95
  1. 我想从数据库中获取记录到DataTable.
  2. 然后将其DataTable转换为 JSON 对象。
  3. 将 JSON 对象返回给我的 JavaScript 函数。

我通过调用使用此代码：

string result = JsonConvert.SerializeObject(DatatableToDictionary(queryResult, "Title"), Newtonsoft.Json.Formatting.Indented);

要将 DataTable 转换为 JSON，它可以正常工作并返回以下内容：

{
    "1": {
    "viewCount": 703,
    "clickCount": 98
    },
    "2": {
    "viewCount": 509,
    "clickCount": 85
    },
    "3": {
    "viewCount": 578,
    "clickCount": 86
    },
    "4": {
    "viewCount": 737,
    "clickCount": 108
    },
    "5": {
    "viewCount": 769,
    "clickCount": 130
    }
}

但我希望它返回以下内容：

{"records":[
{
"Title": 1,
"viewCount": 703,
"clickCount": 98
},
{
"Title": 2,
"viewCount": 509,
"clickCount": 85
},
{
"Title": 3,
"viewCount": 578,
"clickCount": 86
},
{
"Title": 4,
"viewCount": 737,
"clickCount": 108
},
{
"Title": 5,
"viewCount": 769,
"clickCount": 130
}
]}

我怎样才能做到这一点？

4

18 回答 18

205

此代码片段来自 Convert Datatable to JSON String in C#, VB.NET可能会对您有所帮助。它使用System.Web.Script.Serialization.JavaScriptSerializer将内容序列化为 JSON 格式：

public string ConvertDataTabletoString()
{
    DataTable dt = new DataTable();
    using (SqlConnection con = new SqlConnection("Data Source=SureshDasari;Initial Catalog=master;Integrated Security=true"))
    {
        using (SqlCommand cmd = new SqlCommand("select title=City,lat=latitude,lng=longitude,description from LocationDetails", con))
        {
            con.Open();
            SqlDataAdapter da = new SqlDataAdapter(cmd);
            da.Fill(dt);
            System.Web.Script.Serialization.JavaScriptSerializer serializer = new System.Web.Script.Serialization.JavaScriptSerializer();
            List<Dictionary<string, object>> rows = new List<Dictionary<string, object>>();
            Dictionary<string, object> row;
            foreach (DataRow dr in dt.Rows)
            {
                row = new Dictionary<string, object>();
                foreach (DataColumn col in dt.Columns)
                {
                    row.Add(col.ColumnName, dr[col]);
                }
                rows.Add(row);
            }
            return serializer.Serialize(rows);
        }
    }
}
于 2013-07-01T05:44:24.683 回答
83

我们可以通过两种简单的方式完成任务，一种是使用 Json.NET dll，另一种是使用 StringBuilder 类。

使用 Newtonsoft Json.NET

string JSONresult;
JSONresult = JsonConvert.SerializeObject(dt);  
Response.Write(JSONresult);

参考链接：Newtonsoft：在 ASP.Net C# 中将 DataTable 转换为 JSON 对象

使用 StringBuilder

public string DataTableToJsonObj(DataTable dt)
{
    DataSet ds = new DataSet();
    ds.Merge(dt);
    StringBuilder JsonString = new StringBuilder();
    if (ds != null && ds.Tables[0].Rows.Count > 0)
    {
        JsonString.Append("[");
        for (int i = 0; i < ds.Tables[0].Rows.Count; i++)
        {
            JsonString.Append("{");
            for (int j = 0; j < ds.Tables[0].Columns.Count; j++)
            {
                if (j < ds.Tables[0].Columns.Count - 1)
                {
                    JsonString.Append("\"" + ds.Tables[0].Columns[j].ColumnName.ToString() + "\":" + "\"" + ds.Tables[0].Rows[i][j].ToString() + "\",");
                }
                else if (j == ds.Tables[0].Columns.Count - 1)
                {
                    JsonString.Append("\"" + ds.Tables[0].Columns[j].ColumnName.ToString() + "\":" + "\"" + ds.Tables[0].Rows[i][j].ToString() + "\"");
                }
            }
            if (i == ds.Tables[0].Rows.Count - 1)
            {
                JsonString.Append("}");
            }
            else
            {
                JsonString.Append("},");
            }
        }
        JsonString.Append("]");
        return JsonString.ToString();
    }
    else
    {
        return null;
    }
}
于 2015-03-02T07:22:32.620 回答
29

这与接受的答案有类似的方法，但使用 LINQ 将数据表转换为单行代码中的列表。

//convert datatable to list using LINQ. Input datatable is "dt", returning list of "name:value" tuples
var lst = dt.AsEnumerable()
    .Select(r => r.Table.Columns.Cast<DataColumn>()
            .Select(c => new KeyValuePair<string, object>(c.ColumnName, r[c.Ordinal])
           ).ToDictionary(z=>z.Key,z=>z.Value)
    ).ToList();
//now serialize it
var serializer = new System.Web.Script.Serialization.JavaScriptSerializer();
return serializer.Serialize(lst);

这是枚举数据表的一种非常有用的方法，通常需要大量编码！以下是一些变化：

//convert to list with array of values for each row
var list1 = dt.AsEnumerable().Select(r => r.ItemArray.ToList()).ToList();

//convert to list of first column values only
var list2 = dt.AsEnumerable().Select(r => r.ItemArray[0]).ToList();

// parse a datatable with conditions and get CSV string
string MalesOver21 = string.Join(",",
    dt.AsEnumerable()
      .Where(r => r["GENDER"].ToString()=="M" && r.Field<int>("AGE")>21)
      .Select(r => r.Field<string>("FULLNAME"))
 );

这与原始问题无关，但为了完整起见，我会提到，如果您只想从现有数据表中过滤掉行，请参阅此答案

于 2015-06-17T18:45:29.777 回答
19

不使用 javascript 序列化程序的另一种方法：

    public static string DataTableToJSON(DataTable Dt)
            {
                string[] StrDc = new string[Dt.Columns.Count];

                string HeadStr = string.Empty;
                for (int i = 0; i < Dt.Columns.Count; i++)
                {

                    StrDc[i] = Dt.Columns[i].Caption;
                    HeadStr += "\"" + StrDc[i] + "\":\"" + StrDc[i] + i.ToString() + "¾" + "\",";

                }

                HeadStr = HeadStr.Substring(0, HeadStr.Length - 1);

                StringBuilder Sb = new StringBuilder();

                Sb.Append("[");

                for (int i = 0; i < Dt.Rows.Count; i++)
                {

                    string TempStr = HeadStr;

                    for (int j = 0; j < Dt.Columns.Count; j++)
                    {

                        TempStr = TempStr.Replace(Dt.Columns[j] + j.ToString() + "¾", Dt.Rows[i][j].ToString().Trim());
                    }
                    //Sb.AppendFormat("{{{0}}},",TempStr);

                    Sb.Append("{"+TempStr + "},");
                }

                Sb = new StringBuilder(Sb.ToString().Substring(0, Sb.ToString().Length - 1));

                if(Sb.ToString().Length>0)
                Sb.Append("]");

                return StripControlChars(Sb.ToString());

            }
//To strip control characters:

//A character that does not represent a printable character but //serves to initiate a particular action.

            public static string StripControlChars(string s)
            {
                return Regex.Replace(s, @"[^\x20-\x7F]", "");
            }
于 2014-01-11T13:43:19.157 回答
10

这些天很简单..

string json = JsonConvert.SerializeObject(YourDataTable, Formatting.Indented);

现在将您的 Json 转换为 DataTable：

YourDataTable = (DataTable)JsonConvert.DeserializeObject(json, (typeof(DataTable)));

也适用于数据集..

