如何向REST API批量提交数据？解决旧数据被覆盖问题

尝试向REST API一次性提交数据列表时，API会删除之前的数据并仅保存新数据。

提供的JSON数据示例：

[
    {
        "name": "6E72Ln9Y095x8n80Xgk6/435291346914161.jpg",
        "class": "APKZZX",
        "email": "2142103j983H1",
        "roll_number": "6E72Ln9Y095x8n80Xgk6",
        "password": "2023-02-27 21:25:50.347756",
        "image": "",
        "role": "student",
        "school_id": "2142103j983H1",
        "id": "734506"
    },
    {
        "name": "6E72Ln9Y095x8n80Xgk6/435291346914161.jpg",
        "class": "APKZZX",
        "email": "2142103j983H1",
        "roll_number": "6E72Ln9Y095x8n80Xgk6",
        "password": "2023-02-27 21:25:50.347756",
        "image": "",
        "role": "student",
        "school_id": "2142103j983H1",
        "id": "734506"
    }
]

当前Flutter代码：

upload() async {
    try {
      final prefs = await SharedPreferences.getInstance();
      String? schoolId = prefs.getString('school_id');
      var map = {};
      List maps = List.empty(growable: true);
      Users user = Users();
      var value = data.skip(1).toList();
      for (var i = 0; i < value.length; i++) {
        user = Users(
          name: value[i][0].toString(),
          class1: value[i][1].toString(),
          email: value[i][2].toString(),
          roll_number: value[i][3].toString(),
          password: value[i][4].toString(),
          image: '',
          role: 'student',
          school_id: schoolId.toString(),
          id: randomAlphaNumeric(6),
        );
      }
      value.forEach((element) {
        Db.post('users', user.toJson());
      });
      print(user.toJson().toString());
      EasyLoading.showSuccess(user.toJson().toString());
    } catch (e) {
      EasyLoading.showError('Khalad ayaa jira fadlan hubi oo ku celi');
    }
  }

需求：实现向API一次性批量提交多条数据，或逐个提交数据。

代码问题分析 #

1. 循环创建user实例时，每次都会覆盖之前的变量，最终user仅保留最后一条用户数据
2. forEach循环中重复提交同一个用户数据，未实现多用户提交逻辑

解决方案 #

方案一：逐个提交数据 #

可以选择串行提交（等待前一个完成再提交下一个）或并行提交（同时发起所有请求）：

串行提交（避免并发过高）

upload() async {
    try {
      final prefs = await SharedPreferences.getInstance();
      String? schoolId = prefs.getString('school_id');
      var value = data.skip(1).toList();
      
      for (var i = 0; i < value.length; i++) {
        final user = Users(
          name: value[i][0].toString(),
          class1: value[i][1].toString(),
          email: value[i][2].toString(),
          roll_number: value[i][3].toString(),
          password: value[i][4].toString(),
          image: '',
          role: 'student',
          school_id: schoolId.toString(),
          id: randomAlphaNumeric(6),
        );
        // 等待当前请求完成再执行下一个
        await Db.post('users', user.toJson());
        print('已提交第 ${i+1} 个用户');
      }
      
      EasyLoading.showSuccess('所有用户提交完成');
    } catch (e) {
      EasyLoading.showError('提交失败：${e.toString()}');
    }
  }

并行提交（效率更高）

upload() async {
    try {
      final prefs = await SharedPreferences.getInstance();
      String? schoolId = prefs.getString('school_id');
      var value = data.skip(1).toList();
      
      List<Future> submitFutures = [];
      for (var i = 0; i < value.length; i++) {
        final user = Users(
          name: value[i][0].toString(),
          class1: value[i][1].toString(),
          email: value[i][2].toString(),
          roll_number: value[i][3].toString(),
          password: value[i][4].toString(),
          image: '',
          role: 'student',
          school_id: schoolId.toString(),
          id: randomAlphaNumeric(6),
        );
        submitFutures.add(Db.post('users', user.toJson()));
      }
      
      // 等待所有提交请求完成
      await Future.wait(submitFutures);
      EasyLoading.showSuccess('所有用户提交完成');
    } catch (e) {
      EasyLoading.showError('提交失败：${e.toString()}');
    }
  }

方案二：批量提交数据 #

如果后端API支持接收数组格式的请求体（且逻辑为追加数据而非覆盖），可以一次性提交整个用户列表：

upload() async {
    try {
      final prefs = await SharedPreferences.getInstance();
      String? schoolId = prefs.getString('school_id');
      var value = data.skip(1).toList();
      
      // 构建完整用户列表
      List<Map<String, dynamic>> userList = [];
      for (var i = 0; i < value.length; i++) {
        final user = Users(
          name: value[i][0].toString(),
          class1: value[i][1].toString(),
          email: value[i][2].toString(),
          roll_number: value[i][3].toString(),
          password: value[i][4].toString(),
          image: '',
          role: 'student',
          school_id: schoolId.toString(),
          id: randomAlphaNumeric(6),
        );
        userList.add(user.toJson());
      }
      
      // 提交整个列表，注意接口路径需对应后端批量接口
      await Db.post('users/batch', userList);
      EasyLoading.showSuccess('批量提交成功');
    } catch (e) {
      EasyLoading.showError('批量提交失败：${e.toString()}');
    }
  }

额外注意事项 #

由于class是Dart关键字，你的Users类用class1作为属性是正确的，但需要确保后端能识别该字段。可以通过JsonKey注解将class1映射为JSON里的class字段：

class Users {
  // ...其他属性
  @JsonKey(name: 'class')
  final String class1;
  
  // 构造函数、toJson等逻辑
}

内容的提问来源于stack exchange，提问作者abdifatah nour