UIWebView在Safari中打开链接

Question

我有一个非常简单的UIWebView,其中包含我的应用程序包中的内容.我希望Web视图中的任何链接都在Safari中打开,而不是在Web视图中打开.这可能吗？

Answer 1

将其添加到UIWebView委托:

(编辑以检查导航类型.您还可以传递file://相关链接的请求)

- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {
    if (navigationType == UIWebViewNavigationTypeLinkClicked ) {
        [[UIApplication sharedApplication] openURL:[request URL]];
        return NO;
    }

    return YES;
}

Swift版本:

func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
        if navigationType == UIWebViewNavigationType.LinkClicked {
            UIApplication.sharedApplication().openURL(request.URL!)
            return false
        }
        return true
    }

Swift 3版本:

func webView(_ webView: UIWebView, shouldStartLoadWith request: URLRequest, navigationType: UIWebViewNavigationType) -> Bool {
    if navigationType == UIWebViewNavigationType.linkClicked {
        UIApplication.shared.openURL(request.url!)
        return false
    }
    return true
}

更新

正如openURLiOS 10中已弃用的那样:

func webView(_ webView: UIWebView, shouldStartLoadWith request: URLRequest, navigationType: UIWebView.NavigationType) -> Bool {
    guard let url = request.url, navigationType == .linkClicked else { return true }
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
    return false
}

Answer 2

如果有人想知道,Drawnonward的解决方案在Swift中会是这样的:

func webView(webView: UIWebView!, shouldStartLoadWithRequest request: NSURLRequest!, navigationType: UIWebViewNavigationType) -> Bool {
    if navigationType == UIWebViewNavigationType.LinkClicked {
        UIApplication.sharedApplication().openURL(request.URL)
        return false
    }
    return true
}

Answer 3

对user306253的回答快速评论:当你尝试在UIWebView中自己加载某些内容时(即使是代码中),请注意这一点,这种方法可以防止它发生.

你可以做些什么来防止这种情况(感谢韦德):

if (inType == UIWebViewNavigationTypeLinkClicked) {
    [[UIApplication sharedApplication] openURL:[inRequest URL]];
    return NO;
}

return YES;

您可能还想处理UIWebViewNavigationTypeFormSubmitted和UIWebViewNavigationTypeFormResubmitted类型.

Answer 4

其他答案有一个问题:它们依赖于您执行的操作而不依赖于链接本身来决定是在Safari中还是在Webview中加载它.

现在有时这正是你想要的,这很好; 但有些时候,特别是如果你的页面中有锚链接,你真的只想打开Safari中的外部链接,而不是内部链接.在这种情况下,您应该检查URL.host您的请求的属性.

我使用那段代码来检查我是否在正在解析的URL中有一个主机名,或者它是否嵌入了html:

- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {
    static NSString *regexp = @"^(([a-zA-Z]|[a-zA-Z][a-zA-Z0-9-]*[a-zA-Z0-9])[.])+([A-Za-z]|[A-Za-z][A-Za-z0-9-]*[A-Za-z0-9])$";
    NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF MATCHES %@", regexp];

    if ([predicate evaluateWithObject:request.URL.host]) {
        [[UIApplication sharedApplication] openURL:request.URL];
        return NO; 
    } else {
        return YES; 
    }
}

您当然可以调整正则表达式以满足您的需求.

Answer 5

在Swift中,您可以使用以下代码:

extension YourViewController: UIWebViewDelegate {
    func webView(_ webView: UIWebView, shouldStartLoadWith request: URLRequest, navigationType: UIWebView.NavigationType) -> Bool {
        if let url = request.url, navigationType == UIWebView.NavigationType.linkClicked {
            UIApplication.shared.open(url, options: [:], completionHandler: nil)
            return false
        }
        return true
    }

}

确保检查URL值和navigationType.