我想用列表对象中的嵌套对象创建json结构。
这是我的班级：
public class Employee
{
public int EmployeeId { get; set; }
public int Skillssetpoints { get; set; }
public string Name { get; set; }
public Nullable<System.DateTime> Date { get; set; }
}
public class EmployeeModel
{
public int EmployeeId { get; set; }
public List<int> Skillssetpoints { get; set; }
public string Name { get; set; }
public Nullable<System.DateTime> Date { get; set; }
}
记录是这样的：
EmployeeId SkillssetPoints Date
1 10 4/5/2016 16:12:12
2 12 3/5/2016 17:12:12
3 4 8/5/2016 8:12:12
4 20 1/5/2016 2:12:12
这就是我获取数据的方式：
var data=context.Employee.Tolist();
获取数据后，我想使用EmployeeModel从上面的数据创建此 json结构并返回：
预期输出：
{"Date":"8-5-2016 08:12:12","SkillssetPoints":[4,10,12,20]}
在日期字段中，我将采用最高日期，因此8-5-2016和 SkillssetPoints 将按升序排序。
如何使用 EmployeeModel 类创建此json结构？
答案 0 :(得分：4)
添加对nuget包的引用Newtonsoft.Json然后使用...
string result = JsonConvert.Serialize(data);
看起来您需要先将数据转换为模型格式，然后先进行投影，然后序列化结果......
var groupedData = data
.GroupBy(s => s.EmployeeId)
.OrderBy(s => s.Date)
.Select(g => new EmployeeModel {
EmployeeId = g.Key,
Name = g.First().Name,
Date = g.First().Date,
Skillssetpoints = g.Select(s => s.Skillssetpoints).OrderBy(i => i).ToList()
});
那应该产生这个模型的集合......
public class EmployeeModel
{
public int EmployeeId { get; set; }
public List<int> Skillssetpoints { get; set; }
public string Name { get; set; }
public DateTime? Date { get; set; }
}
......当我这样做时......
var data = new List<EmployeeModel> {
new EmployeeModel { EmployeeId = 1, Name = "Homer Simpson", Skillssetpoints = new List<int> { 1,2,3,4 }, Date = DateTime.Now },
new EmployeeModel { EmployeeId = 2, Name = "Marge Simpson", Skillssetpoints = new List<int> { 1,2,3,4 }, Date = DateTime.Now },
new EmployeeModel { EmployeeId = 3, Name = "Lisa Simpson", Skillssetpoints = new List<int> { 1,2,3,4 }, Date = DateTime.Now },
new EmployeeModel { EmployeeId = 4, Name = "Bart Simpson", Skillssetpoints = new List<int> { 1,2,3,4 }, Date = DateTime.Now }
};
var result = JsonConvert.SerializeObject(data);
我得到了这个输出......
[
{
"EmployeeId": 1,
"Skillssetpoints": [1,2,3,4],
"Name": "Homer Simpson",
"Date": "2016-04-05T11:42:09.9126748+01:00"
},
{
"EmployeeId": 2,
"Skillssetpoints": [1,2,3,4],
"Name": "Marge Simpson",
"Date": "2016-04-05T11:42:09.9126748+01:00"
},
{
"EmployeeId": 3,
"Skillssetpoints": [1,2,3,4],
"Name": "Lisa Simpson",
"Date": "2016-04-05T11:42:09.9126748+01:00"
},
{
"EmployeeId": 4,
"Skillssetpoints": [1,2,3,4],
"Name": "Bart Simpson",
"Date": "2016-04-05T11:42:09.9126748+01:00"
}
]
答案 1 :(得分：2)
通过对特定字段进行分组来选择所需记录，然后以所需模式准备匿名对象并将其序列化以获得最终结果，
var models = (from em in employeeModels
group em by em.ID into g
select new
{
Id = g.Key,
maxDate = g.Max(p => p.Date)
}).ToList();
var result = new
{
date = prices.Max(p => p.maxDate),
SkillssetPoints = prices.Select(p => p.Id).ToList()
};
var json = JsonConvert.SerializeObject(result);
您将获得类似
的模式中的json{
