Ios 在swift中拨打电话号码
我试着拨打一个号码,不是使用特定的号码,而是一个在变量中呼叫的号码,或者至少告诉它调出你手机中的号码。在变量中调用的这个数字是我通过使用解析器或从网站sql获取的数字。我做了一个按钮,试图调用存储在变量中的电话号码,但没有用。什么都行,谢谢Ios 在swift中拨打电话号码,ios,swift,phone-number,phone-call,Ios,Swift,Phone Number,Phone Call,我试着拨打一个号码,不是使用特定的号码,而是一个在变量中呼叫的号码,或者至少告诉它调出你手机中的号码。在变量中调用的这个数字是我通过使用解析器或从网站sql获取的数字。我做了一个按钮,试图调用存储在变量中的电话号码,但没有用。什么都行,谢谢 func callSellerPressed (sender: UIButton!){ //(This is calls a specific number)UIApplication.sharedApplication().openURL(NSU
func callSellerPressed (sender: UIButton!){
//(This is calls a specific number)UIApplication.sharedApplication().openURL(NSURL(string: "tel://######")!)
// This is the code I'm using but its not working
UIApplication.sharedApplication().openURL(NSURL(scheme: NSString(), host: "tel://", path: busPhone)!)
}
试试看:
if let url = NSURL(string: "tel://\(busPhone)") where UIApplication.sharedApplication().canOpenURL(url) {
UIApplication.sharedApplication().openURL(url)
}
假设电话号码在总线电话中
NSURL
的init(字符串:)
返回一个可选值,因此通过使用if let
我们确保url
是NSURL
(而不是init
返回的NSURL?
)
对于Swift 3:
if let url = URL(string: "tel://\(busPhone)"), UIApplication.shared.canOpenURL(url) {
if #available(iOS 10, *) {
UIApplication.shared.open(url)
} else {
UIApplication.shared.openURL(url)
}
}
我们需要检查是否使用iOS 10或更高版本,因为:
iOS 10.0中不推荐使用“openURL”
好吧,我得到了帮助,并且找到了答案。我还设置了一个很好的小警报系统,以防电话号码无效。我的问题是我说得对,但是这个号码有空格和不需要的字符,比如(“123456-7890”)。UIApplication仅在您的号码为(“1234567890”)时有效或接受。因此,通过创建一个新变量来仅提取数字,基本上可以删除空格和无效字符。然后使用UIC应用程序调用这些号码
func callSellerPressed (sender: UIButton!){
var newPhone = ""
for (var i = 0; i < countElements(busPhone); i++){
var current:Int = i
switch (busPhone[i]){
case "0","1","2","3","4","5","6","7","8","9" : newPhone = newPhone + String(busPhone[i])
default : println("Removed invalid character.")
}
}
if (busPhone.utf16Count > 1){
UIApplication.sharedApplication().openURL(NSURL(string: "tel://" + newPhone)!)
}
else{
let alert = UIAlertView()
alert.title = "Sorry!"
alert.message = "Phone number is not available for this business"
alert.addButtonWithTitle("Ok")
alert.show()
}
}
func callSellerPressed(发件人:ui按钮!){
var newPhone=“”
对于(变量i=0;i1){
UIApplication.sharedApplication().openURL(NSURL(字符串:“tel:/”+newPhone)!)
}
否则{
let alert=UIAlertView()
alert.title=“对不起!”
alert.message=“电话号码不适用于此业务”
alert.addButtonWithTitle(“确定”)
alert.show()
}
}
iOS 10中的独立解决方案,Swift 3:
private func callNumber(phoneNumber:String) {
if let phoneCallURL = URL(string: "tel://\(phoneNumber)") {
let application:UIApplication = UIApplication.shared
if (application.canOpenURL(phoneCallURL)) {
application.open(phoneCallURL, options: [:], completionHandler: nil)
}
}
}
您应该能够使用
呼叫号码(“7178881234”)
拨打电话。这是使用Swift 2.0更新@Tom的答案
注意-这是我正在使用的整个CallComposer类
class CallComposer: NSObject {
var editedPhoneNumber = ""
func call(phoneNumber: String) -> Bool {
if phoneNumber != "" {
for i in number.characters {
switch (i){
case "0","1","2","3","4","5","6","7","8","9" : editedPhoneNumber = editedPhoneNumber + String(i)
default : print("Removed invalid character.")
