WhatsApp NSURL 在共享文本时返回 nil - Swift

Question

我正在尝试与 WhatsApp 共享文本，但 NSURL 始终返回 nil，但文本编码正确！看看我的代码：

var msg : NSString = "to the world of none";
var titlewithoutspace = msg.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
var urlWhats = NSString(string: "whatsapp://send?text=\(titlewithoutspace)")
println(urlWhats)
var whatsappURL = NSURL(string: urlWhats as String)
println(whatsappURL)

当我打印结果时，字符串等于：

whatsapp://send?text=Optional("to%20the%20world%20of%20none")

whatsappURL 总是返回 nil：

nil

Answer 1

使用 \(...) 应保留用于写入控制台。如您所见，它在您的字符串周围添加了 Optional(....) 并且在 URL.

中是无效的

强制解包 "whatsapp://send?text=\(titlewithoutspace!)" 或更好，使用

String(format: "whatsapp://send?text=%@", titlewithoutspace)

Answer 2

stringByAddingPercentEscapesUsingEncoding: returns 一个可选的 String 这就是 urlWhats 包含 Optional("") 的原因。为避免这种情况，您只需要像这样打开可选的：

var msg: NSString = "to the world of none";
var titlewithoutspace = msg.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
if let titlewithoutspace = titlewithoutspace {
    var urlWhats = NSString(string: "whatsapp://send?text=\(titlewithoutspace)")
    var whatsappURL = NSURL(string: urlWhats as String)
    println(whatsappURL)
} else {
    // Unwrapping failed because titlewithoutspace is nil (might be because stringByAddingPercentEscapesUsingEncoding failed).
}

此外，我建议您直接使用 String 类型，因为 NSString 对您来说毫无用处（stringByAdding… 除外）：

var msg: NSString = "to the world of none";
var titlewithoutspace = msg.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
if let titlewithoutspace = titlewithoutspace {
    var urlWhats = "whatsapp://send?text=\(titlewithoutspace)"
    var whatsappURL = NSURL(string: urlWhats)
    println(whatsappURL)
} else {
    // Unwrapping failed because titlewithoutspace is nil (might be because stringByAddingPercentEscapesUsingEncoding failed).
}

另请注意，NSURL(string:) 可能会失败，因此它也是 returns 一个可选的 NSURL 对象。要使用它，您可能需要像我对 titlewithoutspace.

所做的那样打开它

Answer 3

对于Swift 2.0/2.1

func openURL(urlString: String, message: String) {
    var formattedURL: String = urlString

    if let strippedMessage = message.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet()) {
        formattedURL = "\(urlString)\(strippedMessage)"
    }

    if let url = NSURL(string: formattedURL) as NSURL? {
        print(url)
        UIApplication.sharedApplication().openURL(url)
    }
}

Answer 4

这是swift

中经过测试的分享代码

let url  = NSURL(string: "whatsapp://send?text=https%3A%2F%2Fitunes.apple.com%2Fin%2Fapp%2Fwhatsapp-messenger%2Fid310633997%3Fmt%3D8")

if UIApplication.sharedApplication().canOpenURL(url!) {
      UIApplication.sharedApplication().openURL(url!)
 }
 else {
      let errorAlert = UIAlertView(title: "Cannot Send Message", message: "Your device is not able to send WhatsApp messages.", delegate: self, cancelButtonTitle: "OK")
      errorAlert.show()
      }

注意：您只需将 url 转换为编码格式

WhatsApp NSURL 在共享文本时返回 nil - Swift

WhatsApp NSURL returning nil When sharing text - Swift

ios

whatsapp

swift