C# ADO.NET 两个sql对同一个数据库操作 连接开关上有有关问题
求教:C# ADO.NET 两个sql对同一个数据库操作 连接开关上有问题
SqlConnection conn = new SqlConnection(Connection.GetConnection());
public PsdChange()
{
InitializeComponent();
}
private void button2_Click(object sender, EventArgs e)
{
this.Close();
}
void Change_key()
{
if (txtB2.Text == txtB3.Text)
{
conn.Open();
string sql1 = "update Password set psd='" + txtB2.Text + "'";
SqlCommand cmd1 = new SqlCommand(sql1, conn);
int count = cmd1.ExecuteNonQuery();
if (count == 0)
{ MessageBox.Show("Key错误!"); }
else
{
MessageBox.Show("修改成功!", "Key修改提示", MessageBoxButtons.OK, MessageBoxIcon.Information);
}
}
else
{
MessageBox.Show("Two keys are different!");
}
}
private void button1_Click(object sender, EventArgs e)
{
conn.Open();
string sql = "select * from Password where psd='"+txtB1.Text+"'";
SqlCommand cmd = new SqlCommand(sql, conn);
SqlDataReader reader = cmd.ExecuteReader();
try
{
if (reader.Read() == true)
{
Change_key();
}
else
{
MessageBox.Show("The first key is error!");
}
}
catch (Exception ex)
{
MessageBox.Show(ex.ToString());
}
finally
{
reader.Close();
conn.Close();
}
}
public PsdChange()
{
InitializeComponent();
}
private void button2_Click(object sender, EventArgs e)
{
this.Close();
}
void Change_key()
{
if (txtB2.Text == txtB3.Text)
{
conn.Open();
string sql1 = "update Password set psd='" + txtB2.Text + "'";
SqlCommand cmd1 = new SqlCommand(sql1, conn);
int count = cmd1.ExecuteNonQuery();
if (count == 0)
{ MessageBox.Show("Key错误!"); }
else
{
MessageBox.Show("修改成功!", "Key修改提示", MessageBoxButtons.OK, MessageBoxIcon.Information);
}
}
else
{
MessageBox.Show("Two keys are different!");
}
}
private void button1_Click(object sender, EventArgs e)
{
conn.Open();
string sql = "select * from Password where psd='"+txtB1.Text+"'";
SqlCommand cmd = new SqlCommand(sql, conn);
SqlDataReader reader = cmd.ExecuteReader();
try
{
if (reader.Read() == true)
{
Change_key();
}
else
{
MessageBox.Show("The first key is error!");
}
}
catch (Exception ex)
{
MessageBox.Show(ex.ToString());
}
finally
{
reader.Close();
conn.Close();
}
}