如何打開一個UIWebView到另一個UIWebView的鏈接？

Question

我正在構建一個iOS應用。
我在帶有超鏈接的UIWebView具有HTML內容，但是無法打開到另一個UIWebView的鏈接。
我使用UIWebView作為ViewController的子視圖。 這是代碼：

- (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

您只需要通過Web視圖選擇到新的控制器，然后將請求傳遞給它。 所以在您的第一種情況下，

 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.
        }

請注意，performSegueWithIdentifier：sender：中的sender參數是委托方法返回的請求。 將此請求傳遞到您要選擇的視圖控制器中的屬性，

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

最后，使用該請求將Web視圖加載到NextViewController中。