[英]How to execute 2 Sql statements with one button click
我設法使用wamp運行此查詢。
INSERT INTO guest (guestno,familyname)
VALUES(NULL,'Damn');
INSERT INTO reservation (reservationno, guestno)
VALUES(NUll,LAST_INSERT_ID())
但是,如果我分別執行這2條插入語句,則將具有外鍵約束。 我認為這兩個都需要同時執行。
我的問題是:
當用戶按下“添加預訂”時,我希望執行兩個MySQl查詢。
這是我的插入語句:
private void button7_Click(object sender, EventArgs e)
{
string connectionString =
"Server=localhost;" +
"Database=sad;" +
"User ID=root;" +
"Password=root;" +
"Pooling=false";
IDbConnection dbcon;
dbcon = new MySqlConnection(connectionString);
dbcon.Open();
IDbCommand dbcmd = dbcon.CreateCommand();
string sql = "<insert statement>";
dbcmd.CommandText = sql;
IDataReader reader = dbcmd.ExecuteReader();
reader.Read();
}
更新版本(無效)
string connectionString =
"Server=localhost;" +
"Database=sad;" +
"User ID=root;" +
"Password=root;" +
"Pooling=false";
Form3 f3 = new Form3();
IDbConnection dbcon;
dbcon = new MySqlConnection(connectionString);
dbcon.Open();
IDbCommand dbcmd = dbcon.CreateCommand();
string sql = "insert into guest (guestno, familyname) values (null, '" + textBox6.Text + "'); insert into reservation (reservationno, guestno) values (null, LAST_INSERT_ID())";
dbcmd.CommandText = sql;
IDataReader reader = dbcmd.ExecuteReader();
reader.Read();
MessageBox.Show("Added Guest Reservation Successfully");
f3.guestList();
f3.reservationList();
更新3號(仍然無效)
string connectionString =
"Server=localhost;" +
"Database=sad;" +
"User ID=root;" +
"Password=root;" +
"Pooling=false";
IDbConnection dbcon;
dbcon = new MySqlConnection(connectionString);
dbcon.Open();
IDbCommand dbcmd = dbcon.CreateCommand();
dbcmd = new MySqlCommand("CreateGuestAndReservation", dbcon);
dbcmd.CommandType = CommandType.StoredProcedure;
dbcmd.Parameters.AddWithValue("familyName", "foo");
dbcmd.ExecuteNonQuery();
enter code here
在給定的MySqlCommand上,您不能執行多個語句。
最好的選擇(可維護性,性能,可讀性)是:
ExecuteNonQuery()
調用存儲的proc。 DELIMITER //
CREATE PROCEDURE CreateGuestAndReservation
(
IN familyName VARCHAR(255)
)
BEGIN
insert into guest (guestno, familyname)
values (null, familyName);
insert into reservation (reservationno, guestno)
values (null, LAST_INSERT_ID());
END//
DELIMITER ;
從您的WinForms代碼中調用它,如下所示:
dbcon.Open();
cmd = new MySqlCommand("CreateGuestAndReservation", dbcon);
cmd.CommandType = CommandType.StoredProcedure;
//cmd.Parameters.AddWithValue("?familyName", "foo");
cmd.Parameters.Add("?familyName", MySqlDbType.VarChar,255).Value = "foo";
cmd.ExecuteNonQuery();
試試這個,它將起作用:
private void button56_Click(object sender, EventArgs e) {
con.Open();
SqlCommand cmd = new SqlCommand("insert into stholidays values('" + dateTimePicker12.Text + "','" + dateTimePicker20.Text + "','" + dateTimePicker13.Text + "','" + mbk + "','" + dateTimePicker14.Text + "','" + dateTimePicker15.Text + "','" + lt + "','" + dateTimePicker16.Text + "','" + dateTimePicker17.Text + "','" + ebk + "','" + dateTimePicker18.Text + "','" + dateTimePicker19.Text + "','" + textBox105.Text + "','" + textBox106.Text + "','" + textBox107.Text + "','" + dd + "','" + textBox104.Text + "')", con);
SqlCommand cmd1 = new SqlCommand("insert into holidays values('" + dd + "','" + ms + "','" + day + "','" + textBox104.Text + "')", con);
cmd.ExecuteNonQuery();
cmd1.ExecuteNonQuery();
con.Close();
}
下面的代碼應該可以工作,但是鑒於您正在尋求幫助,我懷疑您可能已經嘗試過了?
string sql = "INSERT INTO guest (guestno,familyname) VALUES(NULL,'Damn'); INSERT INTO reservation (reservationno, guestno) VALUES(NUll,LAST_INSERT_ID())";
如果需要參數,請嘗試以下操作:
string sql = "INSERT INTO guest (guestno,familyname) VALUES(NULL,?familyName); INSERT INTO reservation (reservationno, guestno) VALUES(NUll,LAST_INSERT_ID())";
...
dbcmd.Parameters.Add("@familyName", MySqlDbType.VarChar, 80).Value = _familyName;
編輯:您可能需要運行2插入命令。 看這里 。
我建議除了依靠自動id生成(例如mysql和sql server的自動增量)之外,還有一種獲取id的方法,這是非常有限的。 如果使用HILO ID生成器,則首先獲取ID,然后在單個事務中執行幾次插入操作就沒問題,因為您事先知道了父ID。
它不會解決您眼前的問題,但是它將在將來對您的應用程序產生巨大的幫助,特別是如果像數據這樣經常存儲父母子女的情況。
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