ios - 如何在单击按钮时打开 instagram 应用程序

Question

1. 大家好，我正在尝试通过单击按钮打开 instagram 应用程序，但我没有我也可以在 plist 中将 Url scheme 设置为 instagram

   NSString *instagramURL = @"instagram://app"; 
   NSURL *ourURL = [NSURLURLWithString:instagramURL]; 
   if ([[UIApplication sharedApplication]canOpenURL:ourURL]) {
       [[UIApplication sharedApplication]openURL:ourURL];
   
   } else {
       //The App is not installed. It must be installed from iTunes code.
       NSString *iTunesLink = @"//Some other Url goes here";
       [[UIApplication sharedApplication] openURL:[NSURL URLWithString:iTunesLink]];
   
       UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"URL error"
             message:[NSString stringWithFormat: @"No custom URL defined for %@", ourURL]
             delegate:self cancelButtonTitle:@"Ok" otherButtonTitles:nil];
       [alert show];
   

   我确实喜欢这个但是应用程序无法打开对 iOS 来说是新的任何帮助都可以赞赏

Answer 1

您可以通过用户名打开 instagram 应用程序，例如，

 NSURL *instagramURL = [NSURL URLWithString:@"instagram://user?username=USERNAME"];
if ([[UIApplication sharedApplication] canOpenURL:instagramURL]) {
    [[UIApplication sharedApplication] openURL:instagramURL];
}

可以引用iPhone Hooks of Instagram更多关于api的方法和细节!!!

更新:

替换下面一行，

 NSString *instagramURL = @"instagram://app"; 

与

   NSURL *instagramURL = [NSURL URLWithString:@"instagram://app"];

您正在直接分配字符串作为 url!!