how can return json using response.senderror

Question

In my app,I use springMVC and tomcat,my controller return object,but when something wrong,I only want return some string message with content tye json,so I use response.error, but it not work,the return is a html. my controller:

@RequestMapping(value = "{id}/{name}" ,method=RequestMethod.POST,produces = MediaType.APPLICATION_JSON_VALUE)

public @ResponseBody UserBean  login(@PathVariable String id,@PathVariable("name") String userName,
        @RequestHeader(value = "User-Agent") String user_agen,
        @CookieValue(required = false) Cookie userId,
        HttpServletRequest request,HttpServletResponse response,@RequestBody UserBean entity
        ) throws IOException {
     System.out.println("dsdsd");
     System.out.print(userName);

     response.setContentType( MediaType.APPLICATION_JSON_VALUE);
     response.sendError(HttpServletResponse.SC_BAD_REQUEST, "somethind wrong");
     return  null;

Community · Accepted Answer · 2017-05-23 11:51:34Z

3

According to the Javadoc for the HttpServletReponse#sendError method:

Sends an error response to the client using the specified status. The server defaults to creating the response to look like an HTML-formatted server error page containing the specified message, setting the content type to "text/html", leaving cookies and other headers unmodified...

So sendError will generate an HTML error page using the message that you supplied and will override the content type to text/html.

Since the client end is expecting a JSON response, you may be better to manually set the response code and the message yourself using fields on your UserBean - assuming it can support it. That will then be serialized to a JSON response that your clientside Javascript can evaluate.

@RequestMapping(value = "{id}/{name}" ,method=RequestMethod.POST,produces = MediaType.APPLICATION_JSON_VALUE)
public @ResponseBody UserBean  login(@PathVariable String id,@PathVariable("name") String userName,
        @RequestHeader(value = "User-Agent") String user_agen,
        @CookieValue(required = false) Cookie userId,
        HttpServletRequest request,HttpServletResponse response,@RequestBody UserBean entity
        ) throws IOException {
     System.out.println("dsdsd");
     System.out.print(userName);

     response.setContentType( MediaType.APPLICATION_JSON_VALUE);
     response.setStatus(HttpServletResponse.SC_BAD_REQUEST);

     UserBean userBean = new UserBean();
     userBean.setError("something wrong"); // For the message
     return userBean;

There is also the option of using the Tomcat property org.apache.coyote. USE_CUSTOM_STATUS_MSG_IN_HEADER which will place the message into a custom response header. See this post and the Tomcat docs for more info.

edited May 23, 2017 at 11:51

CommunityBot

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answered Dec 2, 2013 at 10:07

Will Keeling

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2 Comments

user3023874 Over a year ago

thanks,this is a great solution .But the problem is all the response bean must hava a field name error ,it is not so friendly. well,it still a successful resolution.

Will Keeling Over a year ago

Perhaps there is a base class that UserBean (and other beans) inherit from that could contain the error property? That would save on duplication. Alternatively there's also the possibility of the USE_CUSTOM_STATUS_MSG_IN_HEADER Tomcat specific property that I mentioned in the answer. Then you wouldn't need the error property.

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