繁体   English   中英

在C#关系中使用json.net反序列化Json var

[英]Deserializing Json var with json.net in c# relations

我有一个json字符串,我在其中尝试为EmployeeId,EmployeeName和dates创建一个foreach:

{"$id":"1","Employee":{"$id":"2","RelationshipManager":{"$id":"3","_owner":{"$ref":"2"},"_relationships":[[{"$id":"4","Calendarbookings":null,"Id":1,"MId":1,"Startdate":"2011-01-14T00:00:00","Enddate":"2011-01-14T00:00:00","StartTime":"1899-12-30T10:00:00","Endtime":"1899-12-30T16:00:00","EmployeeId":1,"DebtorId":0,"Subject":"Tidevand Odense","Description":null,"AllDay":0,"Wwwaddress":"","Mailaddress":"","IsTeam":1,"IsRessource":0,"RessourceType":0,"Participants":1,"Department":0,"IsRepeating":0,"TaskType":0,"Categorytype":0,"Price":495.0,"WaitinglistIsEnabled":1,"Deadline":"2011-02-01T00:00:00","DeadlineForCancellation":"1899-12-30T00:00:00","Ending":"2011-01-14T00:00:00","OnlineEnabled":1,"Revenueaccount":1502,"Maxbooking":20,"Employee":{"$ref":"2"}}]]},"Id":1,"SearchKey":"","EmployeeName":"Erik Steen Larsen","Address1":"","Address2":"","Zip":"","City":"","Phone":"","CashSuper":"0","ActiveInCalendar":"1","Startdate":"1899-12-30T00:00:00","Enddate":"1899-12-30T00:00:00","Mobile":"30125744","Description":null,"Events":[{"$ref":"4"}]},"ErrorCode":0,"Status":0,"StatusString":"Okay","Message":null,"ApiVersion":"1.0.0.8"}

怎么做???

_jsonContent = JsonConvert.DeserializeObject(JSONSTRING);

foreach (var item in _jsonContent.Employee)
{
   littext.Text += item.EmployeeName;
   littext.Text += item.Id;
}

谢谢

您可以使用动态对象来实现所需的功能,例如:

        dynamic astring = "{\"$id\":\"1\",\"Employee\":{\"$id\":\"2\",\"RelationshipManager\":{\"$id\":\"3\",\"_owner\":{\"$ref\":\"2\"},\"_relationships\":[[{\"$id\":\"4\",\"Calendarbookings\":null,\"Id\":1,\"MId\":1,\"Startdate\":\"2011-01-14T00:00:00\",\"Enddate\":\"2011-01-14T00:00:00\",\"StartTime\":\"1899-12-30T10:00:00\",\"Endtime\":\"1899-12-30T16:00:00\",\"EmployeeId\":1,\"DebtorId\":0,\"Subject\":\"Tidevand Odense\",\"Description\":null,\"AllDay\":0,\"Wwwaddress\":\"\",\"Mailaddress\":\"\",\"IsTeam\":1,\"IsRessource\":0,\"RessourceType\":0,\"Participants\":1,\"Department\":0,\"IsRepeating\":0,\"TaskType\":0,\"Categorytype\":0,\"Price\":495.0,\"WaitinglistIsEnabled\":1,\"Deadline\":\"2011-02-01T00:00:00\",\"DeadlineForCancellation\":\"1899-12-30T00:00:00\",\"Ending\":\"2011-01-14T00:00:00\",\"OnlineEnabled\":1,\"Revenueaccount\":1502,\"Maxbooking\":20,\"Employee\":{\"$ref\":\"2\"}}]]},\"Id\":1,\"SearchKey\":\"\",\"EmployeeName\":\"Erik Steen Larsen\",\"Address1\":\"\",\"Address2\":\"\",\"Zip\":\"\",\"City\":\"\",\"Phone\":\"\",\"CashSuper\":\"0\",\"ActiveInCalendar\":\"1\",\"Startdate\":\"1899-12-30T00:00:00\",\"Enddate\":\"1899-12-30T00:00:00\",\"Mobile\":\"30125744\",\"Description\":null,\"Events\":[{\"$ref\":\"4\"}]},\"ErrorCode\":0,\"Status\":0,\"StatusString\":\"Okay\",\"Message\":null,\"ApiVersion\":\"1.0.0.8\"}";

        dynamic jsonContent = JsonConvert.DeserializeObject(astring);

        foreach (var item in jsonContent.Employee)
        {
            /*.,..*/
        }

你也可以使用匿名对象

var template = new { EmployeName = "", EmplyeeId = 0 };
var jsonContent = JsonConvert.DeserializeAnonymousType(JSONSTRING, template);

 foreach (var item in jsonContent.Employee)
        {
            /*.,..*/
        }

json中的Employee不是数组。 您还需要将json反序列化为适当的容器。

JObject:

var jObject = JsonConvert.DeserializeObject<JObject>(json);

var employeeId = (int) jObject["Employee"]["Id"];
var employeeName = (string) jObject["Employee"]["EmployeeName"];

动态:

var jObject = JsonConvert.DeserializeObject<dynamic>(json);

var employeeId = (int) jObject.Employee.Id;
var employeeName = (string) jObject.Employee.EmployeeName;

POCO:

public class Record
{
    [JsonProperty("Employee")]
    public Employee Employee { get; set; }
}

public class Employee
{
    [JsonProperty("Id")]
    public int Id { get; set; }

    [JsonProperty("EmployeeName")]
    public string EmployeeName { get; set; }
}

var record = JsonConvert.DeserializeObject<Record>(json);

var employeeId = record.Employee.Id;
var employeeName = record.Employee.EmployeeName;

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM