从 URL 和传递参数打开 iOS 应用程序 [英] Open iOS app from URL AND Pass Parameters

问题描述

一个链接应该可以打开应用.我有这个工作.我只想知道如何传递参数.假设 url 是addappt://?code=abc".当视图控制器弹出时，代码字段应该填充文本 - 等号后面的字母.我有一部分工作要做.我使用以下 (在 app delegate.m 中):

A link should open the app. I've got that to work. I just want to know how to pass a parameter. Let's say the url is "addappt://?code=abc". When a view controller pops up, a code field should have populated text - the letters after the equals to sign. I've got part of this to work. I use the following (in app delegate.m):

NSArray *elements = [url.query componentsSeparatedByString:@"="];
NSString *key = [[elements objectAtIndex:0] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
          val = [[elements objectAtIndex:1] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

(顺便说一句:val 在 appdelegate.h 中声明

(BTW: val is declared in appdelegate.h

我还可以将 val 传递给视图控制器.我唯一的问题是填充名为 'code' 的文本字段.如何在通过链接打开应用后立即填充代码?

I am also able to pass val to the view controller. My only problem is populating the textfield, named 'code'. How can you populate code as soon as the app is opened by the link?

感谢帮助.

推荐答案

这里有一个关于 在 iOS 中使用自定义 URL Scheme

在教程中，您应该解析 URL 参数并将它们存储在此方法中的应用程序中使用:

As in the tutorial, you should parse the URL parameters and store them to use in the app in this method:

- (BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url {
  // Do something with the url here
}

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