于 2020-04-25T00:38:06.260 回答
7

您可以使用 Alireza Maddah 指定的相同方式，如果您想使用两个数据表到一个 json 数组中，方法如下：

public string ConvertDataTabletoString()
{
DataTable dt = new DataTable();
DataTable dt1 = new DataTable();
using (SqlConnection con = new SqlConnection("Data Source=SureshDasari;Initial Catalog=master;Integrated Security=true"))
{
    using (SqlCommand cmd = new SqlCommand("select title=City,lat=latitude,lng=longitude,description from LocationDetails", con))
    {
        con.Open();
        SqlDataAdapter da = new SqlDataAdapter(cmd);
        da.Fill(dt);
        System.Web.Script.Serialization.JavaScriptSerializer serializer = new System.Web.Script.Serialization.JavaScriptSerializer();
        List<Dictionary<string, object>> rows = new List<Dictionary<string, object>>();
        Dictionary<string, object> row;
        foreach (DataRow dr in dt.Rows)
        {
            row = new Dictionary<string, object>();
            foreach (DataColumn col in dt.Columns)
            {
                row.Add(col.ColumnName, dr[col]);
            }
            rows.Add(row);
        }
        SqlCommand cmd1 = new SqlCommand("_another_query_", con);
                SqlDataAdapter da1 = new SqlDataAdapter(cmd1);
                da1.Fill(dt1);
                System.Web.Script.Serialization.JavaScriptSerializer serializer1 = new System.Web.Script.Serialization.JavaScriptSerializer();
                Dictionary<string, object> row1;
                foreach (DataRow dr in dt1.Rows) //use the old variable rows only
                {
                    row1 = new Dictionary<string, object>();
                    foreach (DataColumn col in dt1.Columns)
                    {
                        row1.Add(col.ColumnName, dr[col]);
                    }
                    rows.Add(row1); // Finally You can add into old json array in this way
                }
        return serializer.Serialize(rows);
    }
}
}

相同的方式可以用于任意数量的数据表。

于 2014-03-13T05:53:50.457 回答
5

使用 C#.net 将数据表转换为 JSON

 public static object DataTableToJSON(DataTable table)
    {
        var list = new List<Dictionary<string, object>>();

        foreach (DataRow row in table.Rows)
        {
            var dict = new Dictionary<string, object>();

            foreach (DataColumn col in table.Columns)
            {
                dict[col.ColumnName] = (Convert.ToString(row[col]));
            }
            list.Add(dict);
        }
        JavaScriptSerializer serializer = new JavaScriptSerializer();

        return serializer.Serialize(list);
    }
于 2015-04-02T07:49:26.603 回答
3

试试这个自定义功能。 

    public static string DataTableToJsonObj(DataTable dt)
    {
        DataSet ds = new DataSet();
        ds.Merge(dt);
        StringBuilder jsonString = new StringBuilder();

        if (ds.Tables[0].Rows.Count > 0)
        {
            jsonString.Append("[");
            for (int rows = 0; rows < ds.Tables[0].Rows.Count; rows++)
            {
                jsonString.Append("{");
                for (int cols = 0; cols < ds.Tables[0].Columns.Count; cols++)
                {
                    jsonString.Append(@"""" + ds.Tables[0].Columns[cols].ColumnName + @""":");

                    /* 
                    //IF NOT LAST PROPERTY

                    if (cols < ds.Tables[0].Columns.Count - 1)
                    {
                        GenerateJsonProperty(ds, rows, cols, jsonString);
                    }

                    //IF LAST PROPERTY

                    else if (cols == ds.Tables[0].Columns.Count - 1)
                    {
                        GenerateJsonProperty(ds, rows, cols, jsonString, true);
                    }
                    */

                    var b = (cols < ds.Tables[0].Columns.Count - 1)
                        ? GenerateJsonProperty(ds, rows, cols, jsonString)
                        : (cols != ds.Tables[0].Columns.Count - 1)
                          || GenerateJsonProperty(ds, rows, cols, jsonString, true);
                }
                jsonString.Append(rows == ds.Tables[0].Rows.Count - 1 ? "}" : "},");
            }
            jsonString.Append("]");
            return jsonString.ToString();
        }
        return null;
    }

    private static bool GenerateJsonProperty(DataSet ds, int rows, int cols, StringBuilder jsonString, bool isLast = false)
    {

