如何启动打开我的自定义网址方案的应用程序？ [英] How can I launch back the app that opened my custom URL scheme?

问题描述

我正在使用一个管理我自己的URL方案的应用程序，因此我实现回调：

   - （BOOL）应用程序：（UIApplication *）应用程序didFinishLaunchingWithOptions（NSDictionary *）launchOptions 
 {
 //获取我们的启动URL 
 if（launchOptions！= nil）
 {
 //启动字典有数据
 NSURL * launchURL = [launchOptions objectForKey：UIApplicationLaunchOptionsURLKey]; 
 
 //解析URL 
 NSString * hostString = [launchURL host]; 
 
 blah blah blah ... 
  

以启动调用者应用程序（即打开URL的应用程序）。

我一直在玩 UIApplicationLaunchOptionsSourceApplicationKey ，但我无法启动通过应用程序捆绑ID退回应用程序。我可以吗？

我也尝试过 launchApplicationWithIdentifier： UIApplication ，但我需要一个真正的解决方案，似乎解决方法只适用于模拟器。

任何想法？谢谢！

解决方案

唯一的方法是让这两个应用程序都支持自定义网址方案。然后将调用者URL嵌入到其他应用程序的URL中。

例如，让App2想调用App1，这样App1就可以回调到App2。它将创建并打开如下所示的URL：

  app1：//？caller = app2％3A％2F％2Fblabla 
  

当您解码调用者字符串 app2：// blabla ，然后您可以再次使用 openURL： p>

I'm working on an app that manages my own URL scheme so I implement the callback:

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions (NSDictionary *)launchOptions
{
    // Get our launch URL
    if (launchOptions != nil)
    {
        // Launch dictionary has data
        NSURL* launchURL = [launchOptions objectForKey: UIApplicationLaunchOptionsURLKey];

        // Parse the URL
        NSString* hostString = [launchURL host];

        blah blah blah...

It works very nice but I need to launch the caller application (i.e. the app that opened the URL). So my question here is, is it possible?

I have been playing with UIApplicationLaunchOptionsSourceApplicationKey but I can't launch back the app by its application Bundle ID. Can I?

I have also tried the undocumented launchApplicationWithIdentifier: of UIApplication, but I need a real solution and it seems that workaround only works in the Simulator.

Any ideas? Thank you!

解决方案

The only way would be to have both apps each support a custom URL scheme. Then you embed the caller URL in the URL of the other app.

For example, let's say App2 wants to call App1 in a way so that App1 could then "call back" to App2. It would create and open an URL like this:

app1://?caller=app2%3A%2F%2Fblabla

When you decode the caller part you would get back the string app2://blabla which you could then again open with openURL: to "call back".

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