}
}
let phone = "tel://" + editedPhoneNumber
let url = NSURL(string: phone)
if let url = url {
UIApplication.sharedApplication().openURL(url)
} else {
print("There was an error")
}
} else {
return false
}
return true
}
}
我在我的应用程序中使用了这个方法,它工作得很好。我希望这对你也有帮助
func makeCall(phone: String) {
let formatedNumber = phone.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet).joinWithSeparator("")
let phoneUrl = "tel://\(formatedNumber)"
let url:NSURL = NSURL(string: phoneUrl)!
UIApplication.sharedApplication().openURL(url)
}
Swift 3.0解决方案:
let formatedNumber = phone.components(separatedBy: NSCharacterSet.decimalDigits.inverted).joined(separator: "")
print("calling \(formatedNumber)")
let phoneUrl = "tel://\(formatedNumber)"
let url:URL = URL(string: phoneUrl)!
UIApplication.shared.openURL(url)
在Swift 3中
if let url = URL(string:"tel://\(phoneNumber)"), UIApplication.shared.canOpenURL(url) {
UIApplication.shared.openURL(url)
}
上述答案部分正确,但“tel://”只有一个问题。通话结束后,它将返回主屏幕,而不是我们的应用程序。因此最好使用“telprompt://”,它将返回到应用程序
var url:NSURL = NSURL(string: "telprompt://1234567891")!
UIApplication.sharedApplication().openURL(url)
openURL()在iOS 10中已被弃用。以下是新语法:
if let url = URL(string: "tel://\(busPhone)") {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
Swift 3.0和ios 10或更高版本
func phone(phoneNum: String) {
if let url = URL(string: "tel://\(phoneNum)") {
if #available(iOS 10, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url as URL)
}
}
}
这里有一种替代方法,可以使用
扫描仪将电话号码缩减为有效组件
let number = "+123 456-7890"
let scanner = Scanner(string: number)
let validCharacters = CharacterSet.decimalDigits
let startCharacters = validCharacters.union(CharacterSet(charactersIn: "+#"))
var digits: NSString?
var validNumber = ""
while !scanner.isAtEnd {
if scanner.scanLocation == 0 {
scanner.scanCharacters(from: startCharacters, into: &digits)
} else {
scanner.scanCharacters(from: validCharacters, into: &digits)
}
scanner.scanUpToCharacters(from: validCharacters, into: nil)
if let digits = digits as? String {
validNumber.append(digits)
}
}
print(validNumber)
// +1234567890
我正在使用带有数字验证的swift 3解决方案
var validPhoneNumber = ""
phoneNumber.characters.forEach {(character) in
switch character {
case "0"..."9":
validPhoneNumber.characters.append(character)
default:
break
}
}
if UIApplication.shared.canOpenURL(URL(string: "tel://\(validNumber)")!){
UIApplication.shared.openURL(URL(string: "tel://\(validNumber)")!)
}
Swift 3.0和iOS 10+
UIApplication.shared.openURL(url)
改为
UIApplication.shared.open(url:url,选项:[:],completionHandler completion:nil)
选项和完成处理程序是可选的,呈现:
UIApplication.shared.open(url)
Swift 3,iOS 10
func call(phoneNumber:String) {
let cleanPhoneNumber = phoneNumber.components(separatedBy: CharacterSet.decimalDigits.inverted).joined(separator: "")
let urlString:String = "tel://\(cleanPhoneNumber)"
if let phoneCallURL = URL(string: urlString) {
if (UIApplication.shared.canOpenURL(phoneCallURL)) {
UIApplication.shared.open(phoneCallURL, options: [:], completionHandler: nil)
}
}
}
对于Swift 3.1和向后兼容的方法,请执行以下操作:
@IBAction func phoneNumberButtonTouched(_ sender: Any) {
if let number = place?.phoneNumber {
makeCall(phoneNumber: number)
}
}
func makeCall(phoneNumber: String) {
let formattedNumber = phoneNumber.components(separatedBy:
NSCharacterSet.decimalDigits.inverted).joined(separator: "")
let phoneUrl = "tel://\(formattedNumber)"
let url:NSURL = NSURL(string: phoneUrl)!
if #available(iOS 10, *) {
UIApplication.shared.open(url as URL, options: [:], completionHandler:
nil)
} else {
UIApplication.shared.openURL(url as URL)
}
}
适用于swift 3.0
if let url = URL(string: "tel://\(number)"), UIApplication.shared.canOpenURL(url) {
if #available(iOS 10, *) {
UIApplication.shared.open(url)
} else {
UIApplication.shared.openURL(url)
}
}
else {
print("Your device doesn't support this feature.")
