blocks|key|3348256|text|不，不是。除了文档中的URL处理程序之外，没有办法与/启动另一个应用程序进行通信。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3348257|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

No it's not. Besides the documented URL handlers, there's no way to communicate with/launch another app.

blocks|key|3348318|text|正如Kevin所指出的，URL方案是应用程序之间通信的唯一方式。所以，不，不可能启动任意的应用程序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3348319|但是可以启动任何注册URL方案的应用程序，无论它是苹果的、你的，还是其他开发人员的。文档在这里：|3348320|Defining+a+Custom+URL+Scheme+for+Your+App|offset|length|3348321|至于启动手机，看起来你的tel:链接至少需要有三位数才能启动手机。所以你不能不拨一个号码就直接进入这款应用。|style|CODE|3348322|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/documentation/xcode/allowing_apps_and_websites_to_link_to_your_content/defining_a_custom_url_scheme_for_your_app^0|0|0|0|15|0|0|C|4|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@$F|W|G|X|1|Y]]|A|$]]|$1|H|3|I|5|6|7|Z|8|@$F|10|G|11|J|K]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

As Kevin points out, URL Schemes are the only way to communicate between apps. So, no, it's not possible to launch arbitrary apps.
But it is possible to launch any app that registers a URL Scheme, whether it's Apple's, yours, or another developer's. The docs are here:
<a href="https://developer.apple.com/documentation/xcode/allowing_apps_and_websites_to_link_to_your_content/defining_a_custom_url_scheme_for_your_app" rel="nofollow noreferrer">Defining a Custom URL Scheme for Your App</a>
As for launching the phone, looks like your <code>tel:</code> link needs to have least three digits before the phone will launch. So you can't just drop into the app without dialing a number.

blocks|key|3135496|text|您只能启动已注册URL方案的应用程序。然后，就像您使用sms:打开SMS应用程序一样，您将能够使用他们的URL方案打开应用程序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3135497|在文档中有一个非常好的例子，叫做LaunchMe，它演示了这一点。|3135498|截至2017年11月6日的LaunchMe+sample+code。|offset|length|3135499|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/content/samplecode/LaunchMe/Introduction/Intro.html#//apple_ref/doc/uid/DTS40007417^0|0|0|D|K|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@$F|S|G|T|1|U]]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

You can only launch apps that have registered a URL scheme. Then just like you open the SMS app by using sms:, you'll be able to open the app using their URL scheme.

There is a very good example available in the docs called LaunchMe which demonstrates this.

<a href="https://developer.apple.com/library/content/samplecode/LaunchMe/Introduction/Intro.html#//apple_ref/doc/uid/DTS40007417" rel="nofollow noreferrer">LaunchMe sample code</a> as of 6th Nov 2017.

blocks|key|3135531|text|以下是从另一个应用程序中启动应用程序的很好的教程：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3135532|iOS+SDK:+Working+with+URL+Schemes|offset|length|3135533|而且，不可能启动任意应用程序，但可以启动注册了URL+Schemes的本地应用程序。|3135534|entityMap|0|LINK|mutability|MUTABLE|url|http://mobile.tutsplus.com/tutorials/iphone/ios-sdk-working-with-url-schemes/|1|http://developer.apple.com/library/ios/documentation/iPhone/Conceptual/iPhoneOSProgrammingGuide/AdvancedAppTricks/AdvancedAppTricks.html#//apple_ref/doc/uid/TP40007072-CH7-SW50^0|0|0|X|0|0|N|B|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@$D|X|E|Y|1|Z]]|A|$]]|$1|H|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]|P|$5|K|L|M|A|$N|Q]]]]

