Flutter登录遇FormatException错误，寻求技术解决帮助

问题描述 #

开发Flutter应用（基于Dart语言，搭配MySQL数据库）时，登录环节持续触发以下错误：

I/flutter ( 7573): Error ::FormatException: Unexpected character (at character 1)
I/flutter ( 7573): <br />
I/flutter ( 7573): ^

同时编写login.php时，输入fetch_assoc()无代码提示，怀疑该现象与登录报错相关。

相关代码文件 #

login.php代码

<?php

include '../connection.php';

$userEmail = $_POST['EmailUser'];
$userPassword = md5($_POST['PasswordUser']);

$sqlQuery = "SELECT * FROM user_table WHERE EmailUser = '$userEmail' AND PasswordUser = '$userPassword'";

$resultOfQuery = $connectNow->query($sqlQuery);

if($resultOfQuery->num_rows > 0)//allow user to login (record found)
{
    $userRecord = array(); 
    while($rowFound = $resultOfQuery->fetch_assoc()) 
    {
        $userRecord[] = $rowFound;
    }

    echo json_decode( 
        array( 
            "success" =>true,
            "userData" =>$userRecord[0],
            )
    );
}
else// not allow if wrong either one
{
    echo json_encode(array("success"=>false));
}

登录服务代码（Dart）

loginUserNow() async {
    try {
      var res = await http.post(
        Uri.parse(API.login),
        body: {
          "EmailUser": emailcontroller.text.trim(),
          "PasswordUser": passwordcontroller.text.trim(),
        },
      );

      if (res.statusCode == 200) //from flutter app to server connection suc
      {
        var resBodyOfLogin = jsonDecode(res.body);
        if (resBodyOfLogin['success'] == true) {
          Fluttertoast.showToast(msg: " you are Logged in Successfully.");

          User userInfo = User.fromJson(resBodyOfLogin["userData"]);
          //save user info to local storage usin the Shared Preference

          await RememberUserPrefs.saveRememberUser(userInfo);

          //send user to dashboard or default screen
          // ignore: prefer_const_constructors
          Get.to(DashboardOfFragments());
        } else {
          Fluttertoast.showToast(
              msg: "Incorrect Credentials. Please try again.");
        }
      }
    } catch (errorMsg) {
      // ignore: avoid_print, prefer_interpolation_to_compose_strings
      print("Error ::$errorMsg");
    }
  }

用户偏好设置代码（Dart）

import 'dart:convert';

import 'package:kezooapp/users/model/user.dart';
import 'package:shared_preferences/shared_preferences.dart';

class RememberUserPrefs {
  //save-remember User-info
  static Future<void> saveRememberUser(User userInfo) async {
    SharedPreferences preferences = await SharedPreferences.getInstance();
    String userJsonData = jsonEncode(userInfo.toJson());

    await preferences.setString("currentUser", userJsonData);
  }
}

User模型代码（Dart）

class User {
  int IdUser;
  String NameUser;
  String EmailUser;
  String PasswordUser;

  User(
    this.IdUser,
    this.NameUser,
    this.EmailUser,
    this.PasswordUser,
  );

  factory User.fromJson(Map<String, dynamic> json) => User(
        int.parse(json["IdUser"]),
        json["NameUser"],
        json["EmailUser"],
        json["PasswordUser"],
      );

  Map<String, dynamic> toJson() => {
        'IdUser': IdUser.toString(),
        'NameUser': NameUser,
        'EmailUser': EmailUser,
        'PasswordUser': PasswordUser,
      };
}

解决方案 #

1. 修复PHP端核心错误 #

错误日志中的<br />是HTML错误提示，说明PHP输出了非JSON内容。问题出在成功登录分支的json_decode()调用——该函数用于解析JSON字符串，而非生成JSON，正确应使用json_encode()：

// 修改后的成功分支代码
if($resultOfQuery->num_rows > 0)
{
    $userRecord = array(); 
    while($rowFound = $resultOfQuery->fetch_assoc()) 
    {
        $userRecord[] = $rowFound;
    }

    // 替换json_decode为json_encode
    echo json_encode( 
        array( 
            "success" => true,
            "userData" => $userRecord[0],
            )
    );
}

2. 解决PHP代码提示问题 #

fetch_assoc()无代码提示是因为编辑器无法识别$resultOfQuery的类型，可通过添加类型注释解决：

// 在$resultOfQuery声明前添加类型注释
/** @var mysqli_result $resultOfQuery */
$resultOfQuery = $connectNow->query($sqlQuery);

同时确保编辑器安装了PHP语法提示插件（如VSCode的PHP Intelephense），开启代码智能提示功能。

3. 修复SQL注入风险（可选但关键） #

当前PHP代码直接拼接用户输入到SQL语句中，存在严重安全隐患，建议使用预处理语句替代：

// 替换原SQL查询与执行逻辑
$userEmail = $_POST['EmailUser'];
$userPassword = md5($_POST['PasswordUser']);

// 使用预处理语句
$sqlQuery = "SELECT * FROM user_table WHERE EmailUser = ? AND PasswordUser = ?";
$stmt = $connectNow->prepare($sqlQuery);
// 绑定参数（"ss"表示两个字符串类型参数）
$stmt->bind_param("ss", $userEmail, $userPassword);
$stmt->execute();
$resultOfQuery = $stmt->get_result();

4. Flutter端优化建议 #

• 在loginUserNow()中添加响应内容打印，方便排查问题：

if (res.statusCode == 200)
{
    print("Server Response: ${res.body}"); // 添加此行
    var resBodyOfLogin = jsonDecode(res.body);
    // ... 后续逻辑
}

• 补充非200状态码的错误处理，覆盖服务器异常场景。

内容的提问来源于stack exchange，提问作者Yasmine