将Json字符串嵌套到DataTable

本文关键字：DataTable 嵌套 字符串 Json | 更新日期: 2023-09-27 17:57:58

我需要将以下Json字符串转换为DataTable。

{  
   "pnr":"1234567890",
   "train_num":"12311",
   "train_name":"HWH DLIKLK MAI",
   "doj":"23-12-2013",
   "from_station":
   {  
      "code":"DLI",
      "name":"Delhi"
   },
   "to_station":
   {  
      "code":"KLK",
      "name":"Kalka"
   }
}

在DataTable中，我需要显示

train_num
train_name
doj
from_station(name only)
to_station(name only)

到目前为止，我只有

public class Train
{
public string train_num { get; set; }
public string train_name { get; set; }
public string doj { get; set; }
public from_station from_station { get; set; }
public to_station to_station { get; set; }
}
public class from_station
{
public string code { get; set; }
public string name { get; set; }
}
public class to_station
{
public string code { get; set; }
public string name { get; set; }
}
public static DataTable ToDataTable(Train data)
{
    PropertyDescriptorCollection props =
    TypeDescriptor.GetProperties(typeof(Train));
    DataTable table = new DataTable();
    for (int i = 0; i < props.Count; i++)
    {
        PropertyDescriptor prop = props[i];
        table.Columns.Add(prop.Name, prop.PropertyType);
    }
    object[] values = new object[props.Count];
        for (int i = 0; i < values.Length; i++)
        {
            values[i] = props[i].GetValue(data);
        }
        table.Rows.Add(values);
    return table;
}
var data = JsonConvert.DeserializeObject<Train>(JsonString);
    dt = ToDataTable(data);
    ui_grdVw_EmployeeDetail1.DataSource = dt;
    ui_grdVw_EmployeeDetail1.DataBind();

我在数据表中只得到三列

train_num
train_name
doj

将Json字符串嵌套到DataTable

您需要调整DataTable转换方法，使其更通用。然后把你想要的形状的数据传给它。

public static DataTable ToDataTable<T>( IList<T> data)
{
    PropertyDescriptorCollection props =
        TypeDescriptor.GetProperties(typeof(T));
    DataTable table = new DataTable();
    for (int i = 0; i < props.Count; i++)
    {
        PropertyDescriptor prop = props[i];
        table.Columns.Add(prop.Name, prop.PropertyType);
    }
    object[] values = new object[props.Count];
    foreach (T item in data)
    {
        for (int i = 0; i < values.Length; i++)
        {
            values[i] = props[i].GetValue(item);
        }
        table.Rows.Add(values);
    }
    return table;
}

注意：以下方法可用于将任何List转换为DataTable。

用法：

var data = JsonConvert.DeserializeObject<Train>(JsonString);

var shapedData = Enumerable.Range(0, 1).Select(x =>
                    new
                    {
                        train_num = data.train_num,
                        train_name = data.train_name,
                        doj = data.doj,
                        from_station = data.from_station.name,
                        to_station = data.to_station.name
                    }).ToList();
DataTable dt = ToDataTable(shapedData);
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