FOSUser創建用戶的AJAX請求

-2

爲了測試FOSUser，我模擬了發送包含用戶名，他的電子郵件和密碼的JSON數據的AJAX請求。FOSUser創建用戶的AJAX請求

JavaScript端（遠從完美的Ajax請求，但我會在實際開發中使用角）：

function ajaxPost(url, data, callback, isJson) { 
    let req = new XMLHttpRequest(); 
    req.open("POST", url); 
    req.addEventListener("load", function() { 
     if (req.status >= 200 && req.status < 400) {    
      callback(req.responseText); 
     } else { 
      console.error(req.status + " " + req.statusText + " " + url); 
     } 
    }); 
    req.addEventListener("error", function() { 
     console.error("Network error at " + url); 
    }); 
    if (isJson) { 
     req.setRequestHeader("Content-Type", "application/json"); 
     data = JSON.stringify(data); 
    } 
    req.send(data); 
} 

// Simulates an AJAX request to create a user from a form: 
user = prompt("Username:"); 
email = prompt("Email:"); 
password = prompt("Password:"); 
data = { user, email, password }; 
ajaxPost("http://domain.tld/app_dev.php/createuser", data, console.log, true);

Symfony的一面：

/** 
* @Route("/createuser", name="createUser") 
*/ 
public function createUserAction($data, UserManager $userManager) 
{ 
    $data = json_decode($data); 
    if (!$userManager->findUserByUsername($data->username) && !$userManager->findUserByEmail($data->email)) { 
     $user = $userManager->createUser(); 
     $user->setUsername($data->username); 
     $user->setEmail($data->email); 
     $password = password_hash($data->password, PASSWORD_DEFAULT); 
     $user->setPassword($password); 
     $userManager->updateUser($user); 
     return new Response("<html><body>New user created!</body></html>"); 
    } else { 
     return new Response("<html><body>This user or email already exists.</body></html>"); 
    } 
}

當我直接去/ CREATEUSER並通過模擬數據，它的工作原理;但是當我發送我的ajax請求時，它始終返回500內部服務器錯誤：

[2017-06-17 08:31:49] request.CRITICAL: Uncaught PHP Exception Symfony\Component\Debug\Exception\ContextErrorException: "Notice: Trying to get property of non-object" at C:\wamp64\www\MyWebsite\src\AppBundle\Controller\DefaultCont‌roller.php line 50 {"exception":"[object] (Symfony\Component\Debug\Exception\ContextErrorException‌(code: 0): Notice: Trying to get property of non-object at C:\wamp64\www\MyWebsite\src\AppBundle\Controller\Defa‌ultController.php:50‌)"} []" That's: if (!$userManager->findUserByUsername($data->userna

任何想法？ :)

來源

2017-06-17 Joubarbe

你能提供有關錯誤的一些信息？從日誌中提取它 – manuelbcd

我忘了看看它們...還不習慣Symfony：[2017-06-17 08:31:49] request.CRITICAL：未捕獲到PHP異常Symfony \ Component \ Debug \ Exception \ ContextErrorException ：「注意：嘗試獲取非對象的屬性」在C：\ wamp64 \ www \ MyWebsite \ src \ AppBundle \ Controller \ DefaultController.php第50行{「exception」：「[object]（Symfony \\ Component \\ Debug \\ Exception \\ ContextErrorException（code：0）：注意：試圖在C：\\ wamp64 \\ www \\ MyWebsite \\ src \\ AppBundle \\ Controller \\ DefaultController.php中獲取非對象的屬性：在$ data = json_decode（$ data）之後，如果（！$ userManager-> findUserByUsername（$ data-> userna ... – Joubarbe

）可以執行dump（$ data）'; '？ – kRicha

好吧，我很笨。 $數據不算什麼;我必須使用$請求 - >的getContent（）...

public function createUserAction(Request $request, UserManager $userManager) 
    { 
     $data = json_decode($request->getContent());

感謝所有您的意見:)

來源

2017-06-17 09:31:10 Joubarbe

FOSUser創建用戶的AJAX請求

回答

相關問題