无法在 Swift 应用程序上解析服务器 PHP

Question

我正在尝试在我的 iOS 应用上实现注册/登录功能。 为此，我编写了一个简单的服务器端 php 脚本，只是为了检查它是否返回一个值。 这是我的php代码：

<?php

if(isset($_POST["username"])){
$username = $_POST["username"];
$password = $_POST["password"];
if($username == "admin" && $password == "admin")
{
$details;
$details['success'] = $username;
echo json_encode($details);
}
else{
echo "invalid credential";
}
};
?>

这是登录屏幕中的代码：

let myUrl = NSURL(string: "http://localhost/app/index.php")  
    let request = NSMutableURLRequest(URL:myUrl!)
    request.HTTPMethod = "POST" 
    request.addValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
    request.addValue("application/json", forHTTPHeaderField: "Accept") 
    //print("\(usernameField.text!), \(passwordField.text!), \(emailField.text!)")  
    //let params = ["user":"\(usernameField.text!)","pass":"\(passwordField.text!)"] as NSDictionary  
    let bodyData = "username=\(usernameField.text)&password=\(passwordField.text)"  
    request.HTTPBody = bodyData.dataUsingEncoding(NSUTF8StringEncoding)

    let task = NSURLSession.sharedSession().dataTaskWithRequest(request, completionHandler: {
            data, response, error in
            do {
                let json = try NSJSONSerialization.JSONObjectWithData(data!, options: []) as? NSDictionary
                print("json is : \(json!)")
                if let parseJSON = json {
                    let firstNameValue = parseJSON["success"] as? String
                    print("Username: \(firstNameValue!)")
                }
            } catch {
                print("errorrrrr: \(error)")
            }  
            if error != nil {
                print("error=\(error!)")
                return
            }        
            // You can print out response object
            print("Response = \(response!)")
            // Print out reponse body
            let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
            print("response Data = \(responseString!)")


        })
        task.resume()

}

但是，我收到以下错误：

Error Domain=NSCocoaErrorDomain Code=3840 "JSON text did not start with array or object and option to allow fragments not set." UserInfo={NSDebugDescription=JSON text did not start with array or object and option to allow fragments not set.}

我的响应数据是“无效凭据”。

Answer 1

您的 php 脚本对于 web 服务不完整。它没有返回 json。您必须先完成您的网络服务。我不是一个php人。但我仍然尝试改进脚本：

<?php
if(isset($_POST["username"])){
$username = $_POST["username"];
$password = $_POST["password"];
if($username == "admin" && $password == "admin")
{
    $details;
    $details['success'] = $username;
    return json_encode($details);
}
else{
    return json_encode( array("response" => "invalid credential") );
}
};
?>

更新：在迭代 alamofire 库后使用以下代码访问 web 服务：

let parameters = ["username": usernameField.text! ,"password" : passwordField.text! ]

Alamofire.request(.POST, "http://localhost/app/index.php", parameters: parameters)
     .responseString { response in
         print("Response String: \(response.result.value)")
     }
     .responseJSON { response in
         print("Response JSON: \(response.result.value)")
       //Handle the json response here
      if let parseJSON = response.result.value {
                let firstNameValue = parseJSON["success"] as? String
                print("Username: \(firstNameValue!)")
            }
     }