blocks|key|1010331|text|要返回原始应用程序，您可以使用telprompt://而不是tel://+-+tell提示符将首先提示用户，但当调用完成后，它将返回到您的应用程序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1010332|NSString+*phoneNumber+=+[@"telprompt://"+stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication+sharedApplication]+openURL:[NSURL+URLWithString:phoneNumber]];|code-block|syntax|javascript|1010333|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|$]]

To go back to original app you can use telprompt:// instead of tel:// - The tell prompt will prompt the user first, but when the call is finished it will go back to your app:

<pre><code>NSString *phoneNumber = [@"telprompt://" stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
</code></pre>

blocks|key|1010355|text|如果您正在使用Xamarin开发iOS应用程序，下面是在您的应用程序中进行电话呼叫的C#等价物：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1010356|string+phoneNumber+=+"1231231234";
NSUrl+url+=+new+NSUrl(string.Format(@"telprompt://{0}",+phoneNumber));
UIApplication.SharedApplication.OpenUrl(url);|code-block|syntax|javascript|1010357|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|$]]

If you are using Xamarin to develop an iOS application, here is the C# equivalent to make a phone call within your application:

<pre><code>string phoneNumber = "1231231234";
NSUrl url = new NSUrl(string.Format(@"telprompt://{0}", phoneNumber));
UIApplication.SharedApplication.OpenUrl(url);
</code></pre>

blocks|key|1010390|text|Java+RoboVM的等价物：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1010391|public+void+dial(String+number)
{
++NSURL+url+=+new+NSURL("tel://"+%2B+number);
++UIApplication.getSharedApplication().openURL(url);
}|code-block|syntax|javascript|1010392|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|$]]

The Java RoboVM equivalent:

<pre><code>public void dial(String number)
{
 NSURL url = new NSURL("tel://" + number);
 UIApplication.getSharedApplication().openURL(url);
}
</code></pre>

blocks|key|1010437|text|这里的答案是完美的。我只是将Craig+Mellon+answer转换为Swift。如果有人来寻求快速的答案，这将帮助他们。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1010438|+var+phoneNumber:+String+=+"telprompt://".stringByAppendingString(titleLabel.text!)+//+titleLabel.text+has+the+phone+number.
++++++++UIApplication.sharedApplication().openURL(NSURL(string:phoneNumber)!)|code-block|syntax|javascript|1010439|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|$]]

The answers here are perfectly working. I am just converting Craig Mellon answer to Swift. If someone comes looking for swift answer, this will help them.

<pre><code> var phoneNumber: String = "telprompt://".stringByAppendingString(titleLabel.text!) // titleLabel.text has the phone number.
 UIApplication.sharedApplication().openURL(NSURL(string:phoneNumber)!)
</code></pre>

blocks|key|1010853|text|合并@Cristian+Radu和@Craig+Mellon的答案，以及@joel.d的评论，你应该这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1010854|NSURL+*urlOption1+=+[NSURL+URLWithString:[@"telprompt://"+stringByAppendingString:phone]];
NSURL+*urlOption2+=+[NSURL+URLWithString:[@"tel://"+stringByAppendingString:phone]];
NSURL+*targetURL+=+nil;

if+([UIApplication.sharedApplication+canOpenURL:urlOption1])+{
++++targetURL+=+urlOption1;
}+else+if+([UIApplication.sharedApplication+canOpenURL:urlOption2])+{
++++targetURL+=+urlOption2;
}

