小贝子编程

如何将JSON对象反序列化为DataTable



我有JSON数据如下。我想知道如何反序列化JSON对象到数据表。请检查下面的JSON字符串，并建议如何做到这一点。

{
"response":
{
"uri":"/api/junaid@khindia.com/Gloves",
"action":"EXPORT",
"result":
{
"column_order":["AttendanceDate","DivCode","Division","DeptCode","Department","Section","CostType","Gender","Religion","EmpLevel","DegLevel","CurrentSection","EmployeeCode","EmployeeName","Strength","Presents","Absents","LayOffs","Weaving","Wallet","OutSide","PresentIn","PresentOut","AbsentIn","AbsentOut","LayoffIn","LayoffOut","Attendance","Working","LastUpDatedDate","Stage","AreaName","AreaDirection","AreaSerial","ImageURL","ShiftCode","ShiftTime","CurrentDepartment","CurrentDivision","C%_StaffOrWorker","C%_MajorDivision","M_Division","Worked_Hours","Worked_OtHours","Lateby","Earlyby","OverTime1","OverTime2","C%_OtherDivision","C%_ShiftFrom"],
"rows":[
["05/05/2022 00:00:00","AM","Leather Goods","MC","Bags","Production","Direct","Female","Hindus","Worker","Upto X","Arind","12519","SANGEETHA. M","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","","Arappakam","A         ","-2","http://103.76.188.138:85/EmployeePhotos/12519.JPG","30","07:00 - 15:30","Bags","Bags","Worker","Bags","Leather Goods","6.28","2.5","103","0","150","0","Bags","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AM","Leather Goods","MB","Wallet","Production","Direct","Male","Muslims","Worker","Degree","Arind","12548","MD NAIMUDDEEN.R","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","","Kaspa","V         ","17","http://103.76.188.138:85/EmployeePhotos/12548.JPG","30","07:00 - 15:30","Wallet","Wallet","Worker","Wallet","Leather Goods","7.08","1.5","55","0","90","0","Wallet","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AM","Leather Goods","MB","Wallet","Production","Direct","Female","Hindus","Worker","Upto XII","Arind","12567","SWETHA.P","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","","Melakuppam","A         ","-4","http://103.76.188.138:85/EmployeePhotos/12567.JPG","30","07:00 - 15:30","Wallet","Wallet","Worker","Wallet","Leather Goods","7.13","1.5","52","0","90","0","Wallet","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AM","Leather Goods","MB","Wallet","Production","Direct","Male","Muslims","Worker","Upto V","Arind","12605","IRFAN AHMED.C","1","0","1","0","0","0","0","0","0","1","0","0","0","A","Gloves","05/05/2022 20:15:00","","Pernambet","V         ","0","http://103.76.188.138:85/EmployeePhotos/12605.JPG","30","07:00 - 15:30","Wallet","Wallet","Worker","Wallet","Leather Goods","0","0","0","0","0","0","Wallet","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AM","Leather Goods","MD","Belts","Edge Inking","Direct","Male","Hindus","Worker","Upto X","Arind","12633","SIVAKUMAR.S","1","0","1","0","0","0","0","0","0","1","0","0","0","A","Gloves","05/05/2022 20:15:00","","Perumugai","V         ","0","http://103.76.188.138:85/EmployeePhotos/12633.JPG","30","07:00 - 15:30","Belts","Belts","Worker","Belts","Leather Goods","0","0","0","0","0","0","Belts","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AF","Assembly","FA","Pre-Sewing","Applique Stitch","Direct","Male","Hindus","Worker","Upto X","Arind","12940","KULOTHUNGAN. G","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","PreSewing","Sathuvachari","V         ","6","http://103.76.188.138:85/EmployeePhotos/12940.JPG","32","08:00 - 16:30","Pre-Sewing","Gloves","Worker","Gloves","Gloves","8","0","0","0","0","0","Gloves","2022-05-05 08:00:00"],
["05/05/2022 00:00:00","AM","Leather Goods","MD","Belts","Preparation","Direct","Female","Hindus","Worker","Upto X","Arind","11830","ASWINI","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","","Pillaiyar Kuppam","A         ","-1","http://103.76.188.138:85/EmployeePhotos/11830.JPG","30","07:00 - 15:30","Belts","Belts","Worker","Belts","Leather Goods","8","0","0","0","0","0","Belts","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AP","Training","PA","Training","Training","Direct","Female","Hindus","Worker","Upto X","Arind","12957","NALINI.R","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","","Melvisharam","A         ","-9","http://103.76.188.138:85/EmployeePhotos/12957.JPG","30","07:00 - 15:30","Training","Training","Worker","Training","Training","7.1","1.5","54","0","90","0","Training","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AM","Leather Goods","MA","Weaving","Weaving","Direct","Female","Hindus","Worker","Upto XII","Arind","12818","HEMALATHA.R","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","","Melakuppam","A         ","-4","http://103.76.188.138:85/EmployeePhotos/12818.JPG","30","07:00 - 15:30","Weaving","Weaving","Worker","Weaving","Leather Goods","6.17","2.5","110","0","150","0","Weaving","2022-05-05 07:00:00"],
["05/05/2022 00:00:00","AM","Leather Goods","MD","Belts","Sample","Indirect","Female","Hindus","Worker","Upto X","Arind","11869","KALPANA.K","1","1","0","0","0","0","0","1","0","0","0","0","0","P","Gloves","05/05/2022 20:15:00","","Arappakam","A         ","-2","http://103.76.188.138:85/EmployeePhotos/11869.JPG","30","07:00 - 15:30","Belts","Belts","Worker","Belts","Leather Goods","8","0","0","0","0","0","Belts","2022-05-05 07:00:00"]
]
}
}
}
  1. 反序列化json
  2. 从response["response"]["result"]["column_order"]中获取colNames数组
  3. 从response["response"]["result"]["rows"]获取rows数组
  4. 用colNames配置列创建Datatable
  5. 基于colNames序列添加rowsDataRow
using System.Data;
using Newtonsoft.Json;
using Newtonsoft.Json.Linq;
JObject response = JObject.Parse(json);
JArray colNames = response["response"]["result"]["column_order"] as JArray;
JArray rows = response["response"]["result"]["rows"] as JArray;

