spring mvc rest response json和xml

Question

我要求将数据库中的结果作为xml-structure中的字符串或json-structure返回.我有一个解决方案,但我不知道,如果这是解决这个问题的最佳方法.我有两种方法:

@RequestMapping(value = "/content/json/{ids}", method = RequestMethod.GET)
public ResponseEntity<String> getContentByIdsAsJSON(@PathVariable("ids") String ids)
{
  String content = null;
  StringBuilder builder = new StringBuilder();
  HttpHeaders responseHeaders = new HttpHeaders();
  responseHeaders.add("Content-Type", "text/html; charset=utf-8");
  // responseHeaders.add("Content-Type", "application/json; charset=utf-8");

  List<String> list = this.contentService.findContentByListingIdAsJSON(ids);
  if (list.isEmpty())
  {
     content = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><error>no data  found</error>";
     return new ResponseEntity<String>(content, responseHeaders, HttpStatus.CREATED);
  }
  for (String json : list)
  {
     builder.append(json + "\n");
  }
  content = builder.toString();
  return new ResponseEntity<String>(content, responseHeaders, HttpStatus.CREATED);
}

@RequestMapping(value = "/content/{ids}", method = RequestMethod.GET)
public ResponseEntity<String> getContentByIdsAsXML(@PathVariable("ids") String ids)
{
  HttpHeaders responseHeaders = new HttpHeaders();
  responseHeaders.add("Content-Type", "application/xml; charset=utf-8");

  String content = this.contentService.findContentByListingIdAsXML(ids);
  if (content == null)
  {
     content = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><error>no data found</error>";
     return new ResponseEntity<String>(content, responseHeaders, HttpStatus.CREATED);
  }
  return new ResponseEntity<String>(content, responseHeaders, HttpStatus.CREATED);
}

对于第一种方法,我需要一个更好的解决方案,我已经在这里问过: spring mvc rest mongo dbobject response

接下来就是,我在配置中插入了一个json转换器:

<bean id="jsonHttpMessageConverter"
    class="org.springframework.http.converter.json.MappingJacksonHttpMessageConverter">
   <property name="supportedMediaTypes" value="application/json"/>
</bean>

当我将第一个方法的内容类型更改为"application/json"时,它可以正常工作,但是xml响应不再起作用,因为json转换器想要将xml字符串转换为json-structure.

我该怎么做,那个spring识别出一个方法应该返回一个json类型而另一个方法是一个普通的xml作为字符串的区别？我用accept标志尝试了它:

@RequestMapping(value = "/content/json/{ids}", method = RequestMethod.GET, headers = "Accept=application/json")

但这不起作用.我收到以下错误:

org.springframework.web.util.NestedServletException: Handler processing failed; nested exception is java.lang.StackOverflowError

我希望有人可以帮助我.

Answer 1

如果您使用的是Spring 3.1,那么您可以利用注释produces上的新元素@RequestMapping来确保您可以根据需要生成XML或JSON,即使在同一个应用程序中也是如此.

我在这里写了一篇关于此的帖子:

http://springinpractice.com/2012/02/22/supporting-xml-and-json-web-service-endpoints-in-spring-3-1-using-responsebody/

Answer 2

哇......当你和Spring一起工作时,假设其他人遇到了同样的问题.您可以转储所有服务器端JSON生成,因为您需要做的就是:

1. 在您的应用中包含Jackson JSON JAR
2. 将RequestMapping返回类型设置为@ResponseBody(yourObjectType)

Spring会自动将您的对象转换为JSON.真.像魔术一样工作.

Doc for @ResponseBody:http://static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/mvc.html#mvc-ann-responsebody