        // IF LAST PROPERTY THEN REMOVE 'COMMA'  IF NOT LAST PROPERTY THEN ADD 'COMMA'
        string addComma = isLast ? "" : ",";

        if (ds.Tables[0].Rows[rows][cols] == DBNull.Value)
        {
            jsonString.Append(" null " + addComma);
        }
        else if (ds.Tables[0].Columns[cols].DataType == typeof(DateTime))
        {
            jsonString.Append(@"""" + (((DateTime)ds.Tables[0].Rows[rows][cols]).ToString("yyyy-MM-dd HH':'mm':'ss")) + @"""" + addComma);
        }
        else if (ds.Tables[0].Columns[cols].DataType == typeof(string))
        {
            jsonString.Append(@"""" + (ds.Tables[0].Rows[rows][cols]) + @"""" + addComma);
        }
        else if (ds.Tables[0].Columns[cols].DataType == typeof(bool))
        {
            jsonString.Append(Convert.ToBoolean(ds.Tables[0].Rows[rows][cols]) ? "true" : "fasle");
        }
        else
        {
            jsonString.Append(ds.Tables[0].Rows[rows][cols] + addComma);
        }

        return true;
    }
于 2016-05-11T12:18:47.120 回答
3

所有这些答案都非常适合移动数据！他们失败的地方是保留要移动的数据的列类型。当您想要执行诸如合并看起来相同的数据表之类的事情时，这会成为一个问题。JsonConvert会查看第一行数据来判断列数据类型，可能会猜错。

为了解决这个问题；

  • 在单独的响应对象中序列化DataTable和定义。DataColumn
  • 在读取表之前反序列化DataColumn响应中的定义。
  • 反序列化并合并DataTable忽略 Json 定义的模式。

这听起来很多，但它只有三行额外的代码。

// Get our Column definitions and serialize them using an anoymous function.
var columns = dt.Columns.Cast<DataColumn>().Select(c => new { DataPropertyName = c.ColumnName, DataPropertyType = c.DataType.ToString()});
resp.ObjSchema = JsonConvert.SerializeObject(columns);
resp.Obj = JsonConvert.SerializeObject(dt);

resp.ObjSchema变成；

[
  {
    "DataPropertyName": "RowId",
    "DataPropertyType ": "System.Int32"
  },
  {
    "DataPropertyName": "ItemName",
    "DataPropertyType ": "System.String"
  }
]

dt = JsonConvert.DeserializeObject<DataTable>(response)我们可以使用 LINQ自己定义列，而不是让 Json 定义列定义resp.ObjSchema。我们将使用MissingSchemaAction.Ignore忽略 Json 提供的模式。

// If your environment does not support dynamic you'll need to create a class for with DataPropertyName and DataPropertyType.
JsonConvert.DeserializeObject<List<dynamic>>(response.ObjSchema).ForEach(prop =>
{
    dt.Columns.Add(new DataColumn() { ColumnName = prop.DataPropertyName, DataType = Type.GetType(prop.DataPropertyType.ToString()) });
});
// Merge the results ignoring the JSON schema.
dt.Merge(JsonConvert.DeserializeObject<DataTable>(response.Obj), true, MissingSchemaAction.Ignore);
于 2020-08-27T17:39:13.937 回答
2

要在 Json 方法中访问转换数据表值，请按照以下步骤操作：

$.ajax({
        type: "POST",
        url: "/Services.asmx/YourMethodName",
        data: "{}",
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        success: function (data) {
            var parsed = $.parseJSON(data.d);
            $.each(parsed, function (i, jsondata) {
            $("#dividtodisplay").append("Title: " + jsondata.title + "<br/>" + "Latitude: " + jsondata.lat);
            });
        },
        error: function (XHR, errStatus, errorThrown) {
            var err = JSON.parse(XHR.responseText);
            errorMessage = err.Message;
            alert(errorMessage);
        }
    });
于 2013-12-06T05:26:14.977 回答
1

我有简单的函数将数据表转换为 json 字符串。

我使用 Newtonsoft 生成字符串。我不使用 Newtonsoft 来完全序列化 Datatable。请注意这一点。

也许这会很有用。

 private string DataTableToJson(DataTable dt) {
  if (dt == null) {
   return "[]";
  };
  if (dt.Rows.Count < 1) {
   return "[]";
  };

  JArray array = new JArray();
  foreach(DataRow dr in dt.Rows) {
   JObject item = new JObject();
   foreach(DataColumn col in dt.Columns) {
    item.Add(col.ColumnName, dr[col.ColumnName]?.ToString());
   }
   array.Add(item);
  }

  return array.ToString(Newtonsoft.Json.Formatting.Indented);
 }
于 2017-09-07T12:06:08.633 回答
1

