'How to return CREATED status (201 HTTP) in ResponseEntity

There is a Spring-MVC application. In controllers, when returning the results of methods, I return via ResponseEntity<>. On success, I return (200 statutes) the OK-method. But when creating something, I would like to return the CREATED-method (201 status). I just can’t understand what kind of URL to ask in parentheses when calling through CREATED. How can this be implemented?

Now I have such an implementation:

@PostMapping("/create/dish")
    ResponseEntity<Dish> createDish(@Valid @RequestBody DishDTO dishDTO) {

        return ResponseEntity.ok(cookService.createDish(dishDTO.getDishName(), dishDTO.getAboutDish(), dishDTO.getDishType(),
                dishDTO.getCookingTime(), dishDTO.getWeight(),
                dishDTO.getDishCost(), dishDTO.getCooksId()));
    }

I want to remake it like this to make it work(now it not work):

@PostMapping("/create/dish")
    ResponseEntity<Dish> createDish(@Valid @RequestBody DishDTO dishDTO) {

        return ResponseEntity.created(cookService.createDish(dishDTO.getDishName(), dishDTO.getAboutDish(), dishDTO.getDishType(),
                dishDTO.getCookingTime(), dishDTO.getWeight(),
                dishDTO.getDishCost(), dishDTO.getCooksId()));
    }

enter image description here

P.S. I don’t have a frontend at all. All through Swagger or PostMan.

javaspring-bootresthttpspring-mvc


Solution 1:[1]

Just return this way:

return new ResponseEntity<Dish>(cookService.createDish(...), HttpStatus.CREATED)

Making sure you have imported org.springframework.http.HttpStatus

Solution 2:[2]

You can use

return new ResponseEntity(cookService.createDish(...), HttpStatus.CREATED);

Read more here

Solution 3:[3]

If you want to create a 201 (CREATED) response without a body, then use:

ResponseEntity.status(HttpStatus.CREATED).build()

Solution 4:[4]

You can use

    ResponseEntity.created(ServletUriComponentsBuilder.fromCurrentRequest().path("/{id}").buildAndExpand(savedObjectId).toUri()).build()

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 t0r0X
Solution 2 t0r0X
Solution 3 Rajib Biswas
Solution 4 UmutSalihBayrak

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