This small project will help you to understand how can you handle image processing in your VB.net Window Application. There are many times when we want to save images in our project software for instance in school software Saving student image. So you can achieved this through following approach
- Saving Image into database
- Saving image into a folder
First approach is not so much popular because image get more size than normal data So your database saved thousands of byte for an image which lead decline in record retrieval. if you want to follow this approach click on Following Link
Save Image into MS Access database
This post deal with second approach because its good practice to save information into database and image in a folder . you can save image name into database or rename uploaded image as your table ID (PK) field value. I am saving image name into database which is easy to implement but renaming image is more suitable because its decrease database storage.
Screen Shot
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| Image Add Screen |
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| Show Image |
Image Add Form Code :
Imports System.Data.OleDb
Public Class frmAdd
Private Sub btnSave_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnSave.Click
If txtName.Text = "" Or txtcity.Text = "" Or OpenFileDialog1.FileName = "OpenFileDialog1" Then
MsgBox("Please provide all value")
If txtcity.Text = "" Then
txtcity.Focus()
ElseIf txtName.Text = "" Then
txtName.Focus()
ElseIf OpenFileDialog1.FileName = "OpenFileDialog1" Then
Button1.Focus()
End If
Else
Dim con As New OleDbConnection("Provider=Microsoft.ACE.OLEDB.12.0;Data Source=girfa.accdb")
Dim com As New OleDbCommand
Try
com.Connection = con
con.Open()
com.CommandText = "insert into stu (sname,city,pic) values('" & txtName.Text & "','" & txtcity.Text & "','" & System.IO.Path.GetFileName(OpenFileDialog1.FileName) & "')"
com.ExecuteNonQuery()
FileCopy(OpenFileDialog1.FileName, Application.StartupPath & "\image\" & System.IO.Path.GetFileName(OpenFileDialog1.FileName))
MsgBox("Record Saved")
txtName.Text = ""
txtName.Focus()
txtcity.Text = ""
Catch ex As Exception
MsgBox(ex.Message)
Finally
con.Close()
End Try
End If
End Sub
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
OpenFileDialog1.ShowDialog()
End Sub
End Class
Show Image Form Code :
Imports System.Data.OleDb
Public Class frmShow
Private Sub frmShow_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
'
Dim con As New OleDbConnection("Provider=Microsoft.ACE.OLEDB.12.0;Data Source=girfa.accdb")
Dim com As New OleDbCommand("select * from stu", con)
Try
com.CommandType = CommandType.Text
Dim da As New OleDbDataAdapter(com)
Dim ds As New DataSet()
da.Fill(ds)
DataGridView1.DataSource = ds.Tables(0)
Catch ex As Exception
MsgBox(ex.Message)
End Try
End Sub
Private Sub DataGridView1_CellContentClick(ByVal sender As System.Object, ByVal e As System.Windows.Forms.DataGridViewCellEventArgs)
End Sub
Private Sub DataGridView1_RowHeaderMouseClick(ByVal sender As System.Object, ByVal e As System.Windows.Forms.DataGridViewCellMouseEventArgs)Handles DataGridView1.RowHeaderMouseClick
Dim dr As DataGridViewSelectedRowCollection
dr = DataGridView1.SelectedRows()
'lblid.Text = dr.Item(0).Cells(1).Value.ToString()
lblroll.Text = dr.Item(0).Cells(0).Value.ToString
lblName.Text = dr.Item(0).Cells(1).Value.ToString
lblCity.Text = dr.Item(0).Cells(2).Value.ToString
PictureBox1.Image = Image.FromFile(Application.StartupPath & "/image/" & dr.Item(0).Cells(3).Value.ToString)
End Sub
End Class
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