Here is a good tutorial for launching application from within another app: 
<a href="http://mobile.tutsplus.com/tutorials/iphone/ios-sdk-working-with-url-schemes/" rel="noreferrer">iOS SDK: Working with URL Schemes</a> 
And, it is not possible to launch arbitrary application, but the native applications which registered the <a href="http://developer.apple.com/library/ios/documentation/iPhone/Conceptual/iPhoneOSProgrammingGuide/AdvancedAppTricks/AdvancedAppTricks.html#//apple_ref/doc/uid/TP40007072-CH7-SW50" rel="noreferrer">URL Schemes</a>.

blocks|key|351327|text|我之前也尝试过(Launch+iPhone+Application+with+Identifier)，但肯定没有记录的方法来做这件事。:)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|351328|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/7374540/launch-iphone-application-with-identifier^0|8|15|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

I also tried this a while ago (<a href="https://stackoverflow.com/questions/7374540/launch-iphone-application-with-identifier">Launch iPhone Application with Identifier</a>), but there definitely is no DOCUMENTED way to do this. :)

blocks|key|354004|text|我发现，编写一个可以打开另一个应用程序的应用程序很容易。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|354005|假设我们有两个名为FirstApp和SecondApp的应用程序。当我们打开FirstApp时，我们希望能够通过单击按钮来打开SecondApp。执行此操作的解决方案是：|offset|length|style|CODE|354006|SecondApp中的|354007|ordered-list-item|354008|354009|转到SecondApp的plist文件，您需要添加一个包含字符串iOSDevTips的URL+Schemes+(当然，您也可以编写另一个字符串，这取决于您)。|BOLD|354010|​|354011|📷|atomic|354012|354013|2.在FirstApp中|354014|使用以下操作创建一个按钮：|354015|-+(void)buttonPressed:(UIButton+*)button
{
++NSString+*customURL+=+@"iOSDevTips://";

++if+([[UIApplication+sharedApplication]+canOpenURL:[NSURL+URLWithString:customURL]])
++{
++++[[UIApplication+sharedApplication]+openURL:[NSURL+URLWithString:customURL]];
++}
++else
++{
++++UIAlertView+*alert+=+[[UIAlertView+alloc]+initWithTitle:@"URL+error"
++++++++++++++++++++++++++++++message:[NSString+stringWithFormat:@"No+custom+URL+defined+for+%25@",+customURL]
++++++++++++++++++++++++++++++delegate:self+cancelButtonTitle:@"Ok"+
++++++++++++++++++++++++++++++otherButtonTitles:nil];
++++[alert+show];
++}

}|code-block|syntax|javascript|354016|就这样。现在，当您可以单击FirstApp中的按钮时，它应该会打开SecondApp。|354017|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/fbb311a88a25e04ea557998a925a021a.png|imageAlt^0|0|9|8|I|9|0|0|0|0|W|A|17|B|0|0|0|1|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1G|8|@$D|1H|E|1I|F|G]|$D|1J|E|1K|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1L|8|@]|9|@]|A|$]]|$1|J|3|-4|5|K|7|1M|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1O|8|@$D|1P|E|1Q|F|O]|$D|1R|E|1S|F|O]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1T|8|@]|9|@]|A|$]]|$1|R|3|S|5|T|7|1U|8|@]|9|@$D|1V|E|1W|1|1X]]|A|$]]|$1|U|3|Q|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|20|8|@]|9|@]|A|$]]|$1|Z|3|10|5|11|7|21|8|@]|9|@]|A|$12|13]]|$1|14|3|15|5|6|7|22|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|23|8|@]|9|@]|A|$]]]|17|$18|$5|19|1A|1B|A|$1C|1D|1E|-4]]]]

I found that it's easy to write an app that can open another app. 

