[英]object reference not set to an instance of an object asp.net,
我有两个问题:
“你调用的对象是空的。
而且我不知道如何解决它。 请有人看看吗?
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);
}
}
}
Object reference not set to an instance of an object
意味着您试图获取属性/调用null
。
这意味着- txtEmp_num
-您的变量之一为null: txtEmp_num
, txtEmp_fn
, txtEmp_ln
, txtEmp_phone
, txtEmp_email
,当您尝试在AddWithValue
方法中调用Text
属性时,它会引发异常。
也许您在FindControl("txt<smth>")
定义的值存在一些问题
尝试如下
System.Web.UI.WebControls.TextBox txtEmp_num=(System.Web.UI.WebControls.TextBoDataGridView1.Rows[0].Cells[0].FindControl("txtEmp_num");
System.Web.UI.WebControls.TextBox txtEmp_fn = (System.Web.UI.WebControls.TextBox)DataGridView1.Rows[0].Cells[1].FindControl("txtEmp_fn");
System.Web.UI.WebControls.TextBox txtEmp_ln = (System.Web.UI.WebControls.TextBox)DataGridView1.Rows[0].Cells[2].FindControl("txtEmp_ln");
System.Web.UI.WebControls.TextBox txtEmp_phone = (System.Web.UI.WebControls.TextBox)DataGridView1.Rows[0].Cells[3].FindControl("txtEmp_phone");
System.Web.UI.WebControls.TextBox txtEmp_email = (System.Web.UI.WebControls.TextBox)DataGridView1.Rows[0].Cells[4].FindControl("txtEmp_email");
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