使用Cinchoo ETL - 一个开源库，您可以通过几行代码轻松地将 DataTable 导出为 JSON

StringBuilder sb = new StringBuilder();
string connectionstring = @"Data Source=(localdb)\MSSQLLocalDB;Initial Catalog=Northwind;Integrated Security=True";
using (var conn = new SqlConnection(connectionstring))
{
    conn.Open();
    var comm = new SqlCommand("SELECT * FROM Customers", conn);
    SqlDataAdapter adap = new SqlDataAdapter(comm);

    DataTable dt = new DataTable("Customer");
    adap.Fill(dt);

    using (var parser = new ChoJSONWriter(sb))
        parser.Write(dt);
}

Console.WriteLine(sb.ToString());

输出：

{
  "Customer": [
    {
      "CustomerID": "ALFKI",
      "CompanyName": "Alfreds Futterkiste",
      "ContactName": "Maria Anders",
      "ContactTitle": "Sales Representative",
      "Address": "Obere Str. 57",
      "City": "Berlin",
      "Region": null,
      "PostalCode": "12209",
      "Country": "Germany",
      "Phone": "030-0074321",
      "Fax": "030-0076545"
    },
    {
      "CustomerID": "ANATR",
      "CompanyName": "Ana Trujillo Emparedados y helados",
      "ContactName": "Ana Trujillo",
      "ContactTitle": "Owner",
      "Address": "Avda. de la Constitución 2222",
      "City": "México D.F.",
      "Region": null,
      "PostalCode": "05021",
      "Country": "Mexico",
      "Phone": "(5) 555-4729",
      "Fax": "(5) 555-3745"
    }
  ]
}
于 2018-06-29T20:17:20.930 回答
1

试试这个（扩展方法）：

public static string ToJson(this DataTable dt)
{
    List<Dictionary<string, object>> lst = new List<Dictionary<string, object>>();
    Dictionary<string, object> item;
    foreach (DataRow row in dt.Rows)
    {
            item = new Dictionary<string, object>();
                foreach (DataColumn col in dt.Columns)
                {
                    item.Add(col.ColumnName, (Convert.IsDBNull(row[col]) ? null : row[col]));       
        }
        lst.Add(item);
    }
        return Newtonsoft.Json.JsonConvert.SerializeObject(lst);
}

并使用：

DataTable dt = new DataTable();
.
.
.
var json = dt.ToJson();
于 2019-06-30T03:43:16.630 回答
1

根据clamchoda 的答案（保留数据类型），我为此创建了一个类，可以像下面这样使用：

转换DataTable为Json：

JsonDataTable j = JsonDataTable.FromDataTable(myDataTable);

转换Json回DataTable：

DataTable myDataTable = j.ToDataTable();

public class JsonDataTable
    {
        public string Schema { get; set; }
        public string Table { get; set; }

        public static JsonDataTable FromDataTable(DataTable dt)
        {
            JsonDataTable j = new JsonDataTable();
            var columns = dt.Columns.Cast<DataColumn>().Select(c => new { DataPropertyName = c.ColumnName, DataPropertyType = c.DataType.ToString() });
            j.Schema = JsonConvert.SerializeObject(columns);
            j.Table = JsonConvert.SerializeObject(dt);
            return j;
        }


        public DataTable ToDataTable()
        {
            DataTable dt = new DataTable();

            JsonConvert.DeserializeObject<List<dynamic>>(Schema).ForEach(prop =>
            {
                dt.Columns.Add(new DataColumn() { ColumnName = prop.DataPropertyName, DataType = Type.GetType(prop.DataPropertyType.ToString()) });
            });

            dt.Merge(JsonConvert.DeserializeObject<DataTable>(Table), true, MissingSchemaAction.Ignore);

            return dt;
        }

    }
于 2021-05-26T05:26:04.667 回答
0 
public static string ConvertIntoJson(DataTable dt)
{
    var jsonString = new StringBuilder();
    if (dt.Rows.Count > 0)
    {
        jsonString.Append("[");
        for (int i = 0; i < dt.Rows.Count; i++)
        {
            jsonString.Append("{");
            for (int j = 0; j < dt.Columns.Count; j++)
                jsonString.Append("\"" + dt.Columns[j].ColumnName + "\":\"" 
                    + dt.Rows[i][j].ToString().Replace('"','\"') + (j < dt.Columns.Count - 1 ? "\"," : "\""));

