我正在尝试更新表中的单个记录。代码如下:
public void UpdateStudent(int id, int studentClass)
{
SqlConnection myConnection = new SqlConnection();
myConnection.ConnectionString = @"Data Source=PALLAVI-PC\SQLEXPRESS;Initial Catalog=StudentDB;Integrated Security=True;MultipleActiveResultSets=True;";
SqlCommand sqlCmd = new SqlCommand();
sqlCmd.CommandType = CommandType.Text;
sqlCmd.CommandText = "update Tbl_Students set Class="+ studentClass+" where Roll_Number=" + id + ";";
sqlCmd.Connection = myConnection;
myConnection.Open();
int rowUpdated = sqlCmd.ExecuteNonQuery();
myConnection.Close();
}
如何将id和Class传递给fiddler中的URL?
答案 0 :(得分:1)
您可以使用WCF REST服务执行此操作。 实施以下menthod,你可以做你想要的东西而不是提琴手:
[WebInvoke(Method = "PUT", UriTemplate = "api/student/{id}/{studentclass}")]
[OperationContract]
void UpdateEmployee(int id, int studentclass);
请参阅link以获取有关RESTFUL WCF服务的更多详细信息。
答案 1 :(得分:0)
感谢您的帮助。我能够通过稍微改变我的代码来解决问题。 代码如下:
[HttpPut]
[ActionName("UpdateStudent")]
public void UpdateStudent(int id, [FromBody]Tbl_Students student)
{
SqlConnection myConnection = new SqlConnection();
myConnection.ConnectionString = @"Data Source=PALLAVI-PC\SQLEXPRESS;Initial Catalog=StudentDB;Integrated Security=True;MultipleActiveResultSets=True;";
SqlCommand sqlCmd = new SqlCommand();
sqlCmd.CommandType = CommandType.Text;
sqlCmd.CommandText = "update Tbl_Students set Class=" + student.Class + " where Roll_Number=" + id + ";";
sqlCmd.Connection = myConnection;
myConnection.Open();
int rowUpdated = sqlCmd.ExecuteNonQuery();
myConnection.Close();
}
在Fiddler中,我将id作为URL的一部分传递,将studentClass作为JSON传递给Request Body。我还添加了我之前忘记的Content-Type:application/json
。我添加了以下图片: