我试图将客户端详细信息插入数据库,但得到了此异常。
连接必须有效且打开。在MySql.Data.MySqlClient.MySqlCommand.Throw(Exception ex)在MySql.Data.MySqlClient.MySqlCommand.CheckState() at MySql.Data.MySqlClient.MySqlCommand.ExecuteReader(CommandBehavior behavior)在MySql.Data.MySqlClient.MySqlCommand.ExecuteReader() at MySql.Data.MySqlClient.MySqlCommand.ExecuteNonQuery() at book_online.saveClient_Click(对象发送者,EventArgs e)
MySqlConnection con = new MySqlConnection();
con.ConnectionString = ConfigurationManager.ConnectionStrings("conio").ConnectionString();
MySqlCommand cmd = new MySqlCommand();
cmd.CommandText = "INSERT INTO clientsDetails (clientName, companyAddress, city, contactPersonName, contactNumber1) Values(@clientName, @companyAddress, @city, @contactPersonName, @contactNumber1)";
cmd.Parameters.AddWithValue("@clientName", companyName.Text);
cmd.Parameters.AddWithValue("@companyAddress", companyAddress.Text);
cmd.Parameters.AddWithValue("@city", companyCity.Text);
cmd.Parameters.AddWithValue("@contactPersonName", concernPersonName.Text);
cmd.Parameters.AddWithValue("@contactNumber1", countryCodeWithNumber);
con.Open();
cmd.ExecuteNonQuery();
con.Close();发布于 2016-09-05 05:16:39
您已经打开了连接并执行了命令,但是您忘记了将连接分配给命令,那么只有代码可以像您预期的那样正常工作。强烈建议您也使用。然后,代码将如下所示:
using (MySqlConnection con = new MySqlConnection())
{
con.ConnectionString = ConfigurationManager.ConnectionStrings("conio").ConnectionString();
string queryText = "INSERT INTO clientsDetails (clientName, companyAddress, city, contactPersonName, contactNumber1) Values(@clientName, @companyAddress, @city, @contactPersonName, @contactNumber1)";
using (MySqlCommand cmd = new MySqlCommand(queryText, con))
{
cmd.Parameters.AddWithValue("@clientName", companyName.Text);
cmd.Parameters.AddWithValue("@companyAddress", companyAddress.Text);
cmd.Parameters.AddWithValue("@city", companyCity.Text);
cmd.Parameters.AddWithValue("@contactPersonName", concernPersonName.Text);
cmd.Parameters.AddWithValue("@contactNumber1", countryCodeWithNumber);
con.Open();
cmd.ExecuteNonQuery();
}
}发布于 2016-09-05 05:55:44
命令缺少连接字符串.
MySqlConnection con = new MySqlConnection();
con.ConnectionString = ConfigurationManager.ConnectionStrings("conio").ConnectionString();
MySqlCommand cmd = new MySqlCommand();
cmd.Connection = con;
cmd.CommandText = "INSERT INTO clientsDetails (clientName, companyAddress, city, contactPersonName, contactNumber1) Values(@clientName, @companyAddress, @city, @contactPersonName, @contactNumber1)";
cmd.Parameters.AddWithValue("@clientName", companyName.Text);
cmd.Parameters.AddWithValue("@companyAddress", companyAddress.Text);
cmd.Parameters.AddWithValue("@city", companyCity.Text);
cmd.Parameters.AddWithValue("@contactPersonName", concernPersonName.Text);
cmd.Parameters.AddWithValue("@contactNumber1", countryCodeWithNumber);
con.Open();
cmd.ExecuteNonQuery();
con.Close();发布于 2016-09-05 05:12:49
您没有为命令指定连接对象:
cmd.Connection = con;另外,最好使用一个using块,当它退出连接时,它会自动关闭连接:
using(MySqlConnection con = new MySqlConnection())
{
con.ConnectionString =
ConfigurationManager.ConnectionStrings("conio").ConnectionString();
MySqlCommand cmd = new MySqlCommand();
cmd.CommandText = @"INSERT INTO clientsDetails
(clientName, companyAddress, city, contactPersonName, contactNumber1)
Values
(@clientName, @companyAddress, @city, @contactPersonName, @contactNumber1)";
cmd.Connection = con;
cmd.Parameters.AddWithValue("@clientName", companyName.Text);
cmd.Parameters.AddWithValue("@companyAddress", companyAddress.Text);
cmd.Parameters.AddWithValue("@city", companyCity.Text);
cmd.Parameters.AddWithValue("@contactPersonName", concernPersonName.Text);
cmd.Parameters.AddWithValue("@contactNumber1", countryCodeWithNumber);
con.Open();
cmd.ExecuteNonQuery();
}为了获取信息,using关键字允许编译为封闭的块生成一个try...finally,并在生成的最终块中对using中的对象调用Dispose (在您的示例中是con)。因此,在所有最终得到保证的情况下,都保证关闭连接)
https://stackoverflow.com/questions/39324258
复制相似问题