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Ios 电话号码验证中的Swift 3转换错误_Ios_Swift3_Xcode8.1 - Fatal编程技术网

Ios 电话号码验证中的Swift 3转换错误

Ios 电话号码验证中的Swift 3转换错误,ios,swift3,xcode8.1,Ios,Swift3,Xcode8.1,下面是我在扩展中使用的Swift2中的代码,但在迁移到swift 3时,它显示了以下错误。你知道怎么解决吗 var isPhoneNumber: Bool { let charcter = NSCharacterSet(charactersIn: "+0123456789").inverted var filtered:NSString! let inputString:NSArray = self.componentsSeparatedByCharactersInS

下面是我在扩展中使用的Swift2中的代码,但在迁移到swift 3时,它显示了以下错误。你知道怎么解决吗

var isPhoneNumber: Bool {

    let charcter  = NSCharacterSet(charactersIn: "+0123456789").inverted
    var filtered:NSString!
    let inputString:NSArray = self.componentsSeparatedByCharactersInSet(charcter)
    filtered = inputString.componentsJoined(by: "") as NSString!
    return  self == filtered

}
显示的错误有:

  • 类型“String”的值没有成员“componentsSeparatedByCharactersInSet”

  • 二进制运算符“==”不能应用于“String”和“NSString”类型的操作数


  • 我想我也有同感:

    let charcter  = CharacterSet(charactersIn: "+0123456789").inverted
    let inputString = self.components(separatedBy: charcter).joined(separator: "")
    return  self == inputString
    

    1:2:self的类型是什么?2:self和filtered应该具有相同的类型'String'或'NSString!'