if+(targetURL)+{
++++if+(@available(iOS+10.0,+*))+{
++++++++[UIApplication.sharedApplication+openURL:targetURL+options:@{}+completionHandler:nil];
++++}+else+{
#pragma+clang+diagnostic+push
#pragma+clang+diagnostic+ignored+"-Wdeprecated-declarations"
++++++++[UIApplication.sharedApplication+openURL:targetURL];
#pragma+clang+diagnostic+pop
++++}
}+|code-block|syntax|javascript|1010855|这将首先尝试使用"telprompt://“URL，如果失败，将使用"tel://”URL。如果两者都失败了，您将尝试在iPad或iPod+Touch上拨打电话。|1010856|Swift版本：|1010857|let+phone+=+mymobileNO.titleLabel.text
let+phoneUrl+=+URL(string:+"telprompt://\(phone)"
let+phoneFallbackUrl+=+URL(string:+"tel://\(phone)"
if(phoneUrl+!=+nil+&&+UIApplication.shared.canOpenUrl(phoneUrl!))+{
++UIApplication.shared.open(phoneUrl!,+options:[String:Any]())+{+(success)+in
++++if(!success)+{
++++++//+Show+an+error+message:+Failed+opening+the+url
++++}
++}
}+else+if(phoneFallbackUrl+!=+nil+&&+UIApplication.shared.canOpenUrl(phoneFallbackUrl!))+{
++UIApplication.shared.open(phoneFallbackUrl!,+options:[String:Any]())+{+(success)+in
++++if(!success)+{
++++++//+Show+an+error+message:+Failed+opening+the+url
++++}
++}
}+else+{
++++//+Show+an+error+message:+Your+device+can+not+do+phone+calls.
}|1010858|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Merging the answers of @Cristian Radu and @Craig Mellon, and the comment from @joel.d, you should do:

<pre><code>NSURL *urlOption1 = [NSURL URLWithString:[@"telprompt://" stringByAppendingString:phone]];
NSURL *urlOption2 = [NSURL URLWithString:[@"tel://" stringByAppendingString:phone]];
NSURL *targetURL = nil;

if ([UIApplication.sharedApplication canOpenURL:urlOption1]) {
 targetURL = urlOption1;
} else if ([UIApplication.sharedApplication canOpenURL:urlOption2]) {
 targetURL = urlOption2;
}

if (targetURL) {
 if (@available(iOS 10.0, *)) {
 [UIApplication.sharedApplication openURL:targetURL options:@{} completionHandler:nil];
 } else {
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wdeprecated-declarations"
 [UIApplication.sharedApplication openURL:targetURL];
#pragma clang diagnostic pop
 }
} 
</code></pre>

This will first try to use the "telprompt://" URL, and if that fails, it will use the "tel://" URL. If both fails, you're trying to place a phone call on an iPad or iPod Touch.

Swift Version :

<pre><code>let phone = mymobileNO.titleLabel.text
let phoneUrl = URL(string: "telprompt://\(phone)"
let phoneFallbackUrl = URL(string: "tel://\(phone)"
if(phoneUrl != nil &amp;&amp; UIApplication.shared.canOpenUrl(phoneUrl!)) {
 UIApplication.shared.open(phoneUrl!, options:[String:Any]()) { (success) in
 if(!success) {
 // Show an error message: Failed opening the url
 }
 }
} else if(phoneFallbackUrl != nil &amp;&amp; UIApplication.shared.canOpenUrl(phoneFallbackUrl!)) {
 UIApplication.shared.open(phoneFallbackUrl!, options:[String:Any]()) { (success) in
 if(!success) {
 // Show an error message: Failed opening the url
 }
 }
} else {
 // Show an error message: Your device can not do phone calls.
}
</code></pre>

blocks|key|796388|text|let+phone+=+"tel://\("1234567890")";
let+url:NSURL+=+NSURL(string:phone)!;
UIApplication.sharedApplication().openURL(url);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|796389|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>let phone = "tel://\("1234567890")";
let url:NSURL = NSURL(string:phone)!;
UIApplication.sharedApplication().openURL(url);
</code></pre>

blocks|key|1010930|text|在Swift+3.0中，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1010931|static+func+callToNumber(number:String)+{

++++++++let+phoneFallback+=+"telprompt://\(number)"
++++++++let+fallbackURl+=+URL(string:phoneFallback)!

++++++++let+phone+=+"tel://\(number)"
++++++++let+url+=+URL(string:phone)!

