值有时会以 SwiftyJSON 的可选字符串形式返回 [英] Value sometimes returns as optional string with SwiftyJSON

问题描述

我想知道为什么 SwiftyJSON 有时返回值作为 "myString" 和其他时候作为 Optional("myString")

例如:

func getFBUserData(){if((FBSDKAccessToken.currentAccessToken()) != nil){FBSDKGraphRequest(graphPath: "me", parameters: ["fields": "id, email"]).startWithCompletionHandler({ (connection, result, error) -> Void in如果(错误 == 零){让 fbJson = JSON(结果)让 fbId = fbJson["id"].string打印(fbId)}})}}

这里 fbId 将返回:Optional("myString")

下一步:

static func loginWithEmail(facebookId: String, email: String, username: String, response: (token: String?) -> ()) {让 urlString = baseURL + ResourcePath.FbLogin.description让参数:[字符串:AnyObject] = ["fb_id": facebookId,1级，电子邮件":电子邮件，用户名":用户名]Alamofire.request(.POST, urlString, parameters: parameters).responseJSON{ (responseData) ->作废//打印(响应数据)让 json = JSON(responseData.result.value!)让令牌 = json["api_token"].string响应(令牌:令牌)}}

这里 token 将返回 "myString"

那么当两个函数都使用 ".string" 而不是 ".string!" 时，一个函数如何返回 optional() ?

解决方案

SwiftyJSON 有两种用于检索值的getter":可选和非可选.

.string 是值的 String 表示的可选 getter，因此您必须在使用前解开它

if let fbId = fbJson["id"].string {打印(fbId)}

否则字符串本身将包含术语可选".

如果您100% 确定总会有一个值，您可以通过使用非可选的 getter 来使用强制解包"的等效项，并且您不需要 如果让不再:

let fbId = fbJson["id"].stringValue

但要小心，因为如果值为 nil，这会崩溃，就像你用 fbJson["id"].string! 强制解包一样(你永远不应该这样做).>

请注意，SwiftyJSON 对其他类型的工作方式相同:.int 与 .intValue、.array 与 .arrayValue 等

I wonder why the SwiftyJSON sometimes returns a value as "myString" and some other time as Optional("myString")

For example:

func getFBUserData(){
        if((FBSDKAccessToken.currentAccessToken()) != nil){
            FBSDKGraphRequest(graphPath: "me", parameters: ["fields": "id, email"]).startWithCompletionHandler({ (connection, result, error) -> Void in
                if (error == nil){

                    let fbJson = JSON(result)
                    let fbId = fbJson["id"].string
                    print(fbId)
                }
            })
        }
    }

Here fbId will return: Optional("myString")

Next:

static func loginWithEmail(facebookId: String, email: String, username: String, response: (token: String?) -> ()) {
        let urlString = baseURL + ResourcePath.FbLogin.description

        let parameters: [String: AnyObject] = [
            "fb_id": facebookId,
            "level": 1,
            "email": email,
            "username": username
        ]
        Alamofire.request(.POST, urlString, parameters: parameters).responseJSON{ (responseData) -> Void in
            //print(responseData)
            let json = JSON(responseData.result.value!)
            let token = json["api_token"].string
            response(token: token)
        }

    }

Here token will return "myString"

So how can one function return optional() when both functions use ".string" and not ".string!"?

解决方案

SwiftyJSON has two kinds of "getters" for retrieving values: Optional and non-Optional.

.string is the optional getter for the String representation of a value, so you have to unwrap it before use

if let fbId = fbJson["id"].string {
    print(fbId)
}

otherwise the String itself will contain the term "Optional".

If you are 100% sure that there will always be a value, you can use the equivalent of "force unwrap" by using the non-Optional getter and you don't need if let anymore:

let fbId = fbJson["id"].stringValue

But beware because this will crash if the value is nil, exactly as if you were force-unwrapping with fbJson["id"].string! (which you should never do).

Note that SwiftyJSON works the same for the other types: .int vs .intValue, .array vs .arrayValue, etc.

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