}
如果您的电话号码包含空格,请先将其删除!然后你可以使用这个解决方案
let numbersOnly = busPhone.replacingOccurrences(of: " ", with: "")
if let url = URL(string: "tel://\(numbersOnly)"), UIApplication.shared.canOpenURL(url) {
if #available(iOS 10, *) {
UIApplication.shared.open(url)
} else {
UIApplication.shared.openURL(url)
}
}
Swift 4
private func callNumber(phoneNumber:String) {
if let phoneCallURL = URL(string: "telprompt://\(phoneNumber)") {
let application:UIApplication = UIApplication.shared
if (application.canOpenURL(phoneCallURL)) {
if #available(iOS 10.0, *) {
application.open(phoneCallURL, options: [:], completionHandler: nil)
} else {
// Fallback on earlier versions
application.openURL(phoneCallURL as URL)
}
}
}
}
适用于Swift 4.2及以上版本
Swift 5:iOS>=10.0
if let url = URL(string: "tel://\(number)"), UIApplication.shared.canOpenURL(url) {
if #available(iOS 10, *) {
UIApplication.shared.open(url)
} else {
UIApplication.shared.openURL(url)
}
}
else {
print("Your device doesn't support this feature.")
}
此解决方案为零保存
仅适用于物理设备
其他许多答案对Swift 5不起作用。以下是Swift 5的代码更新:
let formattedNumber = phoneNumberVariable.components(separatedBy: NSCharacterSet.decimalDigits.inverted).joined(separator: "")
if let url = NSURL(string: ("tel:" + (formattedNumber)!)) {
if #available(iOS 10.0, *) {
UIApplication.shared.open(url as URL, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url as URL)
}
}
附言:
有了大多数答案,我无法在设备上得到提示。上述代码已成功显示提示
与大多数答案一样,电话之后没有/。而且效果很好
Swift 5
我用了这个,它成功了
@IBAction func btnPhoneClick(_ sender: Any) {
guard let url = URL(string: "telprompt://\(String(describing: self.phoneNumberLbl.text))"),
UIApplication.shared.canOpenURL(url) else {
return
}
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
这是给我一个未解析标识符“url”的错误-指向.openURL(url)谢谢你的帮助!我只是少了一封信。好吧,当我按下按钮时,什么也没发生。这里有更多信息------func parserDidEndDocument(parser:NSXMLParser!){println(busPhone)drawUpdatedView()}您正在电话上测试这个吗?我怀疑打开一个tel://URL只能在实际的iPhone上工作,而不能在模拟器上工作,也不能在iPod或iPad上工作。是的,我正在iPhone 4s上测试它,按钮点击,但它不能从我手机上的sql server上调出busPhone呼叫的号码。当我使用带有电话号码的电话时,它会拨。好的,我想我知道问题出在哪里,但不知道如何解决。从URL中提取的数字完全是这样写的123456-7890,而不是这样写的123456780。我该如何翻译tel://以它可以识别的形式读取URL?我认为您还应该允许使用“+”符号。另外,#是发出请求时使用的字符之一:)它不会贬值。语法上的变化是因为您使用的是swift 3。@Hammadzafar,openURl(url)方法确实不受欢迎。它的新版本有不同的签名,也被重命名为open(url:options:completionHandler),这是因为swift 2.3即将发布,编辑将给出如下错误;swift 3的“此应用程序不允许查询方案电话”,这些代码是否在模拟器上工作?该代码在模拟器中不工作。如果你想检查它,使用一个设备。显然这会让你的应用程序也删除“+”和“#”图表。这个问题现在有17个答案,已经超过3年了。你为什么觉得有必要
let formattedNumber = phoneNumberVariable.components(separatedBy: NSCharacterSet.decimalDigits.inverted).joined(separator: "")
if let url = NSURL(string: ("tel:" + (formattedNumber)!)) {
if #available(iOS 10.0, *) {
UIApplication.shared.open(url as URL, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url as URL)
}
}
@IBAction func btnPhoneClick(_ sender: Any) {
guard let url = URL(string: "telprompt://\(String(describing: self.phoneNumberLbl.text))"),
UIApplication.shared.canOpenURL(url) else {
return
}
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}