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?

Retrench

how are you connecting to the host? – Coquetry

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.

Sidman

isValidUrl returns false for the following URL, although the device seems happy enough with it (it's missing the double slash after file:). "file:/storage/emulated/0/Android/data/com.samsung.android.app.pinboard/files/ClipData/Screenshot_NormarAppImage.png" – Fauch

I would not rely on this method, because it doesn't perform deep validation, only most simple WebView cases. See source – Inebriety

Better answer: #5618249 – Inebriety

it seems that as long as it has http:// or https:// it will return true – Lotuseater

Would this help avoid urls like mailto: [email protected] and about:blank#blocked – Aphonia

Answer 2

URLUtil.isValidUrl(url);

If this doesn't work you can use:

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

Pray

Does this not just check the pattern rather than whether the URL is actually active? – Foreshore

As far as i know,if we need to check whether url is actually pinging or not we should manually check for that using HttpURLConnection class. – Pray

Use the second option if you need to check urls without schema. URLUtil.isValidUrl will return false if schema is not added. – Ol

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;
}

Fumigate

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
}

Skyrocket

Answer 5

Just add this line of code:

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

Darbie

Answer 6

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.

Emilioemily

I don't mind handling the exception but how do I prevent the app from crashing due to that very exception? – Retrench

@Retrench Err, catch the exception? What else could 'handling the exception' possibly mean? – Delphine

Answer 7

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;
}

Tips

Answer 8

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;
            }

Peasecod

Answer 9

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

Elysha

Patterns.WEB_URL.matcher(url).matches() is failing on my end too, I think because the link has no wwwand only https. – Kentkenta

Answer 10

You cans validate the URL by following:

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

Continually

Answer 11

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.

Munford

Answer 12

new java.net.URL(String) throws MalformedURLException

Marcellemarcellina