在我的应用程序中,我正在尝试在Web视图上加载特定的URL。网址已正确加载。当事务失败时,从Web视图响应返回的绝对URL是https://abc.xxx.com/check/resources/load.html#failed
并且当它成功时返回的绝对URL是
https://abc.xxx.com/check/resources/load.html#SUCCESSFUL:1048:7771.00:INR:1397312651798
我在获取响应时遇到问题,就像我获得成功的标记一样,如何从响应中获取字符串SUCCESSFUL。如果响应成功或失败,我想执行条件检查。
答案 0 :(得分:0)
您需要在UIWebView
上设置委托。
然后使用UIWebViewDelegate
方法-webView:shouldStartLoadWithRequest:navigationType:
创建Web视图时:
webView.delegate = self;
然后在视图控制器中添加包含Web视图的委托方法。
-(BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType
{
NSURL *url = request.URL;
if ([url.host isEqualToString:@"abc.xxx.com"]) {
NSString *fragment = url.fragment;
// Do something with the fragment which is the bit after the #
// Return NO if you don't want to load the URL
}
return YES;
}