iPhone: Open a url Programmatically

问题:

I am new to iPhone. I want to open an url in my application. How can I do this task? Please suggest me and provide some useful link.

回答1:

Update (2016): The best way to do this nowadays is to instantiate and present an SFSafariViewController. This gives the user the security and speed of Safari, and access to any cookies or Safari features they may already have set without having to leave your app.

If you want to open the URL in Safari (and exit your application) you can use the openURL method of UIApplication

If you'd rather have it handled inside of your app, use WKWebView.

回答2:

Apparently the link given above is outdated. Here is the update link for the UIApplication class.

The quick and simple code snippet is:

// ObjC
[[UIApplication sharedApplication] openURL:[NSURL URLWithString: @"http://www.google.com"]];

// Swift
UIApplication.shared.open(URL(string: "http://www.google.com")!, options: [:], completionHandler: nil)

回答3:

If you would like to open and just get the data from the URL, you could use NSString:

NSString *ans = [NSString stringWithContentsOfURL:url];

If what you are trying to get is an XML from a URL, you can directly use NSXMLParser:

NSURL *url = [[NSURL alloc] initWithString:urlstr];
NSXMLParser *parser = [[NSXMLParser alloc] initWithContentsOfURL:url];
// parse here
[parser release];
[url release];

On the other hand, if by opening you mean, open a URl in an embedded browser, you could use UIWebView class.

回答4:

if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"https://medium.com/the-traveled-ios-developers-guide/swift-3-feature-highlight-c38f94359731#.83akhtihk"]]) {
                [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://medium.com/the-traveled-ios-developers-guide/swift-3-feature-highlight-c38f94359731#.83akhtihk"]];
            }
            else{
                [SVProgressHUD showErrorWithStatus:@"Please enable Safari from restrictions to open this link"];
            }