【发布时间】:2016-08-10 09:34:45
【问题描述】:
我有两个问题:
- 我怎样才能以其他方式使用文本框,以免引发任何错误?
- 我收到一个错误:
"对象引用未设置为对象的实例。
我不知道如何解决它。 请问有人可以看看吗?
protected void DataGridView1_RowCommand(object sender, GridViewCommandEventArgs e)
{
if (e.CommandName.Equals("InsertFooter"))
{
try
{
System.Web.UI.WebControls.TextBox txtEmp_num = (System.Web.UI.WebControls.TextBox)DataGridView1.Controls[0].Controls[0].FindControl("txtEmp_num");
System.Web.UI.WebControls.TextBox txtEmp_fn = (System.Web.UI.WebControls.TextBox)DataGridView1.Controls[0].Controls[0].FindControl("txtEmp_fn");
System.Web.UI.WebControls.TextBox txtEmp_ln = (System.Web.UI.WebControls.TextBox)DataGridView1.Controls[0].Controls[0].FindControl("txtEmp_ln");
System.Web.UI.WebControls.TextBox txtEmp_phone = (System.Web.UI.WebControls.TextBox)DataGridView1.Controls[0].Controls[0].FindControl("txtEmp_phone");
System.Web.UI.WebControls.TextBox txtEmp_email = (System.Web.UI.WebControls.TextBox)DataGridView1.Controls[0].Controls[0].FindControl("txtEmp_email");
MySqlConnection conn = new MySqlConnection(connection);
MySqlCommand cmd = new MySqlCommand();
cmd.Connection = conn;
cmd.CommandText = "insert into employees(Emp_number,First_name,Last_name,Phone_num,Email) values(@1,@2,@3,@4,@5)";
cmd.Parameters.AddWithValue("@1", txtEmp_num.Text);
cmd.Parameters.AddWithValue("@2", txtEmp_fn.Text);
cmd.Parameters.AddWithValue("@3", txtEmp_ln.Text);
cmd.Parameters.AddWithValue("@4", txtEmp_phone.Text);
cmd.Parameters.AddWithValue("@5", txtEmp_email.Text);
conn.Open();
cmd.ExecuteNonQuery();
conn.Close();
Bind();
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
}
}
【问题讨论】:
-
尝试
debug代码,看看哪个控件是null..
标签: c# asp.net gridview textbox