CODEWITHSUNDEEP
javaspringrest
Freephone Panwal
Freephone Panwal

Reputation: 1591

How to make Content-Type header optional?

I've an heartbeat API implemeted using Spring REST service:

@RequestMapping(value = "heartbeat", method = RequestMethod.GET, consumes="application/json")
public ResponseEntity<String> getHeartBeat() throws Exception {
    String curr_time = myService.getCurrentTime();      
    return Util.getResponse(curr_time, HttpStatus.OK);
}

And MyService.java has below method:

public String getCurrentTime() throws Exception {
    String currentDateTime = null;
    MyJson json = new MyJson();
    ObjectMapper mapper = new ObjectMapper().configure(SerializationConfig.Feature.DEFAULT_VIEW_INCLUSION, false);

    try {           
        Date currDate = new Date(System.currentTimeMillis());
        currentDateTime = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss").format(currDate);           
        json.setTime(currentDateTime);                      

        ObjectWriter writer = mapper.writerWithView(Views.HeartBeatApi.class);
        return writer.writeValueAsString(json);                 
    } catch (Exception e) {
        throw new Exception("Excpetion", HttpStatus.BAD_REQUEST);           
    }
}

It works as expected but have 2 issues:

  1. When I invoke this API, Content-Type header is mandatory & I want to know how to make this header optional.

  2. How to add "Accept" header so that it can support other format such as Google Protobuf?

Thanks!

Upvotes: 0

Views: 1260

Answers (1)

Brandon Yarbrough
Brandon Yarbrough

Reputation: 38369

If you don't want to require Content-Type exist and be "application/json", you can just omit the consumes section entirely.

"Accept" is available via the "produces" value, as opposed to "consumes." So if you wanted to support Google Protobuf OR application/json, you could do this:

@Controller
@RequestMapping(value = "/pets/{petId}", method = RequestMethod.GET, produces="application/json")
@ResponseBody
public ResponseEntity<String> getHeartBeat() throws Exception {
    String curr_time = myService.getCurrentTime();      
    return Util.getResponse(curr_time, HttpStatus.OK);
}

Upvotes: 1

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