如何打开一个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具有HTML内容，但是无法打开到另一个UIWebView的链接。
我使用UIWebView作为ViewController的子视图。 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

您只需要通过Web视图选择到新的控制器，然后将请求传递给它。 所以在您的第一种情况下，

 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.
        }

请注意，performSegueWithIdentifier：sender：中的sender参数是委托方法返回的请求。 将此请求传递到您要选择的视图控制器中的属性，

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

最后，使用该请求将Web视图加载到NextViewController中。