我嘗試使用 dart 將數(shù)據(jù)發(fā)布到 php 后端。以某種方式使用 ajax，我可以獲得響應(yīng)，但是使用 dart，它顯示 <b>Notice</b>: Trying to get property 'email' of non-object in <b>D:\xampp\htdocs\PHPBackend\api\login\login_account.php</b> on line <b>23</b><br /><b>Notice</b>: Trying to get property 'password' of non-object in <b>D:\xampp\htdocs\PHPBackend\api\login\login_account.php</b> on line <b>24</b><br />這些是我用于登錄的 php 代碼。// login_account.php$database = new Database();$db = $database->getConnection();$login = new Login($db);$data = json_decode(file_get_contents("php://input"));$login->email = $data->email;$login->password = $data->password;$login->loginAccount();$login_arr = array( "email" => $login->email, "password" => $login->password);print_r(json_encode($login_arr));?>// login.phpfunction loginAccount(){ // query to read single record $query = "SELECT email, password FROM " . $this->table_name . " WHERE email = :email AND password = :password"; // prepare query statement $stmt = $this->conn->prepare( $query ); // sanitize $this->email=htmlspecialchars(strip_tags($this->email)); $this->password=htmlspecialchars(strip_tags($this->password)); // bind id of food to be updated $stmt->bindParam(":email", $this->email); $stmt->bindParam(":password", $this->password); die($this->email); die($this->password); // execute query $stmt->execute(); // get retrieved row $row = $stmt->fetch(PDO::FETCH_ASSOC); // set values to object properties $this->email = $row['email']; $this->password = $row['password']; }}
2 回答
MM們
TA貢獻1886條經(jīng)驗 獲得超2個贊
附加信息
有一個 dart 包為 http 請求提供了一些幫助類。
Github：https : //github.com/Ephenodrom/Dart-Basic-Utils
安裝它：
dependencies:
basic_utils: ^1.5.1
它也是 EZ-Flutter Collection 的一部分：
Github：https : //github.com/Ephenodrom/EZ-Flutter 文檔：https : //ez-flutter.de/docs
dependencies:
ez_flutter: ^0.2.5
用法
Map<String, String> headers = {
"Some": "Header"
};
Map<String, String> queryParameters = {
"Some": "Parameter"
};
String url = "";
Map payload = "{}";
Map<String, dynamic> reaponseBody;
try {
responseBody = await HttpUtils.postForJson(url, json. encode(payload) ,
queryParameters: queryParameters, headers: headers);
} catch (e) {
// Handle exception, for example if response status code != 200-299
}
// do something with the response body
附加信息 ：
這些都是來自 HttpUtils 類的方法。
Future<Map<Response> getForFullResponse(String url,{Map<String, dynamic> queryParameters,Map<String, String> headers});
Future<Map<String, dynamic>> getForJson(String url,{Map<String, dynamic> queryParameters,Map<String, String> headers});
Future<String> getForString(String url,{Map<String, dynamic> queryParameters,Map<String, String> headers});
Future<Map<Response> postForFullResponse(String url, String body,{Map<String, String> queryParameters,Map<String, String> headers});
Future<Map<String, dynamic>> postForJson(String url, String body,{Map<String, String> queryParameters,Map<String, String> headers});
Future<String> postForString(String url, String body,{Map<String, String> queryParameters,Map<String, String> headers});
Future<Response> putForFullResponse(String url, String body,{Map<String, String> queryParameters,Map<String, String> headers});
Future<Map<String, dynamic>> putForJson(String url, String body,{Map<String, String> queryParameters,Map<String, String> headers});
Future<String> putForString(String url, String body,{Map<String, String> queryParameters,Map<String, String> headers});
Future<Response deleteForFullResponse(String url,{Map<String, String> queryParameters,Map<String, String> headers});
Future<Map<String, dynamic>> deleteForJson(String url,{Map<String, String> queryParameters,Map<String, String> headers});
Future<String> deleteForString(String url,{Map<String, String> queryParameters,Map<String, String> headers});
Map<String, dynamic> getQueryParameterFromUrl(String url);
String addQueryParameterToUrl(String url, Map<String, dynamic> queryParameters);
幕布斯6054654
TA貢獻1876條經(jīng)驗 獲得超7個贊
我可以在 Dart 代碼中看到您正在嘗試直接發(fā)送 Map 對象，而不是首先將其轉(zhuǎn)換為例如 JSON。
要轉(zhuǎn)換為 JSON，您可以使用 dart:convert 包和以下方法：
var encodedLoginObj = json.encode(loginObj);
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