Question

我有关注的REST控制器

import ...ApplicationUser
import org.springframework.http.HttpStatus
import org.springframework.web.bind.annotation.*

@RestController
@RequestMapping("/sign-up")
class SignUpController: BaseController() {

    @ResponseStatus(HttpStatus.CREATED)
    @PostMapping
    fun signUp(@RequestBody applicationUser: ApplicationUser) {
        applicationUser.password = bCryptPasswordEncoder.encode(applicationUser.password)
        applicationUserRepository.save(applicationUser)
    }
}

和模型/实体

import java.util.*
import javax.persistence.*

@Entity
class ApplicationUser(
        @Id
        @GeneratedValue(strategy = GenerationType.AUTO)
        var id: Long? = null,

        @Column(unique = true)
        var username: String? = null,

        var password: String? = null
) {
    @Column(unique = true)
    var email: String? = null

    ...
}

因此，我有意地尝试将user保存为现有的email，然后自然而然地得到跟进

{
    "timestamp": "2019-01-29T15:40:45.784+0000",
    "status": 500,
    "error": "Internal Server Error",
    "message": "could not execute statement; SQL [n/a]; constraint [null]; nested exception is org.hibernate.exception.ConstraintViolationException: could not execute statement",
    "path": "/sign-up"
}

但是这个抛出有一个原因，原因是（葡萄牙语，对不起，哈哈）

com.microsoft.sqlserver.jdbc.SQLServerException：已修复唯一键'UK_cb61p28hanadv7k0nx1ec0n5l'。 Nãoépossívelinserir uma 否否'dbo.application_user'。查瓦尔 duplicadaé（admin@email.com）。

那么，如何发送/提交/加入HTTP响应异常的原因？像这样

{
    "timestamp": ...,
    "status": ...,
    "error": ...,
    "message": ...,
    "cause": "here is the cause of all problems",
    "path": ...
}

有可能吗？感谢您的任何建议！

Answer 1

您可以使用ControllerAdvice处理此类异常。您可以配置它以捕获应用程序中的任何异常，然后返回适当的响应代码/消息。您可以详细了解here

如何在HTTP请求异常上引发CAUSE

1 个答案: