spring mvc 休息响应json和xml
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spring mvc rest response json and xml
提问by Simon Hofstadler
I have the requirement to return the result from the database either as a string in xml-structure or as json-structure. I've got a solution, but I don't know, if this one is the best way to solve this. I have two methods here:
我需要以 xml 结构中的字符串或 json 结构中的字符串形式从数据库返回结果。我有一个解决方案，但我不知道，这是否是解决此问题的最佳方法。我这里有两种方法：
@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);
}
for the first method I need a better solution, which I already asked here: spring mvc rest mongo dbobject response
对于第一种方法，我需要一个更好的解决方案，我已经在这里问过了： spring mvc rest mongo dbobject response
The next thing is, that I inserted in the configuration a json converter:
接下来是，我在配置中插入了一个 json 转换器：
<bean id="jsonHttpMessageConverter"
class="org.springframework.http.converter.json.MappingHymansonHttpMessageConverter">
<property name="supportedMediaTypes" value="application/json"/>
</bean>
when I change the content-type at the first method to "application/json", it works, but then the xml response doesn't work anymore, because the json converter wants to convert the xml string to json-structure I think.
当我将第一种方法中的内容类型更改为“application/json”时，它可以工作，但是 xml 响应不再起作用，因为 json 转换器想要将 xml 字符串转换为我认为的 json 结构。
what can I do, that spring identifies the difference that the one method should return a json type and the other one a normal xml as string? I tried it with the accept flag:
我能做什么，那个 spring 确定了一个方法应该返回一个 json 类型而另一个方法应该返回一个普通的 xml 作为字符串的区别？我用接受标志尝试过：
@RequestMapping(value = "/content/json/{ids}", method = RequestMethod.GET, headers = "Accept=application/json")
but this doesn't work. I get the following error:
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
I hope that somebody can help me out.
我希望有人可以帮助我。
回答by
If you are using Spring 3.1, you can take advantage of the new produceselement on the @RequestMappingannotation to ensure that you produce XML or JSON as you wish, even within the same app.
如果您使用的是 Spring 3.1，您可以利用注释produces上的新元素@RequestMapping来确保您根据需要生成 XML 或 JSON，即使在同一个应用程序中也是如此。
I wrote a post about this here:
我在这里写了一篇关于这个的帖子：
回答by atrain
Whoa...when you're working with Spring, assume someone else has come up against the same issue. You can dump all the server-side JSON generation, because all you need to do is:
哇...当您使用 Spring 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：
- Include the Hymanson JSON JARs in your app
- Set the
RequestMappingreturn type to@ResponseBody(yourObjectType)
- 在您的应用程序中包含 Hymanson JSON JAR
- 将
RequestMapping返回类型设置为@ResponseBody(yourObjectType)
Spring will auto-magically convert your object to JSON. Really. Works like magic.
Spring 会自动神奇地将您的对象转换为 JSON。真的。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
文档@ResponseBody：http: //static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/mvc.html#mvc-ann-responsebody
回答by Mark
You can use ContentNegotiatingViewResolver as below:
您可以使用 ContentNegotiatingViewResolver 如下：
<bean class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver">
<property name="defaultContentType" value="application/json" />
<property name="ignoreAcceptHeader" value="true" />
<property name="favorPathExtension" value="true" />
<property name="order" value="1" />
<property name="mediaTypes">
<map>
<entry key="xml" value="application/xml" />
<entry key="json" value="application/json" />
</map>
</property>
<property name="defaultViews">
<list>
<ref bean="xmlView"/>
<ref bean="jsonView"/>
</list>
</property>
</bean>
<bean id="jsonView"
class="org.springframework.web.servlet.view.json.MappingHymansonJsonView">
<property name="contentType" value="application/json;charset=UTF-8"/>
<property name="disableCaching" value="false"/>
</bean>
<bean id="xmlView"
class="org.springframework.web.servlet.view.xml.MarshallingView">
<property name="contentType" value="application/xml;charset=UTF-8"/>
<constructor-arg>
<ref bean="xstreamMarshaller"/>
</constructor-arg>
</bean>
<bean id="xstreamMarshaller" class="org.springframework.oxm.xstream.XStreamMarshaller">
<property name="autodetectAnnotations" value="true" />
<property name="annotatedClass" value="demo.domain.xml.XMLResponse"/>
<property name="supportedClasses" value="demo.domain.xml.XMLResponse"/>
</bean>
In your controller:
在您的控制器中：
@RequestMapping(value = "/get/response.json", method = RequestMethod.GET)
public JSONResponse getJsonResponse(){
return responseService.getJsonResponse();
}
@RequestMapping(value = "/get/response.xml", method = RequestMethod.GET)
public XMLResponse getXmlResponse(){
return responseService.getXmlResponse();
}
回答by Lucky
If someone using Spring Boot for XML response then add the following dependency,
如果有人使用 Spring Boot 进行 XML 响应，则添加以下依赖项，
<dependency>
<groupId>com.fasterxml.Hymanson.dataformat</groupId>
<artifactId>Hymanson-dataformat-xml</artifactId>
</dependency>
And in the model class you are returning add the @XmlRootElement.
在您返回的模型类中添加@XmlRootElement.
@Entity
@Table(name = "customer")
@XmlRootElement
public class Customer {
//... fields and getters, setters
}
and in your controller add produces="application/xml". This will produce the response in xml format.
并在您的控制器中添加produces="application/xml". 这将产生 xml 格式的响应。
回答by Saurabh
Here I wrote method which take XML as request parameter from your request mapping URL
在这里，我编写了将 XML 作为请求映射 URL 中的请求参数的方法
Here I am posting XML to my URL "baseurl/user/createuser/"
在这里，我将 XML 发布到我的 URL“baseurl/user/createuser/”
public class UserController {
@RequestMapping(value = "createuser/" ,
method=RequestMethod.POST, consumes= "application/xml")
@ResponseBody
ResponseEntity<String> createUser(@RequestBody String requestBody ) {
String r = "<ID>10</ID>"; // Put whatever response u want to return to requester
return new ResponseEntity<String>(
"Handled application/xml request. Request body was: "
+ r,
new HttpHeaders(),
HttpStatus.OK);
}
}
I tested it using chrome poster where you can send any xml in content body like:
我使用 chrome 海报对其进行了测试，您可以在其中发送内容正文中的任何 xml，例如：
"<?xml version="1.0" encoding="UTF-8" standalone="yes"?> <userEntity><id>3</id><firstName>saurabh</firstName><lastName>shri</lastName><password>pass</password><userName>[email protected]</userName></userEntity>"
This XML will capture by my createUser method and stored in String requestBody which i can use further
此 XML 将通过我的 createUser 方法捕获并存储在字符串 requestBody 中，我可以进一步使用
回答by Jobin Thomas
Easiest way to do this to get JSON response would be: Using Spring 3.1, you could do as follows
获得 JSON 响应的最简单方法是：使用 Spring 3.1，您可以执行以下操作
In your pom.xml file (hoping you are using a maven project), add maven dependency for Hymanson-mapper (http://mvnrepository.com/artifact/org.codehaus.Hymanson/Hymanson-mapper-asl/1.9.13)
Modify your code as follows and test the endpoint on postman:
@RequestMapping(value = "/content/json/{ids}", method = RequestMethod.GET) public @ResponseBody String getContentByIdsAsJSON(@PathVariable("ids") String ids){ String content = ""; StringBuilder builder = new StringBuilder(); List<String> list = this.contentService.findContentByListingIdAsJSON(ids); if (list.isEmpty()){ content = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><error>no data found</error>"; return content; } for (String json : list){ builder.append(json + "\n"); } content = builder.toString(); return content; }
在您的 pom.xml 文件中（希望您使用的是 maven 项目），为 Hymanson-mapper 添加 maven 依赖项（http://mvnrepository.com/artifact/org.codehaus.Hymanson/Hymanson-mapper-asl/1.9.13）
如下修改您的代码并在邮递员上测试端点：
@RequestMapping(value = "/content/json/{ids}", method = RequestMethod.GET) public @ResponseBody String getContentByIdsAsJSON(@PathVariable("ids") String ids){ String content = ""; StringBuilder builder = new StringBuilder(); List<String> list = this.contentService.findContentByListingIdAsJSON(ids); if (list.isEmpty()){ content = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><error>no data found</error>"; return content; } for (String json : list){ builder.append(json + "\n"); } content = builder.toString(); return content; }
