Question

出于安全考虑，我过去常常在手机游戏中使用匿名者。然而这有点烦人。如果我搜索了某些内容并点击了链接，那么google会返回错误消息。您可以尝试.. http://proxy2974.my-addr.org/myaddrproxy.php/http/www.google.com.au/（搜索某些内容并点击链接 - 如果您在桌面浏览器上进行测试，它会正常运行..但如果您在iPhone或iPhone上试用它，它会返回http://proxy2974.my-addr.org/myaddrproxy.php/http/url模拟器）

所以我决定自己做。我正在做的是从文本字段中获取URL并将其传递给匿名者的链接，如urlString = [NSString stringWithFormat:@"http://proxy2974.my-addr.org/myaddrproxy.php/http/%@", urlString];

然而，我仍然面临同样的问题..当我点击谷歌上的链接时，它会返回一个错误..所以我想要做的是获取点击的链接，停止加载页面（在它返回之前）错误），然后通过匿名者传递它..我怎么能这样做？感谢..

- (BOOL)webView:(UIWebView *) sender shouldStartLoadWithRequest:(NSURLRequest *) request navigationType:(UIWebViewNavigationType) navigationType {

    NSLog(@"req: %@",request.URL.absoluteString);


    return YES;
}

req: http://proxy2974.my-addr.org/url?sa=t&source=web&cd=1&ved=0CFMQFjAA&url=%2Fmyaddrproxy.php%2Fhttp%2Fwww.perthnow.com.au%2Ffun-games%2Fleft-brain-vs-right-brain%2Fstory-e6frg46u-1111114517613&ei=GXAsUMaUA7H44QTys4HYDA&usg=AFQjCNGB_zOrrEZC0SKx813XGHB1xi_AlA

req: http://proxy2974.my-addr.org/myaddrproxy.php?proxy_url_sjla67z78f8viz4=url&sa=t&source=web&cd=1&ved=0CFMQFjAA&url=%2Fmyaddrproxy.php%2Fhttp%2Fwww.perthnow.com.au%2Ffun-games%2Fleft-brain-vs-right-brain%2Fstory-e6frg46u-1111114517613&ei=GXAsUMaUA7H44QTys4HYDA&usg=AFQjCNGB_zOrrEZC0SKx813XGHB1xi_AlA

req: http://proxy2974.my-addr.org/myaddrproxy.php/http/url

Answer 1

实施

- (BOOL)webView:(UIWebView*)webView shouldStartLoadWithRequest:(NSURLRequest*)request navigationType:(UIWebViewNavigationType)navigationType;

在你的uiwebview代表中

Answer 2

实现此委托方法以检索UIWebView中的单击链接：

- (BOOL)webView:(UIWebView*)webView shouldStartLoadWithRequest:(NSURLRequest*)request navigationType:(UIWebViewNavigationType)navigationType
{
  if(navigationType == UIWebViewNavigationTypeLinkClicked)
  {
     NSLog(@"req: %@",request.URL.absoluteString);
  }
}

如何在iOS中检索单击的链接

2 个答案: