85 lines
2.6 KiB
Dart
85 lines
2.6 KiB
Dart
/// ============================================================
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/// 【auth 模块模型】登录相关实体 —— 对照 Vue 项目 src/api/login.ts
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///
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/// LoginRequest → 登录请求体（toJson 提交给服务器）
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/// LoginResponse → 登录响应（token + 用户信息）
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/// UserInfo → 用户信息子模型（info 字段）
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///
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/// 高频坑：
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/// 1. fromJson 全部用 ?? 兜底，字段缺失/类型漂移不会闪崩；
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/// 2. 数字用 (json['x'] as num?)?.toInt() 防 int/double 不一致；
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/// 3. 兼容业务响应壳 {code, msg, data: {token, info}}。
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/// ============================================================
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library;
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class LoginRequest {
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const LoginRequest({required this.account, required this.password});
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final String account;
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final String password;
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Map<String, dynamic> toJson() {
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return {
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'account': account,
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'password': password,
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};
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}
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}
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class LoginResponse {
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const LoginResponse({required this.token, required this.info});
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final String token;
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final UserInfo info;
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factory LoginResponse.fromJson(Map<String, dynamic> json) {
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// 兼容常见包装结构：后端可能直接返回 {token, info}，
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// 也可能包一层 {code, msg, data: {token, info}}（业务响应通用壳）。
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// data 存在就取 data，否则直接用 json，两种都能解析到。
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final payload = (json['data'] as Map<String, dynamic>?) ?? json;
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return LoginResponse(
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token: payload['token'] as String? ?? '',
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info: UserInfo.fromJson(
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(payload['info'] as Map<String, dynamic>?) ?? const {},
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),
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);
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}
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}
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/// 用户信息（对齐 Vue 端 getUserInfo 的字段，未知的用 ?? 兜底）
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class UserInfo {
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const UserInfo({
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this.userId = 0,
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this.userName = '',
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this.account = '',
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this.avatar = '',
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this.phone = '',
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this.isShowReport = false,
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});
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final int userId;
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final String userName;
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final String account;
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final String avatar;
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final String phone;
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final bool isShowReport;
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factory UserInfo.fromJson(Map<String, dynamic> json) {
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return UserInfo(
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// userId 可能是字符串（后端统一风格），不能直接 as num
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userId: _toInt(json['userId']),
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userName: json['userName'] as String? ?? '',
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account: json['account'] as String? ?? '',
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avatar: json['avatar'] as String? ?? '',
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phone: json['phone'] as String? ?? '',
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isShowReport: json['isShowReport'] as bool? ?? false,
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);
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}
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}
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/// 兼容"字符串数字"（后端 userId 返回 "820466383020136"）
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int _toInt(dynamic v) {
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if (v is num) return v.toInt();
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if (v is String) return int.tryParse(v) ?? 0;
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return 0;
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} |