java 24 lines · 6 steps

Posting an order with Java's HttpClient

Serialize an order, POST it over HTTP, branch on the status code, and deserialize the reply.

Explained by highlit

Intermediate http-client json-serialization error-handling rest-api

1public OrderConfirmation submitOrder(Order order) throws IOException, InterruptedException {
  String payload = objectMapper.writeValueAsString(order);
3 
  HttpRequest request = HttpRequest.newBuilder()
          .uri(URI.create(baseUrl + "/api/v1/orders"))
          .timeout(Duration.ofSeconds(10))
          .header("Content-Type", "application/json")
          .header("Accept", "application/json")
          .header("Authorization", "Bearer " + apiToken)
          .POST(HttpRequest.BodyPublishers.ofString(payload, StandardCharsets.UTF_8))
          .build();
12 
  HttpResponse<String> response = httpClient.send(request, HttpResponse.BodyHandlers.ofString());
14 
  int status = response.statusCode();
  if (status == 429) {
      throw new RateLimitedException(response.headers().firstValue("Retry-After").orElse("5"));
  }
  if (status >= 400) {
      throw new OrderApiException("Order submission failed with status " + status + ": " + response.body());
  }
22 
  return objectMapper.readValue(response.body(), OrderConfirmation.class);
24}

01 / 01

STEP 01

Walkthrough

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Three takeaways

1Jackson's ObjectMapper bookends the request: one call to serialize outbound, one to deserialize the response.
2Inspecting the status code before parsing lets you fail fast with meaningful exceptions instead of a broken body.
3The builder pattern on HttpRequest makes headers, timeout, and body read as a single declarative description of the call.

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