Question

我有一个需要保存为sql server表的json文件。这是test.json，其中包含有关学生作业的详细信息。

export class UserInfoService {

  constructor(private employeeService: EmployeeService) {
      this.handleSuccess = this.handleSuccess.bind(this);
      this.processData = this.processData.bind(this);
  }

  getEmployeeInfo(event?: Refresher) {

      // console.log(event);

      return this.employeeService
          .getEmployeeInfo()
          .map(this.handleSuccess)
          .finally(this.handleFinally);

  }

  handleFinally = (event?: Refresher) => {
      console.log(' handle finally ', event);

      if (event != null) {
          console.log(' event ', event);
          event.complete();
      }
  }
}

下面是C＃程序。我不知道如何访问“ smalldesc”和“ collegeCode”。

[{
    "Studentid": "001006360",
    "Grade": "2",
    "ExtraWork": {
        "TopRecommended": ["000133692",
        "102067155",
        "887273865"],
        "OtherCourses": ["228963647",
        "138909237",
        "899791144",
        "216165613",
        "113239563"]
    },
    "Courses": [{
        "smalldesc": "this is a test ",
        "Details": {
            "description": "Summary of the course",
            "collegeCode": "32466"
        }
    },
    {
        "smalldesc": "Second test",
        "Details": {
            "description": "Business- Course Summary",
            "collegeCode": "32469"
        }
    }]
}]

Json对象类定义：

var jsonText = File.ReadAllText(@"C:\test.json");    
var ser = JsonConvert.DeserializeObject<List<RootObject>>(jsonText);
for (int i = 0; i < ser.Count; i++)
{
    string Studentid = ser[i].Studentid;
    string  Grade = ser[i].Grade;

    var result = JsonConvert.DeserializeObject<List<Course>>(jsonText);
    for (int k = 0; k < result.Count; k++)
    {
        string smalldesc = result[k].smalldesc;
        string collegeCode = result[k].Details.collegeCode;
    }
}

保存到sql服务器表的最佳方法是什么。

Answer 1

此行：

@Component({
  selector: 'app-tasklist-items-grid',
  templateUrl: './tasklist-items-grid.component.html',
  styleUrls: ['./tasklist-items-grid.component.scss'],
  providers: [TasklistItemsService, BsModalService]
})
export class TasklistItemsGridComponent implements OnInit {
 taskListItem: TaskListItemViewModel;
 constructor() {
   // here or in ngOnInit
   // this.taskListItem = new TaskListItemViewModel();
 }
// .....
 ngOnInit() {
   // this.taskListItem = new TaskListItemViewModel();
 }
  convertToViewModel = (item: ITaskListItemViewModel) => item;
}

// in another file or same component you can write this class
export class TaskListItemViewModel implements ITaskListItemViewModel{    
    id?: number;
    tasklistId: number;
    typeOfTask: number;
    statusId: number;
    created_at: Date;
    updated_at: Date;
    checked: Boolean;
    rowVersion: number;
  }

interface ITaskListItemViewModel{    
        id: number;
        tasklistId: number;
        typeOfTask: number;
        statusId: number;
        created_at: Date;
        updated_at: Date;
        checked: Boolean;
        rowVersion: number;
      }

已经已经为您完成了所有反序列化，因此无需在循环内再次调用它。

您的代码可以像这样简单：

var ser = JsonConvert.DeserializeObject<List<RootObject>>(jsonText);

仅供参考：在迭代集合时，使用var ser = JsonConvert.DeserializeObject<List<RootObject>>(jsonText); foreach (var s in ser) { string Studentid = s.Studentid; string Grade = s.Grade; foreach(var course in ser.Courses) { string smalldesc = course .smalldesc; string details = course .Details.collegeCode; } }循环要容易得多（假设您的集合类型实现了foreach，其中大多数包含的集合都将实现）。

从C＃访问json节点的问题

1 个答案: