从Safari中的Web应用程序中的链接打开iphone应用程序

我正试图通过safari中的链接打开我的原生iPhone应用程序。我已关注此链接创建一个url架构。我已将 appgo：// 添加为我的网址架构，将com.xxxx.appgo添加为网址类型下的标识符。以下是我在safari网页中的链接：打开iPhone App
但是当我点击链接时，应用程序无法打开并且safari会生成错误警报：Safari无法进入页面，因为地址无效。

I am trying to open my native iPhone application from a link in safari. I have followed this link to create a url schema. I have added appgo:// as my url schema and com.xxxx.appgo as Identifier under URL Types. Following is my link in web page in safari: Open iPhone App But when I click on the link the app doesn't open up and safari produce an alert with error: Safari cannot enter the page as the address is invalid.

我的包标识符：com.xxxxx.abc
注意：我的包标识符与URL类型中的标识符不同。这可能是一个问题吗？

my bundle identifier: com.xxxxx.abc Note: My bundle identifier is different from the identifier in URL types. Can that be an issue?

编辑：
我已将包标识符和网址标识符设为相同。我还在我的app delegate中添加了以下代码：

I have made bundle identifier and url identifier same. I have also added following code in my app delegate:

- (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url{
NSLog(@"application");
BOOL returnValue = NO;
NSString *urlString = [url absoluteString];
if([urlString hasPrefix:@"com.xxxx.appgo"]){
    returnValue = YES;
}
return returnValue;

}

我正在测试它我的iPad。我首先从xcode安装最新版本的应用程序，然后按主页按钮，然后打开safari中的链接。但是我仍然收到警报说Safari无法进入页面，因为地址无效。我在safari中的链接：

I am testing it in my iPad. I first install the latest version of app from xcode and then press on home button and then open the link in safari. But I am still getting the alert saying the Safari cannot enter the page as address is invalid. My link in safari:

<a href="appgo://">Open iPhone App</a>