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Code (C#)
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog open = new OpenFileDialog();
open.InitialDirectory = "C:\\";
open.Filter = "Image File (*.jpg)|*.jpg|All Files(*.*)|*.*";
open.FilterIndex = 1;
if (open.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
if (open.CheckPathExists)
{
OleDbConnection con = new OleDbConnection(@"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\Users\PEERAKIT\source\repos\WindowsFormsApp1\WindowsFormsApp1\imagepath.accdb; Persist Security Info = False; ");
string CorrectFilename = System.IO.Path.GetFileName(open.FileName);
OleDbCommand cmd = new OleDbCommand("INSERT INTO xxx (ID,path) values ((Select isnull(Max(Id),0) + 1 from xxx),'\\image\\" + CorrectFilename + "')", con);
cmd.ExecuteNonQuery();
con.Close();
string path = Application.StartupPath.Substring(0, (Application.StartupPath.Length - 10));
System.IO.File.Move(open.FileName, path + "\\image\\" + CorrectFilename);
MessageBox.Show("Successfully Uploaded");
}
}
}

Error น่าจะเปิดจากการเขียน insert ไม่รู้ว่าผิดตรงส่วนไหน

Type ของ ID = number path = short text
ขอบคุณมากๆครับ
Tag : .NET, Ms Access, Win (Windows App), C#, VS 2017 (.NET 4.x), Windows
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Date :
2018-04-29 18:14:01 |
By :
snowman0020 |
View :
863 |
Reply :
2 |
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