Let's assume that we have two apps called <code>FirstApp</code> and <code>SecondApp</code>. When we open the FirstApp, we want to be able to open the SecondApp by clicking a button. The solution to do this is: 

<ol>
<li>In SecondApp

Go to the plist file of SecondApp and you need to add a URL Schemes with a string iOSDevTips(of course you can write another string.it's up to you).</li>
</ol>

<img src="https://i.stack.imgur.com/DWTsi.png" alt="enter image description here">

2
. In FirstApp

Create a button with the below action:

<pre><code>- (void)buttonPressed:(UIButton *)button
{
 NSString *customURL = @"iOSDevTips://";

 if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:customURL]])
 {
 [[UIApplication sharedApplication] openURL:[NSURL URLWithString:customURL]];
 }
 else
 {
 UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"URL error"
 message:[NSString stringWithFormat:@"No custom URL defined for %@", customURL]
 delegate:self cancelButtonTitle:@"Ok" 
 otherButtonTitles:nil];
 [alert show];
 }

}
</code></pre>

That's it. Now when you can click the button in the FirstApp it should open the SecondApp.

blocks|key|3135687|text|尝试以下代码将帮助您从应用程序启动应用程序|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3135688|注意:用实际的应用程序名称替换fantacy名称|3135689|NSString+*mystr=[[NSString+alloc]+initWithFormat:@"fantacy://location?id=1"];
NSURL+*myurl=[[NSURL+alloc]+initWithString:mystr];
[[UIApplication+sharedApplication]+openURL:myurl];|code-block|syntax|javascript|3135690|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try the following code will help you to Launch an application from your application 

Note: Replace the name fantacy with actual application name 

<pre><code>NSString *mystr=[[NSString alloc] initWithFormat:@"fantacy://location?id=1"];
NSURL *myurl=[[NSURL alloc] initWithString:mystr];
[[UIApplication sharedApplication] openURL:myurl];
</code></pre>

blocks|key|354037|text|李的答案对于8之前的iOS是绝对正确的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|354038|在iOS+9%2B中，您必须将您的应用程序想要在Info.plist中查询的任何URL方案列在LSApplicationQueriesSchemes键(字符串数组)下：|354039|​|354040|📷|atomic|offset|length|354041|354042|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/f705c6cbaf72ef07251fbccf6c2d0d59.png|imageAlt^0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@$I|Y|J|Z|1|10]]|A|$]]|$1|K|3|E|5|6|7|11|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S|T|-4]]]]

The lee answer is absolutely correct for iOS prior to 8.

In iOS 9+ you must whitelist any URL schemes your App wants to query in Info.plist under the LSApplicationQueriesSchemes key (an array of strings):

<a href="https://i.stack.imgur.com/txMUq.png" rel="noreferrer"><img src="https://i.stack.imgur.com/txMUq.png" alt="enter image description here"></a>

blocks|key|3135728|text|Swift中的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3135729|以防有人想要快速复制和粘贴。|offset|length|style|BOLD|3135730|if+let+url+=+URL(string:+"app://"),+UIApplication.shared.canOpenURL(url)+{
++++++++++++UIApplication.shared.open(url,+options:+[:],+completionHandler:+nil)
++++++++}+else+if+let+itunesUrl+=+URL(string:+appLink),+UIApplication.shared.canOpenURL(itunesUrl)+{
++++++++++++UIApplication.shared.open(itunesUrl,+options:+[:],+completionHandler:+nil)
++++++++}|code-block|syntax|javascript|3135731|entityMap^0|0|0|E|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In Swift
Just in case someone was looking for a quick Swift copy and paste.
<pre><code>if let url = URL(string: &quot;app://&quot;), UIApplication.shared.canOpenURL(url) {
 UIApplication.shared.open(url, options: [:], completionHandler: nil)
 } else if let itunesUrl = URL(string: appLink), UIApplication.shared.canOpenURL(itunesUrl) {
 UIApplication.shared.open(itunesUrl, options: [:], completionHandler: nil)
 }
</code></pre>

blocks|key|352755|text|Swift+3快速粘贴版本的@vily393回答：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|352756|if+UIApplication.shared.canOpenURL(openURL)+{
++++UIApplication.shared.openURL(openURL)
}+else+if+UIApplication.shared.canOpenURL(installUrl)+
++++UIApplication.shared.openURL(installUrl)
}|code-block|syntax|javascript|352757|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Swift 3 quick and paste version of @villy393 answer for :

