Spring boot @ExceptionHandler 将响应返回为 html
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【中文标题】Spring boot @ExceptionHandler 将响应返回为 html【英文标题】：Spring boot @ExceptionHandler return the response as html 【发布时间】：2018-11-07 07:03:58 【问题描述】：我对 Spring boot 更感兴趣当抛出异常时，我得到 html 的响应，而我需要它作为 JSON。
服务响应
HTTP/1.1 500
Content-Type: text/html;charset=UTF-8
Content-Language: en-US
Content-Length: 345
Date: Mon, 28 May 2018 16:13:06 GMT
Connection: close
<html><body><h1>Whitelabel Error Page</h1><p>This application has no explicit mapping for /error, so you are seeing this as a fallback.</p><div id='created'>Mon May 28 19:13:06 EEST 2018</div><div>There was an unexpected error (type=Internal Server Error, status=500).</div><div>The environment must be QAT2,PSQA or DEVSTAGE4</div></body></html>
这是异常环境必须是 QAT2、PSQA 或 DEVSTAGE4 我需要它作为 JSON 响应，而无需编写自定义异常处理程序类，如下所示：
"timestamp" : 1413313361387,
"exception" : "java.lang.IllegalArgumentException",
"status" : 500,
"error" : "internal server error",
"path" : "/greet",
"message" : "The environment must be QAT2,PSQA or DEVSTAGE4"
它之前按预期工作，但我对此做了一些更改 控制器
package main.controller;
@RestController
@RequestMapping("/api")
public class API
private final APIService apiService;
@Autowired
public API(APIService offersService) this.apiService = offersService;
@ExceptionHandler(IllegalArgumentException.class)
void handleIllegalStateException(IllegalArgumentException e, HttpServletResponse response) throws IOException
response.sendError(HttpStatus.FORBIDDEN.value());
@PostMapping(value = "/createMember", produces = "application/json")
public ResponseEntity createMembers(@Valid @RequestBody APIModel requestBody) throws IllegalArgumentException
validatePrams();
apiService.fillMembersData();
return ResponseEntity.ok(HttpStatus.OK);
private void validatePrams() throws IllegalArgumentException
if (APIModel.getEnvironment() == null || (!APIModel.getEnvironment().equalsIgnoreCase("QAT2")
&& !APIModel.getEnvironment().equalsIgnoreCase("PSQA")
&& !APIModel.getEnvironment().equalsIgnoreCase("DEVSTAGE4")))
throw new IllegalArgumentException("The environment must be QAT2,PSQA or DEVSTAGE4");
型号
package main.model;
@Entity
@Table(name = "APIModel")
public class APIModel
@Id
@Column(name = "environment", nullable = false)
@NotNull
private static String environment;
@Column(name = "country", nullable = false)
@NotNull
private static String country;
@Column(name = "emailTo", nullable = false)
@NotNull
private static String emailTo;
@Column(name = "plan", nullable = false)
@NotNull
private static String plan;
@Column(name = "paymentType", nullable = false)
@NotNull
private static String paymentType;
@Column(name = "numberOfUsers", nullable = false)
@NotNull
private static Integer numberOfUsers;
@Column(name = "program")
private static String program;
public APIModel(String environment, String country, String emailTo, String plan, String paymentType, Integer numberOfUsers, String program)
APIModel.environment = environment;
APIModel.country = country;
APIModel.emailTo = emailTo;
APIModel.plan = plan;
APIModel.paymentType = paymentType;
APIModel.numberOfUsers = numberOfUsers;
APIModel.program = program;
public APIModel()
public static String getEnvironment() return environment;
public void setEnvironment(String environment) APIModel.environment = environment;
public static String getCountry() return country;
public static String getEmailTo() return emailTo;
public static String getPlan() return plan;
public static String getPaymentType() return paymentType;
public static Integer getNumberOfUsers() return numberOfUsers;
public static String getProgram() return program;
public void setProgram(String program) APIModel.program = program;
应用程序
package main;
@SpringBootApplication
public class MainClass
static
SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy/MM/dd'/'hh_mm_ss");
System.setProperty("current.date.time", dateFormat.format(new Date()));
System.setProperty("usr_dir", System.getProperty("user.dir") + "\\src\\logs");
public static void main(String[] args)
SpringApplication.run(MainClass.class, args);
【问题讨论】：
尝试使用此代码response.sendError(HttpStatus.BAD_REQUEST) 而不是response.sendError(HttpStatus.BAD_REQUEST.value())
@Generic 这也不起作用，我在控制台中遇到了异常，但仍然不是同样的问题。
【参考方案1】：
从上面的类范围中删除 @ResponseBody 并注意它们的 url 值应该以 '/' 开头，而不仅仅是输入 url，而不是输入
@RequestMapping(value = "createMember", method = RequestMethod.POST)
这应该是
@RequestMapping(value = "/createMember", method = RequestMethod.POST)
或者如果你像下面这样直接用帖子注释它会更好
@PostMapping(value = "/createMember")
GET 和 PUT 等是一样的
【讨论】：
【参考方案2】：您的处理程序仅捕获 IllegalStateException 而不是抛出的 IllegalArgumentException：
@ExceptionHandler(IllegalStateException.class)
void handleIllegalStateException(IllegalStateException e, HttpServletResponse response) throws IOException
response.sendError(HttpStatus.BAD_REQUEST.value());
这些都是 RuntimeExceptions。要在同一个处理程序中捕获两者，您可以尝试将其替换为：
@ExceptionHandler(RuntimeException.class)
void handleIllegalStateException(IllegalStateException e, HttpServletResponse response) throws IOException
response.sendError(HttpStatus.BAD_REQUEST.value());
实际上，这将捕获此控制器抛出的所有 RuntimeExceptions。
【讨论】：
还是同样的问题，控制器确实捕获了异常，但他将错误作为 HTML 返回，我可以在响应中看到异常，但我需要它作为 json 看看这是否有帮助：@RequestMapping(value = "createMember", method = RequestMethod.POST,produces = "application/json")以上是关于Spring boot @ExceptionHandler 将响应返回为 html的主要内容，如果未能解决你的问题，请参考以下文章
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