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?
How to check if URL is valid in Android
134.4k views Asked by kidalex At
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There are 12 answers
0
On
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;
}
0
On
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;
}
0
On
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;
}
0
On
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.
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.