How to Pass Query Parameters in Java Using HttpClient?

How can I pass query parameters while making HTTP requests in Java using HttpClient?

String url = "https://api.example.com/data?query=value";

Passing query parameters in Java while making HTTP requests can easily be achieved using the HttpClient library. This allows you to construct URLs with needed parameters before sending requests to web servers.

import java.net.URI;
import java.net.http.HttpClient;
import java.net.http.HttpRequest;
import java.net.http.HttpResponse;
import java.net.URLEncoder;
import java.nio.charset.StandardCharsets;

public class HttpClientExample {
    public static void main(String[] args) throws Exception {
        HttpClient client = HttpClient.newHttpClient();
        String queryParam = URLEncoder.encode("value", StandardCharsets.UTF_8);
        String url = "https://api.example.com/data?query=" + queryParam;
        HttpRequest request = HttpRequest.newBuilder()
                .uri(new URI(url))
                .GET()  
                .build();
        HttpResponse<String> response = client.send(request, HttpResponse.BodyHandlers.ofString());
        System.out.println(response.body());
    }
}

Causes

• Not properly encoding URL and parameters
• Forgetting to append parameters when forming the URL
• Using the wrong HTTP method for the request

Solutions

• Use URLEncodedUtils to encode the parameters correctly
• Build the URL with the parameters using StringBuilder or URIBuilder
• Check the documentation to ensure the correct HTTP method is used

Mistake: Incorrectly encoding the query parameters leading to malformed URLs

Solution: Use URLEncoder to safely encode the query parameters.

Mistake: Forgetting to add the '?' character before the first query parameter

Solution: Ensure the URL correctly starts with '?' followed by the parameters.

Mistake: Using GET instead of POST for requests that require sending data in body

Solution: Understand the types of requests required by the API.