Java 11 introduced a new HttpClient API (java.net.HttpClient) that simplifies making HTTP requests and handling responses. It provides a clean and concise way to perform synchronous and asynchronous HTTP operations, and it fully supports RESTful APIs.

Here’s a step-by-step guide with examples for how to use the Java 11 HttpClient to call REST APIs.

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1. Create an HttpClient Instance

The HttpClient is the main entry point for sending HTTP requests and receiving responses. You can create an HttpClient using its builder.

HttpClient client = HttpClient.newHttpClient();

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2. Build an HttpRequest

The HttpRequest represents the HTTP request that you want to send. You can build it using a static builder where you set the HTTP method, headers, URI, etc.

Example: Build a GET Request

HttpRequest request = HttpRequest.newBuilder()
        .uri(URI.create("https://jsonplaceholder.typicode.com/posts/1"))
        .GET() // Use GET method
        .build();

Example: Build a POST Request

For a POST request, you also need to include a body in the request.

HttpRequest request = HttpRequest.newBuilder()
        .uri(URI.create("https://jsonplaceholder.typicode.com/posts"))
        .header("Content-Type", "application/json")
        .POST(HttpRequest.BodyPublishers.ofString("{\"title\": \"foo\", \"body\": \"bar\", \"userId\": 1}"))
        .build();

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3. Send the HttpRequest

The HttpClient provides methods to send HTTP requests. You can choose to execute the request synchronously or asynchronously.

Synchronous Call

Use HttpClient.send() to execute the request synchronously and retrieve an HttpResponse.

HttpResponse<String> response = client.send(request, HttpResponse.BodyHandlers.ofString());
System.out.println("Status Code: " + response.statusCode());
System.out.println("Response Body: " + response.body());

Asynchronous Call

Use HttpClient.sendAsync() to execute the request asynchronously. It returns a CompletableFuture which allows non-blocking operations.

client.sendAsync(request, HttpResponse.BodyHandlers.ofString())
      .thenApply(HttpResponse::body) // Extract body
      .thenAccept(System.out::println); // Print response body

The above code sends the request in the background and processes the response once received.

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4. Handle Response

The HttpResponse provides access to the response details like status code, headers, and body. You can use HttpResponse.BodyHandlers to specify how the body should be processed.

Available BodyHandlers:

• HttpResponse.BodyHandlers.ofString(): Read the response as a String.
• HttpResponse.BodyHandlers.ofFile(): Save the response to a file.
• HttpResponse.BodyHandlers.ofInputStream(): Access the response as an InputStream.

For example:

HttpResponse<String> response = client.send(request, HttpResponse.BodyHandlers.ofString());
System.out.println("Headers: " + response.headers());
System.out.println("Body: " + response.body());

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5. Complete Example for a GET Request

Here’s a complete program that sends a GET request to a REST API and prints the response:

package org.kodejava.net.http;

import java.net.URI;
import java.net.http.HttpClient;
import java.net.http.HttpRequest;
import java.net.http.HttpResponse;

public class HttpClientGetExample {
    public static void main(String[] args) {
        try {
            // Step 1: Create HttpClient
            HttpClient client = HttpClient.newHttpClient();

            // Step 2: Create HttpRequest
            HttpRequest request = HttpRequest.newBuilder()
                    .uri(URI.create("https://jsonplaceholder.typicode.com/posts/1"))
                    .GET()
                    .build();

            // Step 3: Send the request and receive HttpResponse
            HttpResponse<String> response = client.send(request, HttpResponse.BodyHandlers.ofString());

            // Step 4: Process the response
            System.out.println("Status Code: " + response.statusCode());
            System.out.println("Response Body: " + response.body());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

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6. Complete Example for a POST Request

Here’s how you can do a POST request to send JSON data:

package org.kodejava.net.http;

import java.net.URI;
import java.net.http.HttpClient;
import java.net.http.HttpRequest;
import java.net.http.HttpResponse;

public class HttpClientPostExample {
    public static void main(String[] args) {
        try {
            // Step 1: Create HttpClient
            HttpClient client = HttpClient.newHttpClient();

            // Step 2: Build HttpRequest with POST body
            HttpRequest request = HttpRequest.newBuilder()
                    .uri(URI.create("https://jsonplaceholder.typicode.com/posts"))
                    .header("Content-Type", "application/json")
                    .POST(HttpRequest.BodyPublishers.ofString("{\"title\": \"foo\", \"body\": \"bar\", \"userId\": 1}"))
                    .build();

            // Step 3: Send the POST request and receive response
            HttpResponse<String> response = client.send(request, HttpResponse.BodyHandlers.ofString());

            // Step 4: Process the response
            System.out.println("Status Code: " + response.statusCode());
            System.out.println("Response Body: " + response.body());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

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Key Points to Remember:

1. Use java.net.URI for specifying the API endpoint.
2. Always handle exceptions (e.g., IOException and InterruptedException).
3. You may need to set headers (e.g., Content-Type or Authorization) depending on the API requirements.
4. Handle different response codes appropriately.

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This approach is suitable for working with REST APIs and provides both blocking and non-blocking operations for flexibility.