程序跳转到itunes商店
2013-08-23 10:38 三戒1993 阅读(131) 评论(0) 编辑 收藏 举报找到应用程序,点击应用程序下面的小三角图标,再选择”复制链接“,就可以获取此应用的链接了。
比如:
itunes.apple.com/cn/app/bai-du-wen-kuhd/id483064532?mt=8
然后将 https:// 替换为 itms:// 或者 itms-apps://:
然后在程序中写如下代码:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"itms://
itunes.apple.com/cn/app/bai-du-wen-kuhd/id483064532?mt=8"]];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"itunes.apple.com/cn/app/bai-du-wen-kuhd/id483064532?mt=8"]]
"Error Domain=WebKitErrorDomain Code=101 The URL can't be shown."
http://stackoverflow.com/questions/4299403/how-to-handle-app-urls-in-a-uiwebview
http://stackoverflow.com/questions/4299403/how-to-handle-app-urls-in-a-uiwebview
- (void)webView:(UIWebView *)wv didFailLoadWithError:(NSError *)error {
// Give iOS a chance to open it.
NSURL *url = [NSURL URLWithString:[error.userInfo objectForKey:@"NSErrorFailingURLStringKey" ]];
if ([error.domain isEqual:@"WebKitErrorDomain"]
&& error.code == 101
&& [[UIApplication sharedApplication]canOpenURL:url])
{
[[UIApplication sharedApplication]openURL:url];
return;
}
// Normal error handling…
}
- (BOOL)webView:(UIWebView *)wv shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {
// Determine if we want the system to handle it.
NSURL *url = request.URL;
if (![url.scheme isEqual:@"http"] && ![url.scheme isEqual:@"https"]) {
if ([[UIApplication sharedApplication]canOpenURL:url]) {
[[UIApplication sharedApplication]openURL:url];
return NO;
}
}
return YES;
}
- (void)webView:(UIWebView *)wv didFailLoadWithError:(NSError *)error {
// Ignore NSURLErrorDomain error -999.
if (error.code == NSURLErrorCancelled) return;
// Ignore "Fame Load Interrupted" errors. Seen after app store links.
if (error.code == 102 && [error.domain isEqual:@"WebKitErrorDomain"]) return;
// Normal error handling…
}
iphone开发笔记和技巧总结(原址持续更新)转载
http://www.cnblogs.com/zhwl/archive/2012/03/09/2387530.html