我得在游览中打开一个 URL... 在做一些关于游览的工作之后......我必须关闭这个游览...
我的代码如下
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:strPageUrl]];
事实上我必须在我的空间上分享视频。 如果用户分享了视频, 那么我就必须关闭搜索, 并需要再次激活我的应用程序 。
我想做类似FacebookSDK登入 与Safari...
请帮我打开这个
我得在游览中打开一个 URL... 在做一些关于游览的工作之后......我必须关闭这个游览...
我的代码如下
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:strPageUrl]];
事实上我必须在我的空间上分享视频。 如果用户分享了视频, 那么我就必须关闭搜索, 并需要再次激活我的应用程序 。
我想做类似FacebookSDK登入 与Safari...
请帮我打开这个
自定义 URL 计划只能用一种方式运作... 就像您从应用程序中打开了 afari 一样, 您无法回到您的应用程序, 除非第二个应用程序也是您的, 您已经实施了另一个自定义 URL 来返回您的应用程序 。 因为 safari 是苹果做的, 所以它没有关于您应用程序的信息 。
我不知道您想要达到什么目标,但如果有帮助,您可以试一下 UIWebView
的要求。
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