从URL和Pass Parameters打开iOS应用程序 [英] Open iOS app from URL AND Pass Parameters
问题描述
链接应该打开应用程序。我有这个工作。我只是想知道如何传递参数。假设网址是addappt://?code = abc。弹出视图控制器时,代码字段应填充文本 - 等于符号后的字母。我有部分工作要做。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];
(BTW:val在appdelegate.h中声明
(BTW: val is declared in appdelegate.h
我也可以将 val
传递给视图控制器。我唯一的问题是填充文本字段,名为'code'
。如果通过链接打开应用程序,您如何填写代码?
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?
帮助感谢。
推荐答案
这是关于在iOS中使用自定义URL方案
在教程中,您应该解析URL参数并存储它们在此方法中在应用程序中使用:
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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