<pre><code>if UIApplication.shared.canOpenURL(openURL) {
 UIApplication.shared.openURL(openURL)
} else if UIApplication.shared.canOpenURL(installUrl) 
 UIApplication.shared.openURL(installUrl)
}
</code></pre>

blocks|key|3135845|text|为了让您从另一个应用程序打开您的应用程序，您需要在这两个应用程序中进行更改。以下是将Swift+3与iOS+10更新配合使用的步骤：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3135846|1.注册要打开的应用程序|3135847|通过定义您的应用程序的自定义和唯一的URL方案来更新Info.plist。|CODE|3135848|​|3135849|📷|atomic|3135850|3135851|请注意，您的方案名称应该是唯一的，否则，如果您的设备上安装了另一个具有相同URL方案名称的应用程序，则将在运行时确定哪个应用程序被打开。|3135852|URL+2.在主应用程序中包含以前的方案|3135853|您需要指定希望应用程序能够与UIApplication类的canOpenURL:方法一起使用的URL方案。因此，打开主应用程序的Info.plist并将另一个应用程序的URL方案添加到LSApplicationQueriesSchemes。(在iOS+9.0中引入)|3135854|3135855|3135856|3135857|3.实现打开应用程序的操作|3135858|现在一切都设置好了，可以在打开其他应用程序的主应用程序中编写代码了。这应该看起来像这样：|3135859|let+appURLScheme+=+"MyAppToOpen://"

guard+let+appURL+=+URL(string:+appURLScheme)+else+{
++++return
}

if+UIApplication.shared.canOpenURL(appURL)+{

++++if+#available(iOS+10.0,+*)+{
++++++++UIApplication.shared.open(appURL)
++++}
++++else+{
++++++++UIApplication.shared.openURL(appURL)
++++}
}
else+{
++++//+Here+you+can+handle+the+case+when+your+other+application+cannot+be+opened+for+any+reason.
}|code-block|syntax|javascript|3135860|请注意，如果您希望打开现有应用程序(从AppStore安装)，则这些更改需要新版本。如果你想打开一个已经发布到苹果AppStore的应用程序，那么你需要首先上传一个新版本，其中包括你的网址方案注册。|3135861|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/51aa31bc49c3db9d87762b156ac42fd1.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/7d1985712a953f0167bf33199c4550ae.png^0|16|7|1E|6|0|0|7|0|Q|A|0|0|0|1|0|0|0|0|I|2|0|0|3P|E|D|T|B|1S|A|2K|R|0|0|1|0|0|1|0|1|1|0|0|1|0|0|D|0|0|18|0|0|B3|0|0|2R|0^^$0|@$1|2|3|4|5|6|7|1L|8|@$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1Q|8|@$9|1R|A|1S|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1T|8|@$9|1U|A|1V|B|J]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1W|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|1X|8|@]|D|@$9|1Y|A|1Z|1|20]]|E|$]]|$1|P|3|L|5|6|7|21|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|22|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|23|8|@$9|24|A|25|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|26|8|@$9|27|A|28|B|C]|$9|29|A|2A|B|J]|$9|2B|A|2C|B|J]|$9|2D|A|2E|B|J]|$9|2F|A|2G|B|J]]|D|@]|E|$]]|$1|W|3|L|5|6|7|2H|8|@$9|2I|A|2J|B|C]]|D|@]|E|$]]|$1|X|3|N|5|O|7|2K|8|@$9|2L|A|2M|B|C]]|D|@$9|2N|A|2O|1|2P]]|E|$]]|$1|Y|3|L|5|6|7|2Q|8|@$9|2R|A|2S|B|C]]|D|@]|E|$]]|$1|Z|3|10|5|6|7|2T|8|@$9|2U|A|2V|B|C]]|D|@]|E|$]]|$1|11|3|12|5|6|7|2W|8|@$9|2X|A|2Y|B|C]]|D|@]|E|$]]|$1|13|3|14|5|15|7|2Z|8|@$9|30|A|31|B|C]]|D|@]|E|$16|17]]|$1|18|3|19|5|6|7|32|8|@$9|33|A|34|B|C]]|D|@]|E|$]]|$1|1A|3|-4|5|6|7|35|8|@]|D|@]|E|$]]]|1B|$1C|$5|1D|1E|1F|E|$1G|1H|1I|-4]]|1J|$5|1D|1E|1F|E|$1G|1K|1I|-4]]]]

