【发布时间】:2021-11-26 06:32:17
【问题描述】:
我有一个 DropDownList,它由名为 tblVisa 的 SQL Server 表填充。我的问题是从 SQL Server 表中填充的值没有被保存。除了我的 DropDownLists 之外,其他所有内容都会被保存。我试过使用.SelectedValue 和.Text,但还是不行。
这是我的代码
protected void PopulateVisaType()
{
List<ListItem> result = new List<ListItem> { new ListItem("", "") };
SqlCommand cmd = new SqlCommand() { Connection = sqlConn, CommandText = "SELECT VisaType FROM tblVisa ORDER BY VisaType ASC" };
if (sqlConn.State == ConnectionState.Closed)
{
sqlConn.Open();
}
SqlDataReader read = cmd.ExecuteReader();
while (read.Read())
{
result.Add(new ListItem(read["VisaType"].ToString(), read["VisaType"].ToString()));
}
read.Close();
sqlConn.Close();
cmd.Dispose();
DDLVisa.DataSource = result;
DDLVisa.DataValueField = "value";
DDLVisa.DataTextField = "text";
DDLVisa.DataBind();
}
这是我将信息保存到数据库中的代码:
protected void LbSaveProfile_Click(object sender, EventArgs e)
{
SqlCommand cmd = new SqlCommand() { Connection = sqlConn, CommandText = "spSaveNewProviderInformation", CommandType = CommandType.StoredProcedure };
if (sqlConn.State == ConnectionState.Closed)
{
sqlConn.Open();
}
cmd.Parameters.AddWithValue("@EmployeeNumber", TbEmployeeNumber.Text.Trim());
cmd.Parameters.AddWithValue("@SSN", TbSSN.Text.Trim());
cmd.Parameters.AddWithValue("@ContractType", DDLContractType.SelectedItem.Value);
cmd.Parameters.AddWithValue("@Firstname", TbFirstname.Text.Trim());
cmd.Parameters.AddWithValue("@Lastname", TbLastname.Text.Trim());
cmd.Parameters.AddWithValue("@MiddleInitial", TbMiddleInitial.Text.Trim());
cmd.Parameters.AddWithValue("@ContractRenewalDate", TbContractRenewalDate.Text.Trim());
cmd.Parameters.AddWithValue("@Position", DDLPosition.Text.Trim());
cmd.Parameters.AddWithValue("@Specialty", DDLSpecialty.Text.Trim());
cmd.Parameters.AddWithValue("@PrimaryDepartment", DDLPrimaryDepartment.Text.Trim());
cmd.Parameters.AddWithValue("@SecondaryDepartment", DDLSecondaryDepartment.Text.Trim());
cmd.Parameters.AddWithValue("@Gender", DDLGender.Text.Trim());
cmd.Parameters.AddWithValue("@Birthdate", TbBirthdate.Text.Trim());
cmd.Parameters.AddWithValue("@EmailAddress", TbEmailAddress.Text.Trim());
cmd.Parameters.AddWithValue("@PhoneNumber", TbPhoneNumber.Text.Trim());
cmd.Parameters.AddWithValue("@Address", TbAddress.Text.Trim());
cmd.Parameters.AddWithValue("@PassportNumber", TbPassportNumber.Text.Trim());
cmd.Parameters.AddWithValue("@Citizenship", DDLCitizenship.Text.Trim());
cmd.Parameters.AddWithValue("@Visa", DDLVisa.Text.Trim());
cmd.Parameters.AddWithValue("@Status", 1);
cmd.ExecuteNonQuery();
sqlConn.Close();
Alert("Provider Information saved!");
ClearControls();
}
【问题讨论】:
-
您应该查看Can we stop using AddWithValue() already? 并停止使用
.AddWithValue()- 它可能会导致意想不到和令人惊讶的结果... -
你调试过你的代码吗?
DDLVisa.Text和其他 DDL 文本有什么价值? -
试试
(ListItem)DDLVisa.SelectedValue。旁注:您的连接、命令和阅读器需要using块。不不缓存连接 -
@Chetan 是的,我添加了一个断点来帮助我跟踪代码。对于我所有的 DDL 文本,它们都是空字符串。
标签: c# asp.net sql-server