flutter架 构项目基础

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崔宝新
2026-09-04 15:22:20 +08:00
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/// ============================================================
/// 【auth 模块模型】登录相关实体 —— 对照 Vue 项目 src/api/login.ts
///
/// LoginRequest → 登录请求体(toJson 提交给服务器)
/// LoginResponse → 登录响应(token + 用户信息)
/// UserInfo → 用户信息子模型(info 字段)
///
/// 高频坑:
/// 1. fromJson 全部用 ?? 兜底,字段缺失/类型漂移不会闪崩;
/// 2. 数字用 (json['x'] as num?)?.toInt() 防 int/double 不一致;
/// 3. 兼容业务响应壳 {code, msg, data: {token, info}}。
/// ============================================================
library;
class LoginRequest {
const LoginRequest({required this.account, required this.password});
final String account;
final String password;
Map<String, dynamic> toJson() {
return {
'account': account,
'password': password,
};
}
}
class LoginResponse {
const LoginResponse({required this.token, required this.info});
final String token;
final UserInfo info;
factory LoginResponse.fromJson(Map<String, dynamic> json) {
// 兼容常见包装结构:后端可能直接返回 {token, info},
// 也可能包一层 {code, msg, data: {token, info}}(业务响应通用壳)。
// data 存在就取 data,否则直接用 json,两种都能解析到。
final payload = (json['data'] as Map<String, dynamic>?) ?? json;
return LoginResponse(
token: payload['token'] as String? ?? '',
info: UserInfo.fromJson(
(payload['info'] as Map<String, dynamic>?) ?? const {},
),
);
}
}
/// 用户信息(对齐 Vue 端 getUserInfo 的字段,未知的用 ?? 兜底)
class UserInfo {
const UserInfo({
this.userId = 0,
this.userName = '',
this.account = '',
this.avatar = '',
this.phone = '',
this.isShowReport = false,
});
final int userId;
final String userName;
final String account;
final String avatar;
final String phone;
final bool isShowReport;
factory UserInfo.fromJson(Map<String, dynamic> json) {
return UserInfo(
// userId 可能是字符串(后端统一风格),不能直接 as num
userId: _toInt(json['userId']),
userName: json['userName'] as String? ?? '',
account: json['account'] as String? ?? '',
avatar: json['avatar'] as String? ?? '',
phone: json['phone'] as String? ?? '',
isShowReport: json['isShowReport'] as bool? ?? false,
);
}
}
/// 兼容"字符串数字"(后端 userId 返回 "820466383020136")
int _toInt(dynamic v) {
if (v is num) return v.toInt();
if (v is String) return int.tryParse(v) ?? 0;
return 0;
}