            jsonString.Append(i < dt.Rows.Count - 1 ? "}," : "}");
        }
        return jsonString.Append("]").ToString();
    }
    else
    {
        return "[]";
    }
}
public static string ConvertIntoJson(DataSet ds)
{
    var jsonString = new StringBuilder();
    jsonString.Append("{");
    for (int i = 0; i < ds.Tables.Count; i++)
    {
        jsonString.Append("\"" + ds.Tables[i].TableName + "\":");
        jsonString.Append(ConvertIntoJson(ds.Tables[i]));
        if (i < ds.Tables.Count - 1)
            jsonString.Append(",");
    }
    jsonString.Append("}");
    return jsonString.ToString();
}
于 2017-02-03T06:52:14.950 回答
0 
//Common DLL client, server
public class transferDataTable
{
    public class myError
    {
        public string Message { get; set; }
        public int Code { get; set; }
    }

    public myError Error { get; set; }
    public List<string> ColumnNames { get; set; }
    public List<string> DataTypes { get; set; }
    public List<Object> Data { get; set; }
    public int Count { get; set; }
}

public static class ExtensionMethod
{
    public static transferDataTable LoadData(this transferDataTable transfer, DataTable dt)
    {
        if (dt != null)
        {
            transfer.DataTypes = new List<string>();
            transfer.ColumnNames = new List<string>();                
            foreach (DataColumn c in dt.Columns)
            {
                transfer.ColumnNames.Add(c.ColumnName);
                transfer.DataTypes.Add(c.DataType.ToString());
            }

            transfer.Data = new List<object>();
            foreach (DataRow dr in dt.Rows)
            {
                foreach (DataColumn col in dt.Columns)
                {
                    transfer.Data.Add(dr[col] == DBNull.Value ? null : dr[col]);
                }
            }
            transfer.Count = dt.Rows.Count;
        }            
        return transfer;
    }        

    public static DataTable GetDataTable(this transferDataTable transfer, bool ConvertToLocalTime = true)
    {
        if (transfer.Error != null || transfer.ColumnNames == null || transfer.DataTypes == null || transfer.Data == null)
            return null;

        int columnsCount = transfer.ColumnNames.Count;
        DataTable dt = new DataTable();
        for (int i = 0; i < columnsCount; i++ )
        {
            Type colType = Type.GetType(transfer.DataTypes[i]);
            dt.Columns.Add(new DataColumn(transfer.ColumnNames[i], colType));
        }

        int index = 0;
        DataRow row = dt.NewRow();
        foreach (object o in transfer.Data)
        {
            if (ConvertToLocalTime && o != null && o.GetType() == typeof(DateTime))
            {
                DateTime dat = Convert.ToDateTime(o);
                row[index] = dat.ToLocalTime();
            }
            else
                row[index] = o == null ? DBNull.Value : o;

            index++;

            if (columnsCount == index)
            {
                index = 0;
                dt.Rows.Add(row);
                row = dt.NewRow();
            }
        }
        return dt;
    }
}

//Server
    [OperationContract]
    [WebInvoke(Method = "GET", ResponseFormat = WebMessageFormat.Json, BodyStyle = WebMessageBodyStyle.WrappedRequest, UriTemplate = "json/data")]
    transferDataTable _Data();

    public transferDataTable _Data()
    {
        try
        {
            using (SqlConnection con = new SqlConnection(ConfigurationManager.AppSettings["myConnString"]))
            {
                con.Open();
                DataSet ds = new DataSet();
                SqlDataAdapter myAdapter = new SqlDataAdapter("SELECT * FROM tbGalleries", con);
                myAdapter.Fill(ds, "table");
                DataTable dt = ds.Tables["table"];
                return new transferDataTable().LoadData(dt);
            }
        }
        catch(Exception ex)
        {
            return new transferDataTable() { Error = new transferDataTable.myError() { Message = ex.Message, Code = ex.HResult } };
        }
    }

