Related:
- How to store image in ms access
- How to insert image in access database form - Best answers
- Access database images - Best answers
- Where does skype store images - How-To - Skype
- Windows 7 - A shortcut to access the Imaging Devices settings - How-To - Windows 7
- IMediaShare Lite - Access media content stored on your PC via your Android/iOS mobile - How-To - Mobile
- Twitter Adds Accessible Images to Apps - News
- How to access app store on itunes pc - How-To - iTunes
5 replies
Zohaib R
- Posts
- 2401
- Registration date
- Sunday September 23, 2012
- Status
- Moderator
- Last seen
- December 13, 2018
Hi mikeoe2003,
I have designed this small application for you that might suite your need. This application has an Open Button which will help you open any picture file to a PictureBox on the form using OpenFileDialog. You will see the path of the picture file in a disabled TextBox. When you click the update button the Picture's path is saved to an Access Database.
Follow the steps below to create a similar project for yourself:
1. Create a new Visual Basic .net project. Select Windows Forms Application from New Project Dialog Box. Name this application as mikeoe2003PictureApplication.
2. Create the following with below mentioned properties:
a. Form - (Name): mikeoe2003PictureApplication, Text: FormPictureApplication
b. PictureBox - (Name): PictureBox1, SizeMode: StretchImage
c. Button - (Name): ButtonUpdate, Text: &Update
d. Button - (Name): ButtonOpen, Text: &Open
e. TextBoxt - (Name): TextBoxPictureFilePath, Enabled: False
3. Double Click the Form, insert the following code right above Public Class mikeoe2003PictureApplication:
Imports System.Data.OleDb
Imports System.IO
Imports Microsoft.Win32
Double Click ButtonOpen and insert the following code:
Dim img As String
Dim myStream As Stream = Nothing
Dim openFileDialog1 As New OpenFileDialog()
openFileDialog1.InitialDirectory = "c:\"
openFileDialog1.Filter = Nothing
openFileDialog1.FilterIndex = 2
openFileDialog1.RestoreDirectory = True
openFileDialog1.FileName = ""
If openFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then
Try
myStream = openFileDialog1.OpenFile()
If (myStream IsNot Nothing) Then
TextBoxPictureFilePath.Text = ""
img = openFileDialog1.FileName
PictureBox1.Image = System.Drawing.Bitmap.FromFile(img)
TextBoxPictureFilePath.Text = openFileDialog1.FileName
End If
Catch Ex As Exception
MessageBox.Show("Cannot read file from disk. Original error: " & Ex.Message)
Finally
If (myStream IsNot Nothing) Then
myStream.Close()
End If
End Try
End If
4. Create a Microsoft Access Database in your convenient location and name it as Databasemikeoe2003PictureApplication.mdb
5. Create a table with the name Tablemikeoe2003PictureApplication and add following Columns to it:
Id - Datatype: Autonumber
PicturePath - DataType: Memo (as file paths can be considerably long at times)
6. Double Click the UpdateButton and insert the following code:
Try
Dim myConnection As OleDbConnection
Dim myCommand As OleDbCommand
Dim mySQLString As String
myConnection = New OleDbConnection("Provider=Microsoft.Jet.OLEDB.4.0;Data Source=Databasemikeoe2003PictureApplication.mdb;")
myConnection.Open()
mySQLString = "INSERT INTO Tablemikeoe2003PictureApplication (PicturePath) VALUES('" & Replace$(TextBoxPictureFilePath.Text, "'", "''") & "')"
myCommand = New OleDbCommand(mySQLString, myConnection)
myCommand.ExecuteNonQuery()
PictureBox1.Image = Nothing
TextBoxPictureFilePath.Text = ""
Catch ex As Exception
MessageBox.Show(ex.Message & " - " & ex.Source)
End Try
7. Run the application, it should work as desired.
Please revert for clarification.
I have designed this small application for you that might suite your need. This application has an Open Button which will help you open any picture file to a PictureBox on the form using OpenFileDialog. You will see the path of the picture file in a disabled TextBox. When you click the update button the Picture's path is saved to an Access Database.
Follow the steps below to create a similar project for yourself:
1. Create a new Visual Basic .net project. Select Windows Forms Application from New Project Dialog Box. Name this application as mikeoe2003PictureApplication.
2. Create the following with below mentioned properties:
a. Form - (Name): mikeoe2003PictureApplication, Text: FormPictureApplication
b. PictureBox - (Name): PictureBox1, SizeMode: StretchImage
c. Button - (Name): ButtonUpdate, Text: &Update
d. Button - (Name): ButtonOpen, Text: &Open
e. TextBoxt - (Name): TextBoxPictureFilePath, Enabled: False
3. Double Click the Form, insert the following code right above Public Class mikeoe2003PictureApplication:
Imports System.Data.OleDb
Imports System.IO
Imports Microsoft.Win32
Double Click ButtonOpen and insert the following code:
Dim img As String
Dim myStream As Stream = Nothing
Dim openFileDialog1 As New OpenFileDialog()
openFileDialog1.InitialDirectory = "c:\"
openFileDialog1.Filter = Nothing
openFileDialog1.FilterIndex = 2
openFileDialog1.RestoreDirectory = True
openFileDialog1.FileName = ""
If openFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then
Try
myStream = openFileDialog1.OpenFile()
If (myStream IsNot Nothing) Then
TextBoxPictureFilePath.Text = ""
img = openFileDialog1.FileName
PictureBox1.Image = System.Drawing.Bitmap.FromFile(img)
TextBoxPictureFilePath.Text = openFileDialog1.FileName
End If
Catch Ex As Exception
MessageBox.Show("Cannot read file from disk. Original error: " & Ex.Message)
Finally
If (myStream IsNot Nothing) Then
myStream.Close()
End If
End Try
End If
4. Create a Microsoft Access Database in your convenient location and name it as Databasemikeoe2003PictureApplication.mdb
5. Create a table with the name Tablemikeoe2003PictureApplication and add following Columns to it:
Id - Datatype: Autonumber
PicturePath - DataType: Memo (as file paths can be considerably long at times)
6. Double Click the UpdateButton and insert the following code:
Try
Dim myConnection As OleDbConnection
Dim myCommand As OleDbCommand
Dim mySQLString As String
myConnection = New OleDbConnection("Provider=Microsoft.Jet.OLEDB.4.0;Data Source=Databasemikeoe2003PictureApplication.mdb;")
myConnection.Open()
mySQLString = "INSERT INTO Tablemikeoe2003PictureApplication (PicturePath) VALUES('" & Replace$(TextBoxPictureFilePath.Text, "'", "''") & "')"
myCommand = New OleDbCommand(mySQLString, myConnection)
myCommand.ExecuteNonQuery()
PictureBox1.Image = Nothing
TextBoxPictureFilePath.Text = ""
Catch ex As Exception
MessageBox.Show(ex.Message & " - " & ex.Source)
End Try
7. Run the application, it should work as desired.
Please revert for clarification.
Report
cpsingh
thank you
mikeoe2003
- Posts
- 25
- Registration date
- Thursday November 1, 2012
- Status
- Member
- Last seen
- March 27, 2014
solved
