Question

I need generate valid URLs.

Example: I pass the url: google.com . The generator returns http://google.com/ .

Some browsers do this. I tried do my own algorithm, but has fails.

Another example: http://www.yadayadayada.com/../test returns http://www.yadayadayada.com/test/

public String generateValidURL(String url) {
    int pos = 0;        
    try {
        url = url.trim();
        url = url.replaceAll(" ", "%20");
        if (url.startsWith("http") && (!url.substring(4).startsWith("://"))) {
            for (int i = 4; i < 7; i++) {
                if ((url.charAt(i) == '/') || (url.charAt(i) == ':')) {
                    pos = i;
                }
            }
            url = url.substring(pos + 1);
        }
        if(url.startsWith("https")){
            url = url.replace("https", "http");
        }
        if (!url.startsWith("http")) {
            url = "http://" + url;
        }
        if (!url.substring(7).contains("/")) {
            url += "/";
        }
        url = url.replace(",", ".");
        url = url.replace("../", "/");
        url = url.substring(0, 7) + url.substring(7).replace("//", "/");            
        return url;
    } catch (Exception e) {
        System.out.println("Error generating valid URL : " + e);
        return null;
    }
}

Answer 1

Update: now that is is clearer what you want to achieve – I don’t think there’s an utility for that. Your method should do, just debug it.

Original answer:

URL url = new URL("http", domain, "/");
String output = url.toExternalForm();

In fact, you may want to use the URI class instead:

URI uri = new URI("http", "google.com", "/test", null);

You can use uri.resolve("../relativePath") and it will get resolved. But have in mind that your example with /../test == /test is not proper (you’d have to handle this case manually)