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我想在Safari瀏覽器中顯示我的網頁視圖中打開某些鏈接。這是我迄今爲止的代碼。網絡視圖在某些關鍵字的Safari瀏覽器中打開鏈接
-(BOOL) webView:(UIWebView *)inWeb shouldStartLoadWithRequest:(NSURLRequest *)inRequest navigationType:(UIWebViewNavigationType)inType {
if (inType == UIWebViewNavigationTypeLinkClicked) {
[[UIApplication sharedApplication] openURL:[inRequest URL]];
return NO;
}
return YES;
}
問題是我想只打開Safari瀏覽器,如果鏈接包含關鍵字「谷歌」。有關如何做的任何提示?
酷,如果我想的相反說在Safari中打開的所有鏈接除非谷歌的關鍵字我切換返回值的權利。如果([[inRequest.URL absoluteString] rangeOfString:@「google」]。位置== NSNotFound){} – 2013-03-07 22:33:27
)然後從檢查 – nsgulliver 2013-03-07 22:34:13
取出'!' shouldStartLoadWithRequest:(的NSURLRequest *)inRequest navigationType:(UIWebViewNavigationType)法菜單{ 如果(法菜單== UIWebViewNavigationTypeLinkClicked){ 如果([[inRequest.URL absoluteString] rangeOfString:@ 「谷歌」]。位置== NSNotFound){ [[ UIApplication sharedApplication] openURL:[inRequest URL]]; return NO; } } return YES; } – nsgulliver 2013-03-07 22:34:43