每當我按下按鈕時，我都會嘗試在運行時創建一個 DataGridView。
我還想從我的資料庫中的表中顯示資料。
當我按下應該創建 DataGridView 并顯示資料的 Button 時，它什么也不做。它不會創建 DataGridView。
這是我的代碼（該表稱為“供應”）：
public partial class managerpage : Form
{
int idWorkersDB = 0;
int idAccountsDB = 1;
int idSuppliersDB = 2;
int idOtherDB = 3;
public static DataGridView workersView = new DataGridView();
public static DataGridView suppliesView = new DataGridView();
public static DataGridView accountsView = new DataGridView();
int clickedID;
private void picBoxSuppliers_MouseUp(object sender, MouseEventArgs e)
{
suppliesView.Location = new Point(59, 51);
suppliesView.Size = new Size(749, 399);
clickedID = idSuppliersDB;
dataDBPanel.Visible = true;
dataDBPanel.Dock = DockStyle.Fill;
titleDataLbl.Text = "Suppliers Data";
titleDataLbl.Left = (this.Width - titleDataLbl.Width) / 2;
OleDbConnection con = new OleDbConnection(@"Provider=Microsoft.Jet.OLEDB.4.0;Data Source=G:\project\FUCKINGVISUALSTUDIOISSOSHIT\loginData.mdb");
OleDbDataAdapter cmd = new OleDbDataAdapter("SELECT * FROM supplies", con);
con.Open();
DataTable tbl = new DataTable();
cmd.Fill(tbl);
con.Close();
suppliesView.DataSource = tbl;
suppliesView.Show();
}
}
uj5u.com熱心網友回復：
您隱藏了大部分代碼，但我懷疑您從未將suppliesView（datagridview？）添加到表單中。這是一個運行良好的示例代碼：
void Main()
{
DataTable t = new DataTable();
using (var con = new SqlConnection(@"server=.;Database=Northwind;Trusted_Connection=yes"))
using (var cmd = new SqlCommand(@"select * from Customers", con))
{
con.Open();
t.Load(cmd.ExecuteReader());
con.Close();
}
var f = new Form();
var dgv = new DataGridView { Dock = DockStyle.Fill, DataSource=t};
f.Controls.Add(dgv);
f.Show();
}
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