Question

大家好我在快速3 iphone程序上使用alamofire，我的问题是我需要从php页面返回一个值我的问题是返回给我的价值是这个。我如何确保我返回的值是：no prova@email.it

我希望我已经解释了

返回值（不正确）：

import Foundation
import Alamofire

class User{

    //URL to our web service

    var email=""
    var password=""

    func PrintValue(){

       // print(username);
        //print(password);
    }

    func Login() -> String{

         //var ris="";
        var readvalue=""
        let URLString = "http://localhost/test/login_mobile.php"
        let parameters_value: Parameters = [
            "email": email,
            "password": password
        ]


        //Sending http post request
        Alamofire.request(URLString, method: .post, parameters: parameters_value).responseJSON
            {
                response in
                //printing response
                print(response)

                //getting the json value from the server
                if let result = response.result.value {

                    //converting it as NSDictionary
                    let jsonData = result as! NSDictionary

                    //displaying the message in label
                    readvalue = (jsonData.value(forKey: "message") as! String?)!

                    print(readvalue)
                }
        }

        return readvalue
    }

}

} 否可选（＆＃34; prova@email.it"）

SWIFT CODE：

<?php

include 'user.php';

header('Content-Type: application/json');
$email= $_POST['email'];
$password = $_POST['password'];

 $ris['message']="";

$user = new User();

//procedo con il login
if($user->login($email,$password,"MOBILE")==true){
    $ris['message']="yes";
}


else{
  $ris['message']="no $email";
}

echo json_encode($ris);

?>

PHP代码：

{{1}}

Answer 1

刚刚在操场上做到这一点，它按预期工作..

import UIKit

let response: [AnyHashable: Any] = [
    "message": "hello world"
]

class MyCoolClass {
    func login() {
        var readvalue = ""

        let jsonData = response as NSDictionary

        readvalue = jsonData.value(forKey: "message") as! String


        debugPrint("the readvalue is: \(readvalue)")

    }
}

let instance = MyCoolClass()
instance.login()

这将打印：＆＃34; readvalue是：hello world＆＃34;

代码不是很安全......

Answer 2

我认为可以这样做：

if let readvalue = jsonData.value(forKey: "message") as? String {
    print(readvalue)
}

必须打印无选择

Answer 3

您需要使用nil合并来使您的值不可选。例如print(readvalue ?? "nil")。更好的方法是重新设计您的响应处理以正确返回类型值或特定错误。

将Json从Php归还给Swift Alamofire

3 个答案: