What is the idiomatic way to compose a URL or URI in Java?

Question

How do I build a URL or a URI in Java? Is there an idiomatic way, or libraries that easily do this?

I need to allow starting from a request string, parse/change various URL parts (scheme, host, path, query string) and support adding and automatically encoding query parameters.

Answer 1

As of Apache HTTP Component HttpClient 4.1.3, from the official tutorial:

public class HttpClientTest {
public static void main(String[] args) throws URISyntaxException {
    List<NameValuePair> qparams = new ArrayList<NameValuePair>();
    qparams.add(new BasicNameValuePair("q", "httpclient"));
    qparams.add(new BasicNameValuePair("btnG", "Google Search"));
    qparams.add(new BasicNameValuePair("aq", "f"));
    qparams.add(new BasicNameValuePair("oq", null));
    URI uri = URIUtils.createURI("http", "www.google.com", -1, "/search",
                                 URLEncodedUtils.format(qparams, "UTF-8"), null);
    HttpGet httpget = new HttpGet(uri);
    System.out.println(httpget.getURI());
    //http://www.google.com/search?q=httpclient&btnG=Google+Search&aq=f&oq=
}
}

Edit: as of v4.2 URIUtils.createURI() has been deprecated in favor of URIBuilder:

URI uri = new URIBuilder()
        .setScheme("http")
        .setHost("www.google.com")
        .setPath("/search")
        .setParameter("q", "httpclient")
        .setParameter("btnG", "Google Search")
        .setParameter("aq", "f")
        .setParameter("oq", "")
        .build();
HttpGet httpget = new HttpGet(uri);
System.out.println(httpget.getURI());

Answer 2

As the author, I'm probably not the best person to judge if my URL/URI builder is good, but here it nevertheless is: https://github.com/mikaelhg/urlbuilder

I wanted the simplest possible complete solution with zero dependencies outside the JDK, so I had to roll my own.

Answer 3

Apache HTTPClient?

Answer 4

Using HTTPClient worked well.

protected static String createUrl(List<NameValuePair> pairs) throws URIException{

  HttpMethod method = new GetMethod("http://example.org");
  method.setQueryString(pairs.toArray(new NameValuePair[]{}));

  return method.getURI().getEscapedURI();

}

Answer 5

There are plenty of libraries that can help you with URI building (don't reinvent the wheel). Here are three to get you started:

---

Java EE 7

import javax.ws.rs.core.UriBuilder;
...
return UriBuilder.fromUri(url).queryParam(key, value).build();

---

org.apache.httpcomponents:httpclient:4.5.2

import org.apache.http.client.utils.URIBuilder;
...
return new URIBuilder(url).addParameter(key, value).build();

---

org.springframework:spring-web:4.2.5.RELEASE

import org.springframework.web.util.UriComponentsBuilder;
...
return UriComponentsBuilder.fromUriString(url).queryParam(key, value).build().toUri();

---

See also: GIST > URI Builder Tests

Answer 6

Use OkHttp

It's 2022 and there is a very popular library named OkHttp which has been starred 41K times on GitHub. With this library, you can build an 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();

Answer 7

After being lambasted for suggesting the URL class. I will take the commenter's advice and suggest the URI class instead. I suggest you look closely at the constructors for a URI as the class is very immutable once created.
I think this constructor allows you to set everything in the URI that you could need.

URI(String scheme, String userInfo, String host, int port, String path, String query, String fragment)
          Constructs a hierarchical URI from the given components.