How to check if URL is valid in Android

Question

Is there a good way to avoid the "host is not resolved" error that crashes an app? Some sort of a way to try connecting to a host ( like a URL ) and see if it's even valid?

Answer 1

Use URLUtil to validate the URL as below.

 URLUtil.isValidUrl(url)

It will return True if URL is valid and false if URL is invalid.

Answer 2

URLUtil.isValidUrl(url);

If this doesn't work you can use:

Patterns.WEB_URL.matcher(url).matches();

Answer 3

I would use a combination of methods mentioned here and in other Stackoverflow threads:

public static boolean IsValidUrl(String urlString) {
    try {
        URL url = new URL(urlString);
        return URLUtil.isValidUrl(urlString) && Patterns.WEB_URL.matcher(urlString).matches();
    } catch (MalformedURLException ignored) {
    }
    return false;
}

Answer 4

If you are using from kotlin you can create a String.kt and write code bellow:

fun String.isValidUrl(): Boolean = Patterns.WEB_URL.matcher(this).matches()

Then:

String url = "www.yourUrl.com"
if (!url.isValidUrl()) {
    //some code
}else{
   //some code
}

Answer 5

Wrap the operation in a try/catch. There are many ways that a URL can be well-formed but not retrievable. In addition, tests like seeing if the hostname exists doesn't guarantee anything because the host might become unreachable just after the check. Basically, no amount of pre-checking can guarantee that the retrieval won't fail and throw an exception, so you better plan to handle the exceptions.

Answer 6

I have tried a lot of methods.And find that no one works fine with this URL:

Now I use the following and everything goes well.

public static boolean checkURL(CharSequence input) {
    if (TextUtils.isEmpty(input)) {
        return false;
    }
    Pattern URL_PATTERN = Patterns.WEB_URL;
    boolean isURL = URL_PATTERN.matcher(input).matches();
    if (!isURL) {
        String urlString = input + "";
        if (URLUtil.isNetworkUrl(urlString)) {
            try {
                new URL(urlString);
                isURL = true;
            } catch (Exception e) {
            }
        }
    }
    return isURL;
}

Answer 7

import okhttp3.HttpUrl;
import android.util.Patterns;
import android.webkit.URLUtil;

            if (!Patterns.WEB_URL.matcher(url).matches()) {
                error.setText(R.string.wrong_server_address);
                return;
            }

            if (HttpUrl.parse(url) == null) {
                error.setText(R.string.wrong_server_address);
                return;
            }

            if (!URLUtil.isValidUrl(url)) {
                error.setText(R.string.wrong_server_address);
                return;
            }

            if (!url.substring(0,7).contains("http://") & !url.substring(0,8).contains("https://")) {
                error.setText(R.string.wrong_server_address);
                return;
            }

Answer 8

In my case Patterns.WEB_URL.matcher(url).matches() does not work correctly in the case when I type String similar to "first.secondword"(My app checks user input). This method returns true.

URLUtil.isValidUrl(url) works correctly for me. Maybe it would be useful to someone else

Answer 9

You cans validate the URL by following:

Patterns.WEB_URL.matcher(potentialUrl).matches()

Answer 10

public static boolean isURL(String text) {
    String tempString = text;

    if (!text.startsWith("http")) {
        tempString = "https://" + tempString;
    }

    try {
        new URL(tempString).toURI();
        return Patterns.WEB_URL.matcher(tempString).matches();
    } catch (MalformedURLException | URISyntaxException e) {
        e.printStackTrace();
        return false;
    }
}

This is the correct sollution that I'm using. Adding https:// before original text prevents text like "www.cats.com" to be considered as URL. If new URL() succeed, then if you just check the pattern to exclude simple texts like "https://cats" to be considered URL.

Answer 11

Just add this line of code:

Boolean isValid = URLUtil.isValidUrl(url) && Patterns.WEB_URL.matcher(url).matches();       

Answer 12

new java.net.URL(String) throws MalformedURLException