Open Link in Safari Swift 5.

This example to open particular link form your app to safari. That's redirect to app to safari browser. 

Swift 5

guard let url = URL(string: "https://iosdevcenters.blogspot.com/") else {
     return
 }
if UIApplication.shared.canOpenURL(url) {
     UIApplication.shared.open(url, options: [:], completionHandler: nil)
 }

Swift 2.0

let url : NSURL = NSURL(string: "https://iosdevcenters.blogspot.com/")!
if UIApplication.sharedApplication().canOpenURL(url) {
     UIApplication.sharedApplication().openURL(url)
}

Objective C

NSURL *url = [NSURL URLWithString:@"https://iosdevcenters.blogspot.com/"];
if ([[UIApplication sharedApplication] canOpenURL:url]) {
    [[UIApplication sharedApplication] openURL:url];
}

Output:

Thanks.