创建一个弹出窗口并刷新父窗口 [英] Make a popup window and Refresh parent window
问题描述
如果点击按钮,我想打开一个弹出窗口。然后从弹出窗口我插入一些项目。
如果我点击保存按钮然后数据插入到sql server
然后单击一个刷新按钮,数据将显示在dropdownList中父页面。
请帮助我。
I want to open a popup window if click a button. Then from the popup window I insert some items.
If I click save button then data insert into sql server
Then Click a Refresh button and data will show in a dropdownList of the parent page.
Please help me.
推荐答案
亲爱的Mohi,
您需要指定什么样的你想要制作的弹出窗口。 (您可以使用图像的链接来指定它)
要在sql server中插入数据,您可以使用以下代码:
Dear Mo
You need to specify what kind of popup window you want to make. (You can use link of a image to specify it)
And for inserting data in sql server you can use the following code:
private void button2_Click(object sender, EventArgs e)
{
SqlConnection conn = new SqlConnection("Data Source:.;Initial Catalog=database_name;Integrated Security=true");
SqlCommand cmd = new SqlCommand("Insert into table_name values(@a,@b)", conn);
conn.Open();
cmd.Parameters.AddWithValue("@a", textBox1.Text.ToString());
cmd.Parameters.AddWithValue("@b", textBox2.Text.toString());
cmd.ExecuteNonQuery();
MessageBox.Show("You data has been saved successfully");
conn.Close();
}
编码:
首先,在button1的click事件中,我使用SqlConnection创建了与sql sever的连接。我已经通过数据源(服务器名称),初始目录(数据库名称),集成安全性为真(因为我的sql服务器在Windows安全上运行)之后我已经从System.Data.SqlClient创建了sql db usig SqlCommand类的命令。
之后我打开了连接状态并在命令中添加了指定的参数。
然后我执行了查询然后显示了消息,然后关闭了连接。
希望它有所帮助!快乐的编码!!
Akky
Explanation of Coding:
First of all, in the click event of button1 I have created a connection to sql sever using SqlConnection. I have passed Data Source(Server Name),Initial Catalog(Database Name), Integrated Security as true(as my sql server was running on windows security) after that I have created command for sql db usig SqlCommand class from System.Data.SqlClient .
After that I have opened the connection state and added the specified parameters in the command.
Then I have executed the query and then displayed the message and then closed the connection.
Hope it helps! Happy Coding!!
Akky
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