"date": "2016-04-05T16:39:54.8420979+05:30",
"SkillssetPoints": [
1,
2,
3
]
}
答案 2 :(得分：2)
您可以尝试创建一个新对象并按以下序列化：
var result = JsonConvert.Serialize (new {
Date = context.Employee.Max(e => e.Date),
SkillssetPoints = context.Employee.Select(e => e.SkillssetPoints)
}));
答案 3 :(得分：2)
使用newton JSON，它可以在NuGet上使用，代码非常简单。
using Newtonsoft.Json;
var jsonList = JsonConvert.SerializeObject(context.Employee.Tolist());
干杯
答案 4 :(得分：2)
使用Nuget Package中的Newton JSON并尝试此代码
列出lstEmp = new List（）;
for (int i = 1; i <= 4; i++)
{
Employee emp = new Employee();
emp.EmployeeId = i;
emp.Name = "Name" + i;
emp.Skillssetpoints = i + 1;
emp.Date = DateTime.Now.AddDays(i);
lstEmp.Add(emp);
}
var data = lstEmp;
var result = new EmployeeModel
{
Date = data.Max(p => p.Date),
Skillssetpoints = data.Select(p => p.Skillssetpoints).ToList()
};
var JsonData = JsonConvert.SerializeObject(new
{
Date = result.Date,
Skillssetpoints = result.Skillssetpoints
});
干杯
答案 5 :(得分：2)
1）使用NuGet
安装Newtonsoft.Json包2）在顶部添加名称空间
using Newtonsoft.Json;
3）在不希望包含在json转换中的Model类属性上添加[JsonIgnore]
public class Employee
{
public int EmployeeId { get; set; }
public int Skillssetpoints { get; set; }
public string Name { get; set; }
public Nullable<System.DateTime> Date { get; set; }
}
public class EmployeeModel
{
[JsonIgnore]
public int EmployeeId { get; set; }
public List<int> Skillssetpoints { get; set; }
[JsonIgnore]
public string Name { get; set; }
public Nullable<System.DateTime> Date { get; set; }
}
4）最终代码在
之下var data = new List<Employee>();
data.Add(new Employee { EmployeeId = 1, Skillssetpoints = 10, Date = Convert.ToDateTime("4/5/2016 16:12:12") });
data.Add(new Employee { EmployeeId = 2, Skillssetpoints = 12, Date = Convert.ToDateTime("3/5/2016 17:12:12") });
data.Add(new Employee { EmployeeId = 3, Skillssetpoints = 4, Date = Convert.ToDateTime("8/5/2016 8:12:12") });
data.Add(new Employee { EmployeeId = 4, Skillssetpoints = 20, Date = Convert.ToDateTime("1/5/2016 2:12:12") });
var highestDate = data.OrderByDescending(e => e.Date).First().Date;
var skillssetpointsList = data.Select(e => e.Skillssetpoints).ToList();
EmployeeModel employeeModel = new EmployeeModel()
{
Date = highestDate,
Skillssetpoints = skillssetpointsList
};
string jsonString = JsonConvert.SerializeObject(employeeModel);
现在，jsonString = {“Skillssetpoints”：[10,12,4,20]，“Date”：“2016-05-08T08：12：12”}
答案 6 :(得分：1)
我想你可以尝试一下 它适用于我。
public ActionResult GetCitiesWithBranches(int regionID)
{
var cities =
_context.Cities.Where(e => e.RegionCode == regionID)
.Select(e => new { ID = e.CityCode, Name = e.Name })
.ToList();
return Json(new { cities = cities });
}
在我使用的视图中：
var json = { regionID: data };
$.ajax({
type: "POST",
url: '@Url.Action("GetCitiesWithBranches", "Admin")',
data: json,
dataType: "json",
error: function (xhr, status, error) {
//alert("error routine");
},
success: function (res) {
if (res.cities) {
}
}
});
希望这会有所帮助..