In order to let you open your application from another, you'll need to make changes in both applications. Here are the steps using Swift 3 with iOS 10 update:

1. Register your application that you want to open

Update the <code>Info.plist</code> by defining your application's custom and unique URL Scheme.

<a href="https://i.stack.imgur.com/OJM5v.png" rel="noreferrer"><img src="https://i.stack.imgur.com/OJM5v.png" alt="enter image description here"></a>

Note that your scheme name should be unique, otherwise if you have another application with the same URL scheme name installed on your device, then this will be determined runtime which one gets opened.

2. Include the previous URL scheme in your main application

You'll need to specify the URL scheme you want the app to be able to use with the <code>canOpenURL:</code> method of the <code>UIApplication</code> class. So open the main application's <code>Info.plist</code> and add the other application's URL scheme to <code>LSApplicationQueriesSchemes</code>. (Introduced in iOS 9.0)

<a href="https://i.stack.imgur.com/ducnT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ducnT.png" alt="enter image description here"></a>

3. Implement the action that opens your application

Now everything is set up, so you're good to write your code in your main application that opens your other app. This should looks something like this:

<pre><code>let appURLScheme = "MyAppToOpen://"

guard let appURL = URL(string: appURLScheme) else {
 return
}

if UIApplication.shared.canOpenURL(appURL) {

 if #available(iOS 10.0, *) {
 UIApplication.shared.open(appURL)
 }
 else {
 UIApplication.shared.openURL(appURL)
 }
}
else {
 // Here you can handle the case when your other application cannot be opened for any reason.
}
</code></pre>

Note that these changes requires a new release if you want your existing app (installed from AppStore) to open. If you want to open an application that you've already released to Apple AppStore, then you'll need to upload a new version first that includes your URL scheme registration.

blocks|key|3348706|text|要实现这一点，我们需要在两个应用程序中添加几行代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3348707|应用程序A:您要从其他应用程序打开的。(源)|offset|length|style|BOLD|3348708|应用程序B：您要从应用程序B打开应用程序A+(目标)|3348709|应用程序A的代码|3348710|在应用程序A的Plist中添加几个标签打开应用程序A的Plist源代码，并在下面添加|3348711|<key>CFBundleURLTypes</key>
<array>
++++<dict>
++++++++<key>CFBundleURLName</key>
++++++++<string>com.TestApp</string>
++++++++<key>CFBundleURLSchemes</key>
++++++++<array>
++++++++++++<string>testApp.linking</string>
++++++++</array>
++++</dict>
</array>|code-block|syntax|javascript|3348712|在应用程序A的应用程序委托中-在此处获取回调|3348713|-+(BOOL)application:(UIApplication+*)application+openURL:(NSURL+*)url
++sourceApplication:(NSString+*)sourceApplication+annotation:(id)annotation
{
//+You+we+get+the+call+back+here+when+App+B+will+try+to+Open+
//+sourceApplication+will+have+the+bundle+ID+of+the+App+B
//+[url+query]+will+provide+you+the+whole+URL+
//+[url+query]+with+the+help+of+this+you+can+also+pass+the+value+from+App+B+and+get+that+value+here+
}|3348714|现在来看应用程序B代码-|3348715|如果您只想打开不带任何输入参数的App+A|3348716|-(IBAction)openApp_A:(id)sender{

++++if(![[UIApplication+sharedApplication]+openURL:[NSURL+URLWithString:@"testApp.linking://?"]]){
+++++++++UIAlertView+*alert+=+[[UIAlertView+alloc]initWithTitle:@"App+is+not+available!"+message:nil+delegate:self+cancelButtonTitle:@"Ok"+otherButtonTitles:nil,+nil];
++++++++++++[alert+show];

++++++++}
}|3348717|如果您想将参数从应用程序B传递到应用程序A，则使用以下代码|3348718|-(IBAction)openApp_A:(id)sender{
++++if(![[UIApplication+sharedApplication]+openURL:[NSURL+URLWithString:@"testApp.linking://?userName=abe&registered=1&Password=123abc"]]){
+++++++++UIAlertView+*alert+=+[[UIAlertView+alloc]initWithTitle:@"App+is+not+available!"+message:nil+delegate:self+cancelButtonTitle:@"Ok"+otherButtonTitles:nil,+nil];
++++++++++++[alert+show];

++++++++}
}|3348719|注意:你也可以在safari浏览器上输入testApp.linking://?打开应用程序|3348720|entityMap^0|0|0|I|J|3|0|0|5|G|5|M|4|0|0|5|0|1|I|0|0|1|5|0|0|4|7|0|0|0|8|5|G|5|0|0|K|J|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1G|8|@$D|1H|E|1I|F|G]|$D|1J|E|1K|F|G]|$D|1L|E|1M|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|1N|8|@$D|1O|E|1P|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1Q|8|@$D|1R|E|1S|F|G]]|9|@]|A|$]]|$1|N|3|O|5|P|7|1T|8|@]|9|@]|A|$Q|R]]|$1|S|3|T|5|6|7|1U|8|@$D|1V|E|1W|F|G]]|9|@]|A|$]]|$1|U|3|V|5|P|7|1X|8|@]|9|@]|A|$Q|R]]|$1|W|3|X|5|6|7|1Y|8|@$D|1Z|E|20|F|G]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|21|8|@]|9|@]|A|$]]|$1|10|3|11|5|P|7|22|8|@]|9|@]|A|$Q|R]]|$1|12|3|13|5|6|7|23|8|@$D|24|E|25|F|G]|$D|26|E|27|F|G]]|9|@]|A|$]]|$1|14|3|15|5|P|7|28|8|@]|9|@]|A|$Q|R]]|$1|16|3|17|5|6|7|29|8|@$D|2A|E|2B|F|G]]|9|@]|A|$]]|$1|18|3|-4|5|6|7|2C|8|@]|9|@]|A|$]]]|19|$]]

To achieve this we need to add few line of Code in both App

App A: Which you want to open from another App. (Source)

App B: From App B you want to open App A (Destination)

Code for App A

Add few tags into the Plist of App A
Open Plist Source of App A and Past below XML

<pre><code>&lt;key&gt;CFBundleURLTypes&lt;/key&gt;
&lt;array&gt;
 &lt;dict&gt;
 &lt;key&gt;CFBundleURLName&lt;/key&gt;
 &lt;string&gt;com.TestApp&lt;/string&gt;
 &lt;key&gt;CFBundleURLSchemes&lt;/key&gt;
 &lt;array&gt;
 &lt;string&gt;testApp.linking&lt;/string&gt;
 &lt;/array&gt;
 &lt;/dict&gt;
&lt;/array&gt;
</code></pre>

In App delegate of App A - Get Callback here

<pre><code>- (BOOL)application:(UIApplication *)application openURL:(NSURL *)url
 sourceApplication:(NSString *)sourceApplication annotation:(id)annotation
{
// You we get the call back here when App B will try to Open 
// sourceApplication will have the bundle ID of the App B
// [url query] will provide you the whole URL 
// [url query] with the help of this you can also pass the value from App B and get that value here 
}
</code></pre>

