spring - 使用 RestController 接受字符串和 XML 数据

Question

我想创建 REST 服务器，它接受 XML 请求和纯文本到不同的 Controller 中。我尝试实现这个:

@SpringBootApplication
public class Application extends SpringBootServletInitializer implements WebMvcConfigurer {

    @Override
    protected SpringApplicationBuilder configure(SpringApplicationBuilder application) {
        return application.sources(Application.class);
    }
    ..............

    private BasicAuthenticationInterceptor basicAuthenticationInterceptor;

    @Override
    public void extendMessageConverters(List<HttpMessageConverter<?>> converters) {
        converters.removeIf(converter -> converter instanceof MappingJackson2XmlHttpMessageConverter);
        converters.removeIf(converter -> converter instanceof MappingJackson2HttpMessageConverter);
        converters.add(new MappingJackson2XmlHttpMessageConverter(
                ((XmlMapper) createObjectMapper(Jackson2ObjectMapperBuilder.xml()))
                        .enable(ToXmlGenerator.Feature.WRITE_XML_DECLARATION)));
        converters.add(new MappingJackson2HttpMessageConverter(createObjectMapper(Jackson2ObjectMapperBuilder.json())));
    }

    private ObjectMapper createObjectMapper(Jackson2ObjectMapperBuilder builder) {
        builder.indentOutput(true);
        builder.modules(new JaxbAnnotationModule());
        builder.serializationInclusion(JsonInclude.Include.NON_NULL);
        builder.defaultUseWrapper(false);
        return builder.build();
    }

    @Autowired
    public void setBasicAuthenticationInterceptor(BasicAuthenticationInterceptor basicAuthenticationInterceptor) {
        this.basicAuthenticationInterceptor = basicAuthenticationInterceptor;
    }

    @Override
    public void addInterceptors(InterceptorRegistry registry) {
        registry.addInterceptor(basicAuthenticationInterceptor);
    }
}

检查 XML 格式是否正确:

@ControllerAdvice
public class RestExceptionHandler extends ResponseEntityExceptionHandler {

    @Override
    protected ResponseEntity<Object> handleHttpMessageNotReadable(HttpMessageNotReadableException ex,
                                                                  HttpHeaders headers, HttpStatus status, WebRequest request) {
        PaymentTransaction response;
        if (ex.getMessage().contains("Required request body")) {
            response = new PaymentTransaction(PaymentTransaction.Response.failed_response, 350,
                    "Invalid XML message: No XML data received", "XML request parsing failed!");
        } else {
            response = new PaymentTransaction(PaymentTransaction.Response.failed_response, 351,
                    "Invalid XML message format", null);
        }
        return ResponseEntity.badRequest().body(response);
    }
}

Controller 类:

@RestController()
public class HomeController {

    @Autowired
    public HomeController(Map<String, MessageProcessor> processors, Map<String, ReconcileProcessor> reconcileProcessors,
            @Qualifier("defaultProcessor") MessageProcessor defaultProcessor,
            AuthenticationService authenticationService, ClientRepository repository,
            @Value("${request.limit}") int requestLimit) {
        // Here I receive XML 
    }

    @GetMapping(value = "/v1/*")
    public String message() {
        return "REST server";
    }

    @PostMapping(value = "/v1/{token}", consumes = { MediaType.APPLICATION_XML_VALUE,
            MediaType.APPLICATION_JSON_VALUE }, produces = { MediaType.APPLICATION_XML_VALUE,
                    MediaType.APPLICATION_JSON_VALUE })
    public PaymentResponse handleMessage(@PathVariable("token") String token,
            @RequestBody PaymentTransaction transaction, HttpServletRequest request) throws Exception {
        // Here I receive XML 
    }

    @PostMapping(value = "/v1/notification")
    public ResponseEntity<String> handleNotifications(@RequestBody Map<String, String> keyValuePairs) {
         // Here I receive key and value in request body
    }

    @PostMapping(value = "/v1/summary/by_date/{token}", consumes = { MediaType.APPLICATION_XML_VALUE,
            MediaType.APPLICATION_JSON_VALUE }, produces = { MediaType.APPLICATION_XML_VALUE,
                    MediaType.APPLICATION_JSON_VALUE })
    public PaymentResponses handleReconcile(@PathVariable("token") String token, @RequestBody Reconcile reconcile,
            HttpServletRequest request) throws Exception {
         // Here I receive XML 
    }

    @ResponseStatus(value = HttpStatus.UNAUTHORIZED)
    public static class UnauthorizedException extends RuntimeException {
        UnauthorizedException(String message) {
            super(message);
        }
    }
}

正如您所看到的，在某些方法中我收到 XML，而在其他方法中我收到 key=value&.....

