英文:
Better way to add custom error message during Unique Constrains JPA/Hibernate
问题
我认为在Spring Boot/Hibernate中有更好的方法来做这件事。有很多表格，我不想为每个表格都添加用户定义的模型。
我有一个实体：
@Entity
public class Item extends ExtraEntity {
@NotEmpty(message = "Item code can not be Empty !")
@Column(unique = true)
private String code;
@NotEmpty(message = "Item name can not be Empty !")
@Column(unique = true)
private String name;
}
如何在我的REST API中发送自定义消息，指示代码重复或名称重复？
英文:
I think there is a better way to do this in Spring boot/hibernate. There are many tables and I don't want to add User Defined model for every table: here
I have an entity
@Entity
public class Item extends ExtraEntity {
@NotEmpty(message = "Item code can not be Empty !")
@Column(unique = true)
private String code;
@NotEmpty (message = "Item name can not be Empty !")
@Column(unique = true)
private String name;
}
How do I send a custom message for my REST APIs stating if code is duplicate or name is duplicate?
答案1
得分: 2
在最外层捕获异常如下：
try {
newEntity = ourService.createUpdate(entity);
} catch (JpaSystemException jpae) {
if (jpae.getCause().getCause() instanceof ConstraintViolationException) {
if (((ConstraintViolationException)jpae.getCause().getCause()).getConstraintName().equals("SCHEMA.UK_CODE_01")){
throw new DuplicatedCodeException("Message",jpae);
} else if (((ConstraintViolationException)jpae.getCause().getCause()).getConstraintName().equals("SCHEMA.UK_NAME_01")){
throw new DuplicatedNameException("Message",jpae);
}
}
}
为每个唯一键创建自定义异常，如下：
public class DuplicatedNameException extends Exception {
public DuplicatedNameException(String message){
super(message);
}
public DuplicatedNameException(String message, Throwable anException){
super(message, anException);
}
}
定义一个异常处理程序，扩展自ResponseEntityExceptionHandler类，并处理接收到的异常如下：
@ExceptionHandler({ DuplicatedNameException.class })
public ResponseEntity<Object> handleDuplicatedNameException(Exception anException, WebRequest request) {
[...]
return new ResponseEntity<>(anException, new HttpHeaders(), YOUR_HTTP_STATUS_WITH_CUSTOM_CODE_HERE);
}
这个HTTPStatus是你应该在Web层检查以显示错误消息的地方。
英文:
In the outermost layer, catch the exception as follows:
try {
newEntity = ourService.createUpdate(entity);
} catch (JpaSystemException jpae) {
if (jpae.getCause().getCause() instanceof ConstraintViolationException) {
if (((ConstraintViolationException)jpae.getCause().getCause()).getConstraintName().equals("SCHEMA.UK_CODE_01")){
throw new DuplicatedCodeException("Message",jpae);
} else if (((ConstraintViolationException)jpae.getCause().getCause()).getConstraintName().equals("SCHEMA.UK_NAME_01")){
throw new DuplicatedNameException("Message",jpae);
}
}
}
Create a custom exception for each unique key like this:
public class DuplicatedNameException extends Exception {
public DuplicatedNameException(String message){
super(message);
}
public DuplicatedNameException(String message, Throwable anException){
super(message, anException);
}
}
Define an exception handler that extends from the ResponseEntityExceptionHandler class and treats the received exception as follows:
@ExceptionHandler({ DuplicatedNameException.class })
public ResponseEntity<Object> handleDuplicatedNameException(Exception anException, WebRequest request) {
[...]
return new ResponseEntity<>(anException, new HttpHeaders(), YOUR_HTTP_STATUS_WITH_CUSTOM_CODE_HERE);
}
This HTTPStatus is the one that you should check in the web layer to show the error message
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