How to build an absolute URL from a relative URL using Java?

Question

I have a relative url string, know host and protocol. How can I build an absolute url string?

Seems easy? Yes at first look, but until escaped characters coming. I have to build absolute url from 302 code http(s) response Location header.

lets consider an example

protocol: http
host: example.com
location: /path/path?param1=param1Data&param2= " 

First I tried to build url string like:

Sting urlString = protocol+host+location

Constructor of URL class not escapes spaces and double quotes:

new URL(urlString)

Constructors of URI class fail with exception:

new URI(urlString)

URI.resolve method also fails with exception

Then I found URI can escape params in query string, but only with few constructors like for example:

URI uri = new URI("http", "example.com", 
    "/path/path", "param1=param1Data&param2= \"", null);

This constructor needs path and query be a separate arguments, but I have a relative URL, and it not split by path and query parts.

I could consider to check if relative URL contains "?" question sign and think everything before it is path, and everything after it is query, but what if relative url not contain path, but query only, and query contains "?" sign? Then this will not works because part of query will be considered as path.

Now I cannot get how to build absolute url from relative url.

These accepted answers seems just wrong:

• how to get URL using relative path
• Append relative URL to java.net.URL
• Building an absolute URL from a relative URL in Java

It could be nice to consider scenario when relative url was given in relation to url with both host and some path part:

initial url http://example.com/...some path... relative /home?...query here ...

It would be great to get java core solution, though it still possible to use a good lib.

Answer 1

The first ? indicates where the query string begins:

3.4. Query

[...] The query component is indicated by the first question mark (?) character and terminated by a number sign (#) character or by the end of the URI.

A simple approach (that won't handle fragments and assumes that the query string is always present) is as simple as:

String protocol = "http";
String host = "example.com";
String location = "/path/path?key1=value1&key2=value2";

String path = location.substring(0, location.indexOf("?"));
String query = location.substring(location.indexOf("?") + 1);

URI uri = new URI(protocol, host, path, query, null);

A better approach that can also handle fragments could be :

String protocol = "http";
String host = "example.com";
String location = "/path/path?key1=value1&key2=value2#fragment";

// Split the location without removing the delimiters
String[] parts = location.split("(?=\\?)|(?=#)");

String path = null;
String query = null;
String fragment = null;

// Iterate over the parts to find path, query and fragment
for (String part : parts) {
    
    // The query string starts with ?
    if (part.startsWith("?")) {
        query = part.substring(1); 
        continue;
    }
    
    // The fragment starts with #
    if (part.startsWith("#")) {
        fragment = part.substring(1);
        continue;
    }
    
    // Path is what's left
    path = part;
}

URI uri = new URI(protocol, host, path, query, fragment);

Answer 2

The best way seems to be to create a URI object with the multi piece constructors, and then convert it to a URL like so:

URI uri = new URI("https", "sitename.domain.tld", "/path/goes/here", "param1=value&param2=otherValue");
URL url = uri.toURL();