Nicht behandelte Ausnahme: FormatException: Unerwartetes Zeichen (bei Zeichen 1) E/flutter (6084): <br /> E/flutter (6084): ^

Question

php-Datei:

$email=$_POST['email'];

    $passworda=$_POST['passworda'];

$sql="SELECT * FROM user WHERE email='".$email."'AND passworda='".$password."' ";
$result=mysqli_query($db,$sql);
$count=mysqli_num_rows($result);
if($count>=1){
echo json_encode("success");
}
else
{
echo json_encode("error");
}

Anmeldeseite Flutter:

class Login extends StatelessWidget {
  TextEditingController email = TextEditingController();
  TextEditingController password = TextEditingController();

  Future login(BuildContext cont) async {
    if (email.text == "" || password.text == "") {
      Fluttertoast.showToast(
        msg: "please complete!",
        toastLength: Toast.LENGTH_SHORT,
        gravity: ToastGravity.CENTER,
        fontSize: 16.0,
      );
    } else {
      var url = "http://192.168.43.150/v1_flutter/lib/php/connection.php";
      var response = await http.post(Uri.parse(url), body: {
        "email": email.text,
        "pass": password.text,
      }, headers: {"Accept":"applicarion/json"});
      var data = jsonDecode(response.body);

      if (data == "success") {
        Navigator.pop(cont);
        Navigator.pushNamed(cont, "/registre");
      } else {
        Fluttertoast.showToast(
          msg: "The user and password does not exist!",
          toastLength: Toast.LENGTH_SHORT,
          gravity: ToastGravity.CENTER,
          fontSize: 16.0,
        );
      }
    }}

Konsole:

E/flutter (6084): [Fehler: flutter/lib/ui/ui_dart_state.cc(198)] Unbehandelte Ausnahme: FormatException: Unerwartetes Zeichen (at Charakter 1) E/Flutter (6084):

E/Flutter(6084):^ E/Flattern(6084): E/Flutter(6084): #0 _ChunkedJsonParser.fail (dart:convert-patch/convert_patch.dart:1383:5) E/Flutter(6084): #1 _ChunkedJsonParser.parseNumber (dart:convert-patch/convert_patch.dart:1250:9) E/Flutter(6084): #2 _ChunkedJsonParser.parse (dart:convert-patch/convert_patch.dart:915:22) E/Flutter(6084): #3 _parseJson (dart:convert-patch/convert_patch.dart:35:10) E/Flutter (6084): #4 JsonDecoder.convert (dart:convert/json.dart:612:36) E/Flutter (6084): #5 JsonCodec.decode (dart:convert/json.dart:216:41) E/Flutter (6084): #6 jsonDecode (dart:convert/json.dart:155:10) E/Flutter (6084): #7 Login.login (package:mes_v1/pages/Authentification/login.dart:25:18) E/Flattern(6084): E/Flattern (6084):

Answer 1

您的代码看起来确实容易出错，但重点关注您的问题：异常看起来很清楚，您的响应解析正在中断。 （我还想说，你应该用 try/catch 包装你的代码，以防止任何代码破坏和正确捕获问题）。

让我们解决您的问题：

var data = jsonDecode(response.body);

这会将您的字符串 (response.body) 转换为 json，这意味着 data 是 Map 或 List< /code> （其中动态是 Map 或另一个嵌套列表），因此以下内容没有任何意义

if（数据==“成功”）{

现在，让我们看一下您的 php 代码：

echo json_encode("成功");

我不是 php 专家，但从文档来看它应该像下面这样使用：

$response = array("result" => "Success");
echo json_encode($response);

现在让我们回到您的 dart 代码：

var success = false;
    try{

      final baseUrl = "http://192.168.43.150"; // Use final wherever you can
      final url = "$baseUrl/v1_flutter/lib/php/connection.php"; // Improving flexibility

      final body = {
        "email": email.text,
        "pass": password.text,
      }; // Decouple in variables for readability

      final headers = {"Accept":"application/json"}; // Fix typo

      final response = await http.post(Uri.parse(url), body: body, headers: headers);
      final data = jsonDecode(response.body);
      success = data["result"] == "Success";
   } catch(e) {
      print("Catched an error!");
      print(e);
      success = false;
   }

   if(success) {
   ...