如何从字符串值Swift中删除Optional
我想使用不带可选扩展名的String值。我使用以下代码从firebase解析此数据:
Database.database().reference(withPath: "Locations").child("Cities").observe(.value, with: { (snapShot) in
if snapShot.exists() {
let array:NSArray = snapShot.children.allObjects as NSArray
for child in array {
let snap = child as! DataSnapshot
let cityName = snap.key
let cityNameString = "\(cityName)"
if snap.value is NSDictionary {
let data:NSDictionary = snap.value as! NSDictionary
let lat = data.value(forKey: "lat")
let lng = data.value(forKey: "lng")
let radius = data.value(forKey: "radius")
let latstring = "\(lat)"
let lngstring = "\(lng)"
let radiusstring = "\(radius)"
let city = CityObject(name: cityNameString , lat: latstring , lng: lngstring, radius: radiusstring)
print("Value is", laststring)
self.selectCity(cityObject: city)
}
}
}
})
解析此数据后,我尝试打印例如latstring并得到以下结果:
值是可选的(52.523553)
我的CityObject如下所示:
class CityObject{var name: String?
var lat: String?
var lng: String?
var radius: String?
init(name: String?, lat: String?, lng: String?, radius: String?){
self.name = name
self.lat = lat
self.lng = lng
self.radius = radius
}
回答:
就像@GioR所说的那样,该值是Optional(52.523553),因为latstring的类型是隐式的:String?。这是由于let lat =
data.value(forKey:“ lat”)将返回字符串?隐式设置lat的类型。
有关值的文档,请参阅https://developer.apple.com/documentation/objectivec/nsobject/1412591-value(forKey
:)
Swift有多种处理nil的方法。这三个可能对您有帮助,nil合并运算符:
??
如果可选结果为nil,则此运算符将提供默认值:
let lat: String = data.value(forKey: "lat") ?? "the lat in the dictionary was nil!"
警卫声明
guard let lat: String = data.value(forKey: "lat") as? String else { //Oops, didn't get a string, leave the function!
}
Guard语句使您可以将可选变量转换为等效的非可选变量,如果恰好为nil,则可以退出该函数
如果让
if let lat: String = data.value(forKey: "lat") as? String { //Do something with the non-optional lat
}
//Carry on with the rest of the function
希望这可以帮助^^
以上是 如何从字符串值Swift中删除Optional 的全部内容, 来源链接: utcz.com/qa/426041.html