Saving image path to access database and retrieving images
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mikeoe2003
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Updated on Jan 21, 2019 at 10:13 AM
cpsingh - Jan 20, 2019 at 09:30 PM
cpsingh - Jan 20, 2019 at 09:30 PM
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5 responses
Zohaib R
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Nov 13, 2012 at 12:42 AM
Nov 13, 2012 at 12:42 AM
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.
mikeoe2003
Posts
24
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Thursday November 1, 2012
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Member
Last seen
March 27, 2014
2
Jan 11, 2013 at 04:17 AM
Jan 11, 2013 at 04:17 AM
solved
tonywerstein
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1
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Thursday September 5, 2013
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Member
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September 5, 2013
Sep 5, 2013 at 04:57 PM
Sep 5, 2013 at 04:57 PM
You may use access databases by the medium of access mdb repair tool
http://www.access.repairtoolboxx.com/
http://www.access.repairtoolboxx.com/
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meme002
Posts
1
Registration date
Sunday November 30, 2014
Status
Member
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November 30, 2014
Nov 30, 2014 at 04:11 AM
Nov 30, 2014 at 04:11 AM
its useful but can you help me how to view image from ms access database I have try your codes and its working...but how about viewing it?
Jan 20, 2019 at 09:30 PM