如何使用按钮单击事件从数据库 table 中的文本框中获取内容?
how to get a content from textboxes in a database table with a Button click event?
我What i wanna do
所以我尝试了这段代码
但它无法与数据库建立连接。我的连接字符串是正确的
SqlConnection con = new SqlConnection("server=localhost;uid=root;database=menucreator");
con.Open();
SqlCommand cmd = new SqlCommand();
cmd.CommandText = "insert into [kunde]Vorname()values(@nm)";
cmd.Parameters.AddWithValue("@nm", Vorname.Text);
cmd.Connection = con;
SqlCommand cmdd = new SqlCommand();
cmdd.CommandText = "insert into [kunde]Nachname()values(@nmm)";
cmdd.Parameters.AddWithValue("@nmm", Nachname.Text);
cmdd.Connection = con;
int a = cmd.ExecuteNonQuery();
if (a == 1)
{
MessageBox.Show("Dateien bitte");
}
遵循以下代码。它将两列插入 table (tableName)。在 SQL 查询中维护正确的 space,如以下示例代码所示。此外,最好的做法是将代码保存在 try-catch 块中,以捕获数据库操作期间发生的任何错误。
try
{
SqlConnection con = new SqlConnection("server=localhost;uid=root;database=menucreator");
con.Open();
SqlCommand cmd = new SqlCommand();
cmd.CommandText = "insert into tableName(column1,column2) values(@nm,@nmm)";
cmd.Parameters.AddWithValue("@nm", Vorname.Text);
cmd.Parameters.AddWithValue("@nmm", Nachname.Text);
cmd.Connection = con;
int a = cmd.ExecuteNonQuery();
if (a == 1)
{
MessageBox.Show("Dateien bitte");
}
}
catch (Exception ex)
{
MessageBox.Show("Error:"+ex.ToString());
}
我What i wanna do
所以我尝试了这段代码 但它无法与数据库建立连接。我的连接字符串是正确的
SqlConnection con = new SqlConnection("server=localhost;uid=root;database=menucreator");
con.Open();
SqlCommand cmd = new SqlCommand();
cmd.CommandText = "insert into [kunde]Vorname()values(@nm)";
cmd.Parameters.AddWithValue("@nm", Vorname.Text);
cmd.Connection = con;
SqlCommand cmdd = new SqlCommand();
cmdd.CommandText = "insert into [kunde]Nachname()values(@nmm)";
cmdd.Parameters.AddWithValue("@nmm", Nachname.Text);
cmdd.Connection = con;
int a = cmd.ExecuteNonQuery();
if (a == 1)
{
MessageBox.Show("Dateien bitte");
}
遵循以下代码。它将两列插入 table (tableName)。在 SQL 查询中维护正确的 space,如以下示例代码所示。此外,最好的做法是将代码保存在 try-catch 块中,以捕获数据库操作期间发生的任何错误。
try
{
SqlConnection con = new SqlConnection("server=localhost;uid=root;database=menucreator");
con.Open();
SqlCommand cmd = new SqlCommand();
cmd.CommandText = "insert into tableName(column1,column2) values(@nm,@nmm)";
cmd.Parameters.AddWithValue("@nm", Vorname.Text);
cmd.Parameters.AddWithValue("@nmm", Nachname.Text);
cmd.Connection = con;
int a = cmd.ExecuteNonQuery();
if (a == 1)
{
MessageBox.Show("Dateien bitte");
}
}
catch (Exception ex)
{
MessageBox.Show("Error:"+ex.ToString());
}