ios - 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

stringByAddingPercentEscapesUsingEncoding: 返回一个可选的 String，这就是 urlWhats 包含 Optional("") 的原因。为了避免这种情况，你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(除了 stringByAdding...):

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，因此它还会返回一个可选的 NSURL 对象。要使用它，您可能需要解包它，就像我对 titlewithoutspace 所做的那样。