DataTable dt = new DataTable();

foreach (var col in colNames)
{
dt.Columns.Add(col.ToString(), typeof(string)); 
}

foreach (var row in rows)
{
var newRow = dt.NewRow();

for (int i = 0; i < colNames.Count; i++)
{
var col = colNames[i].ToString();
newRow[col] = row[i];
}

dt.Rows.Add(newRow);
}

示例。net提琴演示

如果你也用类似的结构处理数据，你也可以将convert to Datatable逻辑应用为扩展/helper方法:

JObject response = JObject.Parse(json);
List<string> colNames = (response["response"]["result"]["column_order"] as JArray).ToObject<List<string>>();
List<List<string>> rows = (response["response"]["result"]["rows"] as JArray).ToObject<List<List<string>>>();

DataTable dt = DataTableExtensions.ToDataTable(colNames, rows);

public static class DataTableExtensions
{
public static DataTable ToDataTable(List<string> colNames, List<List<string>> rows)
{
DataTable dt = new DataTable();

foreach (var col in colNames)
{
dt.Columns.Add(col, typeof(string));    
}

foreach (var row in rows)
{
var newRow = dt.NewRow();

for (int i = 0; i < colNames.Count; i++)
{
var col = colNames[i];
newRow[col] = row[i];
}

dt.Rows.Add(newRow);
}

return dt;
}
}

示例。net提琴演示(作为扩展/助手)

使用无数转换实用程序之一构建对象模型，甚至可以将JSON作为内建到Visual Studio中的类粘贴。(你有没有试过这么做?)

public class Response
{
public string uri { get; set; }
public string action { get; set; }
public Result result { get; set; }
}
public class Result
{
public List<string> column_order { get; set; }
public List<List<string>> rows { get; set; }
}
public class Root
{
public Response response { get; set; }
}

然后解析数据。假设JSON是一个字符串，这是微不足道的。

var response = System.Text.Json.JsonSeralizer.Deserialize<Root>(jsonString).Response;

然后迭代response.Result的值并填充数据表对象。既然你认为不适合提供这方面的细节，你就不能提供这部分答案的细节。

顺便说一下，你的JSON是无效的，它缺少一个后大括号。

你的对象是:

public class Response
{
public string uri { get; set; }
public string action { get; set; }
public Result result { get; set; }
}
public class Result
{
public List<string> column_order { get; set; }
public List<List<string>> rows { get; set; }
}
public class Root
{
public Response response { get; set; }
}

然后使用Newtonsoft.Json解析如下:

Root root = JsonConvert.DeserializeObject<Root>(jsonString);

创建DataTable:

DataTable dt = new DataTable(); 
dt.Clear();
foreach(var column in root.response.result.column_order) {
dt.Columns.Add(column);
}
foreach(var jsonRow in root.response.result.rows) {
for(int i =0; i < jsonRow.Length; i++) {
DataRow row = dt.NewRow();
row[i] = jsonRow[i];
dt.Rows.Add(row);
}
}

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