++++++++let+shared+=+UIApplication.shared

++++++++if(shared.canOpenURL(fallbackURl)){
++++++++++++shared.openURL(fallbackURl)
++++++++}else+if+(shared.canOpenURL(url)){
++++++++++++shared.openURL(url)
++++++++}else{
++++++++++++print("unable+to+open+url+for+call")
++++++++}

++++}|code-block|syntax|javascript|1010932|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|$]]

In Swift 3.0,

<pre><code>static func callToNumber(number:String) {

 let phoneFallback = "telprompt://\(number)"
 let fallbackURl = URL(string:phoneFallback)!

 let phone = "tel://\(number)"
 let url = URL(string:phone)!

 let shared = UIApplication.shared

 if(shared.canOpenURL(fallbackURl)){
 shared.openURL(fallbackURl)
 }else if (shared.canOpenURL(url)){
 shared.openURL(url)
 }else{
 print("unable to open url for call")
 }

 }
</code></pre>

blocks|key|1010992|text|“openURL：”已被弃用:在iOS+10.0中首次被弃用-请改用openURL:options:completionHandler：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1010993|在Objective-c+iOS+10%2B中使用：|1010994|NSString+*phoneNumber+=+[@"tel://"+stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication+sharedApplication]+openURL:[NSURL+URLWithString:phoneNumber]+options:@{}+completionHandler:nil];|code-block|syntax|javascript|1010995|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|$]]

'openURL:' is deprecated: first deprecated in iOS 10.0 - Please use openURL:options:completionHandler: instead

in Objective-c iOS 10+ use :

<pre><code>NSString *phoneNumber = [@"tel://" stringByAppendingString:mymobileNO.titleLabel.text];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber] options:@{} completionHandler:nil];
</code></pre>

blocks|key|1011038|text|斯威夫特|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1011039|if+let+url+=+NSURL(string:+"tel://\(number)"),+
++++UIApplication.sharedApplication().canOpenURL(url)+{
++++++++UIApplication.shared.open(url,+options:+[:],+completionHandler:+nil)
}|code-block|syntax|javascript|1011040|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|$]]

<h1>Swift</h1>
<pre><code>if let url = NSURL(string: &quot;tel://\(number)&quot;), 
 UIApplication.sharedApplication().canOpenURL(url) {
 UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
</code></pre>

blocks|key|1011084|text|使用openurl。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1011085|要在swift+5.1中进行调用，只需使用以下代码：(我已经在Xcode+11中测试过它)|1011086|let+phone+=+"1234567890"
if+let+callUrl+=+URL(string:+"tel://\(phone)"),+UIApplication.shared.canOpenURL(callUrl)+{
+++++UIApplication.shared.open(callUrl)
}|code-block|syntax|javascript|1011087|编辑:对于Xcode+12.4，swift+5.3，只需使用以下内容：|1011088|UIApplication.shared.open(NSURL(string:+"tel://555-123-1234")!+as+URL)|offset|length|style|CODE|1011089|确保您导入的是UIKit，否则它会提示您在作用域中找不到UIApplication。|1011090|entityMap^0|0|0|0|0|0|1Y|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Use openurl.
For making a call in swift 5.1, just use the following code: (I have tested it in Xcode 11)
<pre><code>let phone = &quot;1234567890&quot;
if let callUrl = URL(string: &quot;tel://\(phone)&quot;), UIApplication.shared.canOpenURL(callUrl) {
 UIApplication.shared.open(callUrl)
}
</code></pre>
Edit: For Xcode 12.4, swift 5.3, just use the following:
<code>UIApplication.shared.open(NSURL(string: &quot;tel://555-123-1234&quot;)! as URL)</code>
<h2>Make sure that you import UIKit, or it will say that it cannot find UIApplication in scope.</h2>

How can I make a phone call programmatically on iPhone? I tried the following code but nothing happened:

<pre><code>NSString *phoneNumber = mymobileNO.titleLabel.text;
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
</code></pre>

Make a phone call programmatically

Xamarin

Swift

Xcode

Java

如何在iPhone上以编程方式拨打电话？我尝试了以下代码，但什么也没有发生：NSString *phoneNumber = mymobileNO.titleLabel.text;[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNu...

问以编程方式进行电话呼叫
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