Build Url in Java

Question

Trying to build . I don't know much about it but I wrote below code from searching from google:

import java.net.MalformedURLException;
import java.net.URI;
import java.net.URL;

public class MyUrlConstruct {

    public static void main(String a[]){

        try {
            String protocol = "http";
            String host = "IP";
            int port = 4567;
            String path = "foldername/1234";
            URL url = new URL (protocol, host, port, path);
            System.out.println(url.toString()+"?");
        } catch (MalformedURLException ex) {
            ex.printStackTrace();
        }
    }
}

I am able to build URL . I am stuck at query part. Please help me to move forward.

Accepted Answer

You can just pass raw spec

new URL("");

Or you can take something like org.apache.http.client.utils.URIBuilder and build it in safe manner with proper url encoding

URIBuilder builder = new URIBuilder();
builder.setScheme("http");
builder.setHost("IP");
builder.setPath("/foldername/1234");
builder.addParameter("abc", "xyz");
URL url = builder.build().toURL();

Answer 2

Use OkHttp

There is a very popular library named OkHttp which has been starred 20K times on GitHub. With this library, you can build the url like below:

import okhttp3.HttpUrl;

URL url = new HttpUrl.Builder()
    .scheme("http")
    .host("example.com")
    .port(4567)
    .addPathSegments("foldername/1234")
    .addQueryParameter("abc", "xyz")
    .build().url();

Or you can simply parse an URL:

URL url = HttpUrl.parse("").url();

Accepted Answer

In general non-Java terms, a URL is a specialized type of URI. You can use the URI class (which is more modern than the venerable URL class, which has been around since Java 1.0) to create a URI more reliably, and you can convert it to a URL with the toURL method of URI:

String protocol = "http";
String host = "example.com";
int port = 4567;
String path = "/foldername/1234";
String auth = null;
String fragment = null;
URI uri = new URI(protocol, auth, host, port, path, query, fragment);
URL url = uri.toURL();

Note that the path needs to start with a slash.

Answer 4

Must Read

If using Spring Framework:

UriComponentsBuilder.newInstance()
  .scheme(scheme)
  .host(host)
  .path(path)
  .build()
  .toUri()
  .toURL();

A new UriComponentsBuilder class helps to create UriComponents instances by providing fine-grained control over all aspects of preparing a URI including construction, expansion from template variables, and encoding.

Know more:

JavaDoc:

Answer 5

If you happen to be using Spring already, I have found the org.springframework.web.util.UriComponentsBuilder to be quite nifty. Here is how you would use it in your case.

final URL myUrl = UriComponentsBuilder
        .fromHttpUrl("")
        .build()
        .toUri()
        .toURL();

Build Url in Java

5 Answers

Use OkHttp

Must Read

If using Spring Framework:

Maya Lin-Takahashi