How to show Alert when UIApplication is not able to open URL?

Question

This method opens up the url in Safari when the website string is not null and it is atleast of length 3. But when I have supplierWebsite=@"www.heritage.com", nothing happens. I know that heritage.com is not valid website and so it is not activating in UIApplication. I would like to display atleast a pop up that would tell user that website is not available. Is there any way i can show Alertview telling that website is not available.

- (IBAction)doWebOpen:(UIButton *)sender {

if (self.provider.supplierWebSite && [self.provider.supplierWebSite length] > 3) {
    NSString *urlString = [self.provider supplierWebSite];
    NSURL *url = [NSURL URLWithString:urlString];
    [[UIApplication sharedApplication] openURL:url];

}else {

    NSError *err = [NSError errorWithDomain:@"com.cantopenweb" code:509 andDescription:@"This supplier does not have a website."];
    [self showErrorAlert:err];
}}

Answer 1

Just use canOpenURL of UIApplication class, like:

if([[UIApplication sharedApplication] canOpenURL:url])
 {
    [[UIApplication sharedApplication] openURL:url];
 }
 else
 {
   //show alert
 }

---

canOpenURL:

Returns whether an application can open a given URL resource.

- (BOOL)canOpenURL:(NSURL *)url

Parameters

url

A URL object that identifies a given resource. The URL’s scheme—possibly a custom scheme—identifies which application can

handle the URL.

Return Value

NO if no application is available that will accept the URL; otherwise, returns YES. Discussion

This method guarantees that that if openURL: is called, another application will be launched to handle it. It does not guarantee that the full URL is valid. Availability

Available in iOS 3.0 and later.

Declared In UIApplication.h

Answer 2

You could use canOpenURL method,

[[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"your website"]];

The method returns a BOOL, so check that for YES or NO.

If YES, it CAN else NO.