Build URL in java

Question

Trying to build http://IP:4567/foldername/1234?abc=xyz. 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 http://IP:port/foldername/1234?. I am stuck at query part. Please help me to move forward.

Answer 1

You can just pass raw spec

new URL("http://IP:4567/foldername/1234?abc=xyz"); 

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("http://example.com:4567/foldername/1234?abc=xyz").url();