.8.jar:2.4.8]
at java.base/java.util.stream.ReferencePipeline$3$1.accept(ReferencePipeline.java:195) ~[na:na]
at java.base/java.util.stream.ReferencePipeline$2$1.accept(ReferencePipeline.java:177) ~[na:na]
at java.base/java.util.Spliterators$ArraySpliterator.forEachRemaining(Spliterators.java:948) ~[na:na]
at java.base/java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:484) ~[na:na]
at java.base/java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:474) ~[na:na]
at java.base/java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:913) ~[na:na]
at java.base/java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234) ~[na:na]
at java.base/java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:578) ~[na:na]
at org.springframework.data.repository.query.parser.PartTree$Predicate.<init>(PartTree.java:384) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.repository.query.parser.PartTree.<init>(PartTree.java:95) ~[spring-data-commons-2.4.8.jar:2.4.8]
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery.<init>(PartTreeJpaQuery.java:89) ~[spring-data-jpa-2.4.8.jar:2.4.8]
... 62 common frames omitted

这么标准的错误是吧？好的，它告诉我，没有为我的实体找到"存储"的属性。有道理，但我似乎不明白为什么，因为我没有定义存储。好吧，这是一个让一团糟的查询

@Repository
public interface StorageItemDao extends JpaRepository<StorageItemEntity, Long> {

// All other methods omitted for sake of keeping it short
public List<StorageItemEntity> findAllByStoredIn(StorageType storedIn);
}

这是StorageItemEntity类

@Entity
@Table(name = "storage_item")
public class StorageItemEntity implements Serializable{
/**
* 
*/
private static final long serialVersionUID = -5373908574936516693L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "id")
private int id;

@NotEmpty(message = "Storage item name can not be empty")
@Size(max = 100, message = "Storage item name can not exceed 100 characters")
@Column(name = "name")
private String name;

@Range(min = 0, message = "Amount in storage can not be less than 0")
@Column(name = "amount")
private int amount;

@Range(min = 0, message = "Amount expected in storage per week can not be less than 0")
@Column(name = "amount_expected")
private int amountExpected;

@NotNull(message = "Storage in type can not be empty")
@Column(name = "stored_in")
private StorageType storedIn;

@Size(max = 500, message = Constants.ADDITIONAL_INFO_EXCEED)
@Column(name = "additional_info")
private String additionalInfo;

@NotNull(message = "You need to have a vendor associated to this item")
@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "vendor_id", referencedColumnName="id")
private VendorEntity vendor;

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "type_of_storage_id", referencedColumnName="id")
private TypeOfStorageEntity typeOfStorage;
// All getters, setters, and constructors are omitted
}

我还想指出，StorageType是一个非常标准的枚举。这不是错误，但这是一个很好的提示，以防万一。那么，为什么我会得到一个"；找不到类型StorageItemEntity"的属性"stored"；如果我从未在我的JPA方法中定义said存储？

谢谢

Answer 1

看起来In是Spring查询派生使用的关键字。因此它期望一个stored属性，它是参数中提供的In值。

https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#appendix.query.method.predicate

您需要使用@Query注释来获得您想要的内容。

Answer 2

我认为这是JPA模型中列名和变量名的问题，如果你想使用findAllByStoredIn，你可以像这样使用@Query注释。

类似的Repository类

@Repository
public interface AccountRepository extends JpaRepository<Account, Long> {

@Query("SELECT a FROM Account a WHERE a.userName = :username")
Account findByUsername(@Param("username") String username);
}

像这样的Model类

@Getter
@Setter
@RequiredArgsConstructor
@Entity
@Table(name = "account")
public class Account {
@Id
@GeneratedValue
@Column(name="user_id")
private Long userId;
@NotNull
@UniqueUsername(message="Username already exists")
@Size(min = 6, max = 255, message = "Username have to be grater than 6 characters")
@Column(name="user_name",unique = true)
private String userName;
}

Answer 3

这就足够解决这个问题了。@存储库公共接口AccountRepository扩展了JpaRepository<帐户，长期>｛

@Query("SELECT a FROM Account a WHERE a.username = :username")
Account findByUsername(@Param("username") String username);

}