CODEWITHSUNDEEP
javaspring-mvcservlets
Steve
Steve

Reputation: 513

Spring REST controller post request

I have this controller in spring 

@RestController
public class GreetingController {

    @RequestMapping(value = "/greeting",  method = RequestMethod.POST)
    public String greeting(@RequestParam("uouo") String uouo) {
        return uouo;
    }
}

and when I testing it 

 curl -k -i -X POST -H "Content-Type:application/json"  -d uouo=test http://192.168.1.104:8080/api/greeting

the result of the testing

HTTP Status 400 - Required String parameter 'uouo' is not present

I tried may thing, but I think @RequestParam can't use for POST it always passed the parameter in URL using GET, I use post only if I had object JSON as parameter using @RequestBody, is there any way to make string parameter send using POST?

Upvotes: 5

Views: 16296

Answers (1)

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 279990

The Servlet container will only provide parameters from the body for POST requests if the content type is application/x-www-form-urlencoded. It will ignore the body if the content type is anything else. This is specified in the Servlet Specification Chapter 3.1.1 When Parameters Are Available

The following are the conditions that must be met before post form data will be populated to the parameter set:

  1. The request is an HTTP or HTTPS request.
  2. The HTTP method is POST.
  3. The content type is application/x-www-form-urlencoded.
  4. The servlet has made an initial call of any of the getParameter family of methods on the request object.

If the conditions are not met and the post form data is not included in the parameter set, the post data must still be available to the servlet via the request object’s input stream. If the conditions are met, post form data will no longer be available for reading directly from the request object’s input stream.

Since you aren't sending any JSON, just set the appropriate content type

curl -k -i -X POST -H "Content-Type:application/x-www-form-urlencoded" -d uouo=test http://192.168.1.104:8080/api/greeting

or let curl infer it

curl -k -i -X POST -d uouo=test http://192.168.1.104:8080/api/greeting?uouo=test

Note that you can still pass query parameters in the URL

curl -k -i -X POST -H "Content-Type:application/json"  http://192.168.1.104:8080/api/greeting?uouo=test

Upvotes: 8

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