Now coming to App B code - 

If you just want to open App A without any input parameter 

<pre><code>-(IBAction)openApp_A:(id)sender{

 if(![[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"testApp.linking://?"]]){
 UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"App is not available!" message:nil delegate:self cancelButtonTitle:@"Ok" otherButtonTitles:nil, nil];
 [alert show];

 }
}
</code></pre>

If you want to pass parameter from App B to App A then use below Code

<pre><code>-(IBAction)openApp_A:(id)sender{
 if(![[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"testApp.linking://?userName=abe&amp;registered=1&amp;Password=123abc"]]){
 UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"App is not available!" message:nil delegate:self cancelButtonTitle:@"Ok" otherButtonTitles:nil, nil];
 [alert show];

 }
}
</code></pre>

Note: You can also open App with just type testApp.linking://? on safari browser

blocks|key|351556|text|在Swift+4.1和Xcode+9.4.1中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|351557|我有两个应用程序1)PageViewControllerExample和2)DelegateExample。现在我想用PageViewControllerExample应用程序打开DelegateExample应用程序。当我点击PageViewControllerExample中的打开按钮时，DelegateExample应用程序将被打开。|351558|为此，我们需要对这两个应用程序的.plist文件进行一些更改。|351559|步骤1|offset|length|style|BOLD|351560|在DelegateExample应用程序中，打开.plist文件并添加网址类型和网址方案。在这里，我们需要添加所需的名称，如"myapp“。|351561|​|351562|📷|atomic|351563|351564|步骤2|351565|在PageViewControllerExample应用程序中，打开.plist文件并添加以下代码|351566|<key>LSApplicationQueriesSchemes</key>
<array>
++++<string>myapp</string>
</array>|code-block|syntax|javascript|351567|现在我们可以通过点击PageViewControllerExample中的按钮来打开DelegateExample应用。|351568|//In+PageViewControllerExample+create+IBAction
@IBAction+func+openapp(_+sender:+UIButton)+{

++++let+customURL+=+URL(string:+"myapp://")
++++if+UIApplication.shared.canOpenURL(customURL!)+{

++++++++//let+systemVersion+=+UIDevice.current.systemVersion//Get+OS+version
++++++++//if+Double(systemVersion)!+>=+10.0+{//10+or+above+versions
++++++++++++//print(systemVersion)
++++++++++++//UIApplication.shared.open(customURL!,+options:+[:],+completionHandler:+nil)
++++++++//}+else+{
++++++++++++//UIApplication.shared.openURL(customURL!)
++++++++//}

++++++++//OR

++++++++if+#available(iOS+10.0,+*)+{
++++++++++++UIApplication.shared.open(customURL!,+options:+[:],+completionHandler:+nil)
++++++++}+else+{
++++++++++++UIApplication.shared.openURL(customURL!)
++++++++}
++++}+else+{
+++++++++//Print+alert+here
++++}
}|351569|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/3f831c6d02e883eb7b5c3cfc6c29aee2.png|imageAlt^0|0|0|0|0|3|0|0|0|0|1|0|0|0|0|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1G|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1H|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1I|8|@$H|1J|I|1K|J|K]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1L|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1M|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1N|8|@]|9|@$H|1O|I|1P|1|1Q]]|A|$]]|$1|S|3|O|5|6|7|1R|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1S|8|@$H|1T|I|1U|J|K]]|9|@]|A|$]]|$1|V|3|W|5|6|7|1V|8|@]|9|@]|A|$]]|$1|X|3|Y|5|Z|7|1W|8|@]|9|@]|A|$10|11]]|$1|12|3|13|5|6|7|1X|8|@]|9|@]|A|$]]|$1|14|3|15|5|Z|7|1Y|8|@]|9|@]|A|$10|11]]|$1|16|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|17|$18|$5|19|1A|1B|A|$1C|1D|1E|-4]]]]

In Swift 4.1 and Xcode 9.4.1

I have two apps 1)PageViewControllerExample and 2)DelegateExample. Now i want to open DelegateExample app with PageViewControllerExample app. When i click open button in PageViewControllerExample, DelegateExample app will be opened.