//Client
        Response = Vossa.getAPI(serviceUrl + "json/data");
        transferDataTable transfer = new JavaScriptSerializer().Deserialize<transferDataTable>(Response);
        if (transfer.Error == null)
        {
            DataTable dt = transfer.GetDataTable();
            dbGrid.ItemsSource = dt.DefaultView;
        }
        else
            MessageBox.Show(transfer.Error.Message, "Error", MessageBoxButton.OK, MessageBoxImage.Error);
于 2017-04-03T07:32:55.720 回答
0

将数据传递给此方法，它将返回 json 字符串。

public DataTable GetTable()
        {
            string str = "Select * from GL_V";
            OracleCommand cmd = new OracleCommand(str, con);
            cmd.CommandType = CommandType.Text;
            DataTable Dt = OracleHelper.GetDataSet(con, cmd).Tables[0];

            return Dt;
        }

        public string DataTableToJSONWithJSONNet(DataTable table)
        {
            string JSONString = string.Empty;
            JSONString = JsonConvert.SerializeObject(table);
            return JSONString;
        }



public static DataSet GetDataSet(OracleConnection con, OracleCommand cmd)
        {
            // create the data set  
            DataSet ds = new DataSet();
            try
            {
                //checking current connection state is open
                if (con.State != ConnectionState.Open)
                    con.Open();

                // create a data adapter to use with the data set
                OracleDataAdapter da = new OracleDataAdapter(cmd);

                // fill the data set
                da.Fill(ds);
            }
            catch (Exception ex)
            {

                throw;
            }
            return ds;
        }
于 2018-04-12T13:05:28.817 回答
0

我正在使用这个函数来描述表。
填充数据表后使用

static public string DataTableToJSON(DataTable dataTable,bool readableformat=true)
        {
            string JSONString="[";
            string JSONRow;
            string colVal;
            foreach(DataRow dataRow in dataTable.Rows)
            {
                if(JSONString!="[") { JSONString += ","; }
                JSONRow = "";
                if (readableformat) { JSONRow += "\r\n"; }
                JSONRow += "{";

                foreach (DataColumn col in dataTable.Columns)
                {
                    colVal = dataRow[col].ToString();
                    colVal = colVal.Replace("\"", "\\\"");
                    colVal = colVal.Replace("'", "\\\'");
                    if(JSONRow!="{"&&JSONRow!="\r\n{") {

                        JSONRow += ",";

                    }
                    JSONRow += "\"" + col.ColumnName + "\":\"" + colVal + "\"";

                }
                JSONRow += "}";
                JSONString += JSONRow;
            }
            JSONString += "\r\n]";
            return JSONString;
        }

MySQL 查询：“DESCRIBE 表名；”；DataTableToJSON(dataTable) 示例输出：

[
{"Field":"id","Type":"int(5)","Null":"NO","Key":"PRI","Default":"","Extra":"auto_increment"},
{"Field":"ad","Type":"int(11) unsigned","Null":"NO","Key":"MUL","Default":"","Extra":""},
{"Field":"soyad","Type":"varchar(20)","Null":"YES","Key":"","Default":"","Extra":""},
{"Field":"ulke","Type":"varchar(20)","Null":"YES","Key":"","Default":"","Extra":""},
{"Field":"alan","Type":"varchar(20)","Null":"YES","Key":"","Default":"","Extra":""},
{"Field":"numara","Type":"varchar(20)","Null":"NO","Key":"","Default":"","Extra":""}
]

用 PHP 测试：

$X='[
{"Field":"id","Type":"int(5)","Null":"NO","Key":"PRI","Default":"","Extra":"auto_increment"},
{"Field":"ad","Type":"int(11) unsigned","Null":"NO","Key":"MUL","Default":"","Extra":""},
{"Field":"soyad","Type":"varchar(20)","Null":"YES","Key":"","Default":"","Extra":""},
{"Field":"ulke","Type":"varchar(20)","Null":"YES","Key":"","Default":"","Extra":""},
{"Field":"alan","Type":"varchar(20)","Null":"YES","Key":"","Default":"","Extra":""},
{"Field":"numara","Type":"varchar(20)","Null":"NO","Key":"","Default":"","Extra":""}
]';
$Y=json_decode($X,true);
echo $Y[0]["Field"];
var_dump($Y);
于 2020-01-12T15:21:08.063 回答