形式的字符串

如何配置 Spring 以接受这两种类型？我还应该将 Rest Controller 拆分为不同的文件吗？

编辑:

示例 XML 请求:

<?xml version="1.0" encoding="UTF-8"?>
<payment_transaction>
  <transaction_type>authorize</transaction_type>
  <transaction_id>2aeke4geaclv7ml80</transaction_id>
  <amount>1000</amount>
  <currency>USD</currency>
  <card_number>22</card_number>
  <shipping_address>
    <first_name>Name</first_name>    
  </shipping_address>
</payment_transaction>

示例 XML 响应:

<?xml version="1.0" encoding="UTF-8"?>
<payment_response>
    <transaction_type>authorize</transaction_type>
    <status>approved</status>
    <unique_id>5f7edd36689f03324f3ef531beacfaae</unique_id>
    <transaction_id>asdsdlddea4sdaasdsdsa4dadasda</transaction_id>
    <code>500</code>
    <amount>101</amount>
    <currency>EUR</currency>
</payment_response>

示例通知请求:

uniqueid=23434&type=sale&status=33

示例通知响应:它应该仅返回 HTTP 状态 OK。

我使用:

    <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>2.0.4.RELEASE</version>
        <relativePath />
    </parent>

Java版本:“10.0.2”2018-07-17

关于我使用的 XML 生成:

@XmlRootElement(name = "payment_transaction")
public class PaymentTransaction {

    public enum Response {
        failed_response, successful_response
    }

    @XmlElement(name = "transaction_type")
    public String transactionType;
    @XmlElement(name = "transaction_id")
    public String transactionId;
    @XmlElement(name = "usage")

POM 配置:https://pastebin.com/zXqYhDH3

Answer 1

对于 Spring boot 2.0.4-RELEASE，看来你不需要做太多事情。

我做了这样的配置:

@Configuration
public class WebConfiguration implements WebMvcConfigurer {

    @Override
    public void configureMessageConverters(List<HttpMessageConverter<?>> converters) {
        //MyRequestBodyHttpMessageConverter converter = new MyRequestBodyHttpMessageConverter();
        FormHttpMessageConverter converter = new FormHttpMessageConverter();
        //MediaType utf8FormEncoded = new MediaType("application","x-www-form-urlencoded", Charset.forName("UTF-8"));
        //MediaType mediaType = MediaType.APPLICATION_FORM_URLENCODED; maybe UTF-8 is not needed
        converter.setSupportedMediaTypes(Arrays.asList(MediaType.APPLICATION_FORM_URLENCODED));
        //converter.setSupportedMediaTypes(Arrays.asList(utf8FormEncoded));
        converters.add(converter);
        MappingJackson2HttpMessageConverter conv1 = new MappingJackson2HttpMessageConverter();
        conv1.getObjectMapper().registerModule(new JaxbAnnotationModule());
        converters.add(conv1);

        MappingJackson2XmlHttpMessageConverter conv = new MappingJackson2XmlHttpMessageConverter();
        // required by jaxb annotations
        conv.getObjectMapper().registerModule(new JaxbAnnotationModule());
        converters.add(conv);
    }
}

我用过你的 DTO:

@XmlRootElement(name = "payment_transaction")
public class PaymentTransaction {

    @XmlElement(name = "transaction_type")
    public String transactionType;
    @XmlElement(name = "transaction_id")
    public String transactionId;

    public String getTransactionType() {
        return transactionType;
    }
    public void setTransactionType(String transactionType) {
        this.transactionType = transactionType;
    }
    public String getTransactionId() {
        return transactionId;
    }
    public void setTransactionId(String transactionId) {
        this.transactionId = transactionId;
    }
    @Override
    public String toString() {
        return "PaymentTransaction [transactionType=" + transactionType
                + ", transactionId=" + transactionId + "]";
    }
}

Controller :

@RestController
public class MyController {

    /**
     * https://stackoverflow.com/questions/34782025/http-post-request-with-content-type-application-x-www-form-urlencoded-not-workin/38252762#38252762
    */

    @PostMapping(value = "/v1/{token}",
            consumes =  MediaType.APPLICATION_XML_VALUE,
            produces = MediaType.APPLICATION_JSON_VALUE)
    public @ResponseBody  PaymentTransaction handleMessage(@PathVariable("token") String token,
            @RequestBody PaymentTransaction transaction, HttpServletRequest request) throws Exception {
           System.out.println("handleXmlMessage");
           System.out.println(transaction);
           PaymentTransaction body = new PaymentTransaction();
           body.setTransactionId(transaction.getTransactionId());
           body.setTransactionType("received: " + transaction.getTransactionType());
           return body;
    }

    @PostMapping(consumes = MediaType.APPLICATION_FORM_URLENCODED_VALUE, value = "/v1/notification")
    public ResponseEntity<String> handleNotifications(@ModelAttribute PaymentTransaction transaction) {
           System.out.println("handleFormMessage");
           System.out.println(transaction);
           return new ResponseEntity<String>(HttpStatus.OK);
    }
 }

唯一要记住的主要事情是，用解析的数据填充 DTO 似乎是通过反射发生的:

供您输入

<payment_transaction>
  <transaction_id>1</transaction_id>
  <transaction_type>name</transaction_type>
</payment_transaction>

我收到此响应(请参阅我的 Controller ):

{
"transactionType": "received: null",
"transactionId": null
}

但是当我更改为 DTO 字段的名称时，它开始工作(根元素并不重要，有趣):

<payment_transaction>
  <transactionId>1</transactionId>
  <transactionType>name</transactionType>
</payment_transaction>

结果:

{
"transactionType": "received: name",
"transactionId": "1"
}

查询字符串也是如此。我不知道要更改什么才能让 spring 使用 @XmlRootElement/@XmlElement 中定义的名称解析 xml。