ios - 如何打开从一个 UIWebView 到另一个 UIWebView 的链接？

Question

我正在构建一个 iOS 应用。
我在带有超链接的 UIWebView 中有 HTML 内容，但我无法打开指向另一个 UIWebView 的链接。
我使用 UIWebView 作为 ViewController 的 subview 。这是代码:

- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType
{
    switch (navigationType) {
        case UIWebViewNavigationTypeLinkClicked: { //this is when a user tap is detected
            //write the handling code here.
            isWebViewLoaded = false; //set this to false only if you open another view  controller.
            return NO; //prevent tapped URL from loading inside the UIWebView.
        }

        // some other typical parameters within a UIWebView. Use what is needed
        case UIWebViewNavigationTypeFormResubmitted: return YES;
        case UIWebViewNavigationTypeReload: return YES;

        //for all other cases, including UIWebViewNavigationTypeOther called when UIWebView is loading for the first time
        default: {
            if (!isWebViewLoaded) { 
                isWebViewLoaded = true; 
                return YES; 
            }
            else 
                return NO;
        } 
    }
} 

Answer 1

您只需要使用 Web View 转至一个新的 Controller ，并将请求传递给它。所以在你的第一个案例中是这样的，

 case UIWebViewNavigationTypeLinkClicked: { //this is when a user tap is detected
            [self performSegueWithIdentifier:@"Next" sender:request];
            isWebViewLoaded = false; //set this to false only if you open another view  controller.
            return NO; //prevent tapped URL from loading inside the UIWebView.
        }

注意 performSegueWithIdentifier:sender: 中的 sender 参数是委托(delegate)方法返回的请求。将此请求传递给您要转到的 View Controller 中的属性，

-(void)prepareForSegue:(UIStoryboardSegue *)segue sender:(NSURLRequest *)sender {
    NextViewController *next = segue.destinationViewController;
    next.request = sender;
}

最后，使用该请求在 NextViewController 中加载 WebView 。