For this we need to make some changes in .plist files for both the apps.

Step 1

In DelegateExample app open .plist file and add URL Types and URL Schemes. Here we need to add our required name like "myapp".

<a href="https://i.stack.imgur.com/gSwM7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gSwM7.png" alt="enter image description here"></a>

Step 2

In PageViewControllerExample app open .plist file and add this code

<pre><code>&lt;key&gt;LSApplicationQueriesSchemes&lt;/key&gt;
&lt;array&gt;
 &lt;string&gt;myapp&lt;/string&gt;
&lt;/array&gt;
</code></pre>

Now we can open DelegateExample app when we click button in PageViewControllerExample.

<pre><code>//In PageViewControllerExample create IBAction
@IBAction func openapp(_ sender: UIButton) {

 let customURL = URL(string: "myapp://")
 if UIApplication.shared.canOpenURL(customURL!) {

 //let systemVersion = UIDevice.current.systemVersion//Get OS version
 //if Double(systemVersion)! &gt;= 10.0 {//10 or above versions
 //print(systemVersion)
 //UIApplication.shared.open(customURL!, options: [:], completionHandler: nil)
 //} else {
 //UIApplication.shared.openURL(customURL!)
 //}

 //OR

 if #available(iOS 10.0, *) {
 UIApplication.shared.open(customURL!, options: [:], completionHandler: nil)
 } else {
 UIApplication.shared.openURL(customURL!)
 }
 } else {
 //Print alert here
 }
}
</code></pre>

blocks|key|351576|text|关于这个话题的旁注...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|351577|未注册的协议有50个请求限制。|351578|苹果公司(+In+this+discussion+)提到，对于一个特定版本的应用程序，你只能查询canOpenUrl的有限次数，而且会在调用50次未声明的方案后失败。我还看到，一旦进入失败状态，如果添加了协议，它仍然会失败。|offset|length|style|CODE|BOLD|351579|要意识到这一点，这可能对某些人有用。|351580|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/documentation/uikit/uiapplication/1622952-canopenurl#discussion^0|0|0|1C|A|1U|D|6|I|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@$F|X|G|Y|H|I]|$F|Z|G|10|H|J]]|9|@$F|11|G|12|1|13]]|A|$]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

A side note regarding this topic...

There is a 50 request limit for protocols that are not registered.

<a href="https://developer.apple.com/documentation/uikit/uiapplication/1622952-canopenurl#discussion" rel="nofollow noreferrer">In this discussion</a> apple mention that for a specific version of an app you can only query the <code>canOpenUrl</code> a limited number of times and will fail after 50 calls for undeclared schemes. I've also seen that if the protocol is added once you have entered this failing state it will still fail.

Be aware of this, could be useful to someone.

Is it possible to launch any arbitrary iPhone application from within another app?, For example in my application if I want the user to push a button and launch right into the Phone app (close the current app, open the Phone app).

would this be possible? I know this can be done for making phone calls with the tel URL link, but I want to instead just have the Phone app launch without dialing any specific number.

Launch an app from within another (iPhone)

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如果我想让用户按一个按钮直接进入手机应用程序(关闭当前应用程序，打开手机应用程序)，是否可以从另一个应用程序中启动任意的iPhone应用程序?例如，我的应用程序中的。这有可能吗？我知道可以通过tel URL链接拨打电话，但我希望在不拨打任何特定号码的情况下启动手机应用程序。

问从另一个应用程序中启动应用程序(iPhone)
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