Question

answer

在结束阅读之前清除textBox是一个错误。

感谢@ muhammad-azim

结束答案

我首先在数据库中插入了2个，第二个在第一个中具有外键。执行第二个INSERT时出现错误：

INSERT语句与FOREIGN KEY约束冲突

这部分出错了

代码：

SqlConnection con1 = new SqlConnection(connectionString);

SqlCommand cmd1 = new SqlCommand(cmdString, con1);
con1.Open();

cmd1.ExecuteNonQuery(); // insert(987,10000,2019-07-27 12:00:00 AM)
con1.Close();

// Some Code

SqlConnection con2 = new SqlConnection(connectionString);

SqlCommand cmd2 = new SqlCommand(cmdString2, con2);
con2.Open()

cmd2.ExecuteNonQuery();//want insert(1,987,2000,2019-07-28 12:00:00 AM) but Got error
con2.Close();

两个查询均为INSERT

我认为Connection不理解数据库已更新并出现此错误。当我停止应用程序并单独运行另一个第二次查询时，它就可以工作了（以前没用过）

所有代码

        PersianCalendar pc = new PersianCalendar();
        string[] dateInfo = txtcreteDate.Text.Split('-');
        DateTime createDate = new DateTime(Convert.ToInt32(dateInfo[0]), Convert.ToInt32(dateInfo[1]), Convert.ToInt32(dateInfo[2]), pc);

        int companyid = 1;
        SqlConnection con = new SqlConnection(connectionString);
        SqlCommand cmd = new SqlCommand("INSERT INTO Factors (compId,createDate, code, duration, value," +
            " discount , payType , payTypeInfo , status)" +
            "VALUES " +
            " (@compId,@createDate, @code, @duration, @value, @discount , @payType , @payTypeInfo , 1) ", con);
        cmd.Parameters.Add("@compId", companyid);
        cmd.Parameters.Add("@createDate", createDate);
        cmd.Parameters.Add("@code", txtCode.Text);
        cmd.Parameters.Add("@duration", Convert.ToInt32(txtDuration.Text));
        cmd.Parameters.Add("@value", Convert.ToDecimal(txtValue.Text));
        cmd.Parameters.Add("@discount", Convert.ToDecimal(txtDiscount.Text));
        cmd.Parameters.Add("@payType", companyid);
        cmd.Parameters.Add("@payTypeInfo", companyid);

        con.Open();
        try
        {
            var insertedItem = cmd.ExecuteNonQuery();
        }
        catch(Exception ez)
        {
            MessageBox.Show("شماره فاکتور تکراری است");
            return;
        }
        ClearTextBoxes();
        if (comboBox1.SelectedItem.ToString() == "نقد" && insertedItem >0)
        {
            for (int counter1 = 0; counter1 < cashDataGridView.Rows.Count; counter1++)
            {
                string status = "پرداخت نشده";
                PersianCalendar pc1 = new PersianCalendar();
                string[] dateInfoPay = cashDataGridView.Rows[counter1].Cells["date"].Value.ToString().Split('-');
                DateTime PayDate = new DateTime(Convert.ToInt32(dateInfoPay[0]), Convert.ToInt32(dateInfoPay[1]), Convert.ToInt32(dateInfoPay[2]), pc1);
                //System.Data.SqlClient.SqlConnection con1 = new System.Data.SqlClient.SqlConnection(connectionString);
                string cmdString = "INSERT INTO Payout (value,date,status,description,fid)";
                cmdString += " VALUES ";
                cmdString += " ( '" + Convert.ToDecimal(cashDataGridView.Rows[counter1].Cells["value"].Value.ToString()).ToString() + "' ,";
                cmdString += " '" + PayDate.ToString() + "' , ";
                cmdString += " '" + status.ToString() + "' , ";
                cmdString += " '" + status.ToString() + "' , ";
                cmdString += " '" + txtCode.Text + "' )";
                System.Data.SqlClient.SqlCommand cmd1 = new System.Data.SqlClient.SqlCommand(cmdString, con);
                MessageBox.Show(cmd1.CommandText.ToString());
                cmd1.ExecuteNonQuery(); // i got error here

            }
            con.Close();

Answer 1

您在这里遇到问题：

SqlConnection con1 = new SqlConnection(connectionString);
SqlCommand cmd1 = new SqlCommand(cmdString, con1);
con1.Open();
cmd1.ExecuteNonQuery();
con1.Close();

ExecuteNonQuery()返回受影响的行数，如果没有受影响的行，则返回值为-1并回滚到该操作。

为了确保插入成功并且两个操作都成功完成，您应该始终使用vlidations：

try
{
  SqlConnection con1 = new SqlConnection(connectionString);
  SqlCommand cmd1 = new SqlCommand(cmdString, con1);
  con1.Open();
  var rowsAffected = cmd1.ExecuteNonQuery();
  con1.Close();
  //Some Code
  if(rowsAffected > 0)
  {
    SqlConnection con2 = new SqlConnection(connectionString);
    SqlCommand cmd2 = new SqlCommand(cmdString2, con2);
    con2.Open()
    cmd2.ExecuteNonQuery();
    con2.Close();
  }
}
catch(Exception ex)
{
  //handle error in insert
}

另请参阅SqlCommand.ExecuteNonQuery() returns -1 when doing Insert / Update / Delete

Answer 2

您不需要使用两个连接对象。

 using(SqlConnection con1 = new SqlConnection(connectionString))
 {
  using(SqlCommand cmd1 = new SqlCommand(cmdString, con1))
   {
    con1.Open();

    //ExecuteNonQuery Method will return 0 or 1.
    bool result = Convert.ToBool(cmd1.ExecuteNonQuery());

   If insert is succsessful, you can try to execute the second query.
   if(result)
    {  
      If doesn't work, you should check your "Some Code" to be sure if you assign the the right parameters. 
      //Some Code
      cmd1.CommandText = cmdString2;
      cmd1.ExecuteNonQuery();
    }
  }
 }

执行查询后刷新数据库并执行另一个查询

2 个答案: