Java 如何使用包含关键字的属性创建 Spring JPA 存储库 findBy 查询？
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How do you create a Spring JPA repository findBy query using a property that contains a keyword?
提问by Ben M.
Here is a simplified example of my problem. I have this repository and entity class.
这是我的问题的简化示例。我有这个存储库和实体类。
public interface ThingRepository extends JpaRepository<ThingEntity, Long> {
ThingEntity findByFooInAndBar(String fooIn, String bar);
}
@Entity
public class ThingEntity {
@Column(name="FOO_IN", nullable=false, length=1)
private String fooIn;
public String getFooIn() {
return fooIn;
}
public setFooIn(String fooIn) {
this.fooIn = fooIn;
}
/* not including bar property for brevity's sake */
}
Spring is throwing the following exception.
Spring 抛出以下异常。
org.springframework.data.mapping.PropertyReferenceException: No property foo found for type ThingEntity!
It looks like Spring is taking the method findByFooInAndBarand thinks that foois my property name and inis a keyword for matching values within a collection.
看起来 Spring 正在采用该方法findByFooInAndBar并认为这foo是我的属性名称，并且in是用于匹配集合中值的关键字。
How do I get it to understand that the property name is fooIn, not foo?
我如何才能理解属性名称是fooIn，不是foo？
采纳答案by Ben M.
To overcome this problem, I've defined the query manually using the @Queryannotation. I'll happily accept anyone else's answer if they find a solution that doesn't require a manual query.
为了克服这个问题，我使用@Query注释手动定义了查询。如果他们找到不需要手动查询的解决方案，我会很乐意接受其他人的回答。
public interface ThingRepository extends JpaRepository<ThingEntity, Long> {
@Query("SELECT t FROM Thing t WHERE t.fooIn = ?1 AND t.bar = ?2")
ThingEntity findByFooInAndBar(String fooIn, String bar);
}
回答by Oscar E. Rodríguez Gómez
Spring is parsing 'In' in your method to create the query check the link for create your query you should change the name of the variable fooIn to fooin or something like that....
Spring 正在您的方法中解析“In”以创建查询检查链接以创建您的查询您应该将变量 fooIn 的名称更改为 fooin 或类似的名称....
回答by Rog Tebs
I observed the code kindly and notice your return is not a listbut an object.
Change that part to List<ThingEntity>.
我友好地观察了代码，并注意到您的返回不是 alist而是object。
将该部分更改为List<ThingEntity>.
