本文介绍了想要根据所选的下拉列表值显示数据库中的数据。的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

大家好,



我想根据选定的下拉值显示数据库中的数据。



我的代码:

.aspx.cs

  protected   void  ddlDisplayName_SelectedIndexChanged( object  sender,EventArgs e)
{
AdminActivityBL ObjAdminActivity = new AdminActivityBL();
ObjAdminActivity.GetDisplayNames_Details();
string SelectedValue = ddlDisplayName.SelectedItem.Value;

}

Bussines图层:

  public  DataSet GetDisplayNamesDetails()
{
AdminActivityDB objAdminActivityDB = new AdminActivityDB();
return objAdminActivityDB.GetDisplayNamesDetails();
}



数据访问层:

  public  DataSet GetDisplayNamesDetails()
{
try
{
Database objDB = new SqlDatabase(ConfigurationHelper.ConnectionString);
使用(DbCommand objCMD = objDB.GetStoredProcCommand(Constants.USP_GET_DISPLAYNAMES_DETAILS))
{
DataSet retDS = objDB.ExecuteDataSet(objCMD) ;
return retDS;
}
}
catch (例外情况)
{
// EventLog objLog = new EventLog();
// objLog.LogError(ex);

throw ex;
}

}



这是对的吗?



: - )

解决方案

Hi everyone,

I would like to display the data from database as per the selected value of drop down.

My code:
.aspx.cs

protected void ddlDisplayName_SelectedIndexChanged(object sender, EventArgs e)
        {
            AdminActivityBL ObjAdminActivity = new AdminActivityBL();
            ObjAdminActivity.GetDisplayNames_Details();
            string SelectedValue = ddlDisplayName.SelectedItem.Value;

        }

Bussines layer:

public DataSet GetDisplayNamesDetails()
        {
            AdminActivityDB objAdminActivityDB = new AdminActivityDB();
            return objAdminActivityDB.GetDisplayNamesDetails();
        }


Data access layer:

public DataSet GetDisplayNamesDetails()
        {
            try
            {
                Database objDB = new SqlDatabase(ConfigurationHelper.ConnectionString);
                using (DbCommand objCMD = objDB.GetStoredProcCommand(Constants.USP_GET_DISPLAYNAMES_DETAILS))
                {
                    DataSet retDS = objDB.ExecuteDataSet(objCMD);
                    return retDS;
                }
            }
            catch (Exception ex)
            {
                //EventLog objLog = new EventLog();
                //objLog.LogError(ex);

                throw ex;
            }

        }


Is this something right?

:-)

解决方案


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08-13 15:11