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.n
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: https://www.baeldung.com/spring-uricomponentsbuilder
JavaDoc: https://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/web/util/UriComponentsBuilder.html
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("http://IP:4567/foldername/1234?abc=xyz")
.build()
.toUri()
.toURL();
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();
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();
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.