Astrazelle
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vb
Imports System.Data.Odbc
Imports System.IO
vb
Dim connectionString As String = "Driver={MySQL ODBC 8.0 Unicode Driver};Server=<server_name>;Database=<database_name>;Uid=<username>;Pwd=<password>;"
OdbcConnection
object to establish a connection:vb
Dim connection As New OdbcConnection(connectionString)
vb
Private Sub Form1_Load(sender As Object, e As EventArgs) Handles MyBase.Load
Dim command As New OdbcCommand("SELECT picture_name FROM your_table", connection)
connection.Open()
Dim reader As OdbcDataReader = command.ExecuteReader()
While reader.Read()
ComboBox1.Items.Add(reader("picture_name").ToString())
End While
reader.Close()
connection.Close()
End Sub
vb
Private Sub ComboBox1_SelectedIndexChanged(sender As Object, e As EventArgs) Handles ComboBox1.SelectedIndexChanged
If ComboBox1.SelectedIndex <> -1 Then
Dim selectedPictureName As String = ComboBox1.SelectedItem.ToString()
Dim command As New OdbcCommand("SELECT picture_data FROM your_table WHERE picture_name = ?", connection)
command.Parameters.AddWithValue("pictureName", selectedPictureName)
connection.Open()
Dim pictureData As Byte() = DirectCast(command.ExecuteScalar(), Byte())
connection.Close()
If pictureData IsNot Nothing Then
Dim ms As New MemoryStream(pictureData)
PictureBox1.Image = Image.FromStream(ms)
End If
End If
End Sub
vb
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
If ComboBox1.SelectedIndex <> -1 Then
Dim selectedPictureName As String = ComboBox1.SelectedItem.ToString()
Dim ms As New MemoryStream()
PictureBox1.Image.Save(ms, PictureBox1.Image.RawFormat)
Dim pictureData As Byte() = ms.ToArray()
Dim command As New OdbcCommand("UPDATE your_table SET picture_data = ? WHERE picture_name = ?", connection)
command.Parameters.AddWithValue("pictureData", pictureData)
command.Parameters.AddWithValue("pictureName", selectedPictureName)
connection.Open()
command.ExecuteNonQuery()
connection.Close()
End If
End Sub
vb
Private Sub Button2_Click(sender As Object, e As EventArgs) Handles Button2.Click
If OpenFileDialog1.ShowDialog() = DialogResult.OK Then
Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName)
Dim fileStream As New FileStream(OpenFileDialog1.FileName, FileMode.Open, FileAccess.Read)
Dim fileData As Byte() = New Byte(fileStream.Length - 1) {}
fileStream.Read(fileData, 0, CInt(fileStream.Length))
fileStream.Close()
Dim command As New OdbcCommand("INSERT INTO your_table (picture_name, picture_data) VALUES (?, ?)", connection)
command.Parameters.AddWithValue("pictureName", fileName)
command.Parameters.AddWithValue("pictureData", fileData)
connection.Open()
command.ExecuteNonQuery()
connection.Close()
ComboBox1.Items.Add(fileName)
End If
End Sub
<server_name>
, <database_name>
, <username>
, and <password>
with your actual database connection details. Also, adjust the table and column names accordingly.Change the update and upload code with try catchTo display a picture from a database to a picture box when selected in a combobox, and to insert, upload, and update pictures to a database using VB.NET and ODBC/ODBCCommand, you can follow these steps:
1. Set up your project:
- Create a new VB.NET project.
- Add a combobox, a picture box, and a button to your form.
- Import the necessary namespaces:
Code:vb Imports System.Data.Odbc Imports System.IO
2. Establish a database connection:
- Create a connection string to connect to your PHPMyAdmin database:
- Create anCode:vb Dim connectionString As String = "Driver={MySQL ODBC 8.0 Unicode Driver};Server=<server_name>;Database=<database_name>;Uid=<username>;Pwd=<password>;"
OdbcConnection
object to establish a connection:
Code:vb Dim connection As New OdbcConnection(connectionString)
3. Populate the combobox with data from the database:
- In the form load event, fetch the data from the database and populate the combobox:
Code:vb Private Sub Form1_Load(sender As Object, e As EventArgs) Handles MyBase.Load Dim command As New OdbcCommand("SELECT picture_name FROM your_table", connection) connection.Open() Dim reader As OdbcDataReader = command.ExecuteReader() While reader.Read() ComboBox1.Items.Add(reader("picture_name").ToString()) End While reader.Close() connection.Close() End Sub
4. Display the selected picture in the picture box:
- Handle the combobox selected index changed event:
Code:vb Private Sub ComboBox1_SelectedIndexChanged(sender As Object, e As EventArgs) Handles ComboBox1.SelectedIndexChanged If ComboBox1.SelectedIndex <> -1 Then Dim selectedPictureName As String = ComboBox1.SelectedItem.ToString() Dim command As New OdbcCommand("SELECT picture_data FROM your_table WHERE picture_name = ?", connection) command.Parameters.AddWithValue("pictureName", selectedPictureName) connection.Open() Dim pictureData As Byte() = DirectCast(command.ExecuteScalar(), Byte()) connection.Close() If pictureData IsNot Nothing Then Dim ms As New MemoryStream(pictureData) PictureBox1.Image = Image.FromStream(ms) End If End If End Sub
5. Insert or update picture data to the database:
- Handle the button click event to insert or update the picture:
Code:vb Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click If ComboBox1.SelectedIndex <> -1 Then Dim selectedPictureName As String = ComboBox1.SelectedItem.ToString() Dim ms As New MemoryStream() PictureBox1.Image.Save(ms, PictureBox1.Image.RawFormat) Dim pictureData As Byte() = ms.ToArray() Dim command As New OdbcCommand("UPDATE your_table SET picture_data = ? WHERE picture_name = ?", connection) command.Parameters.AddWithValue("pictureData", pictureData) command.Parameters.AddWithValue("pictureName", selectedPictureName) connection.Open() command.ExecuteNonQuery() connection.Close() End If End Sub
6. Insert a new picture to the database:
- Handle the button click event to insert a new picture:
Code:vb Private Sub Button2_Click(sender As Object, e As EventArgs) Handles Button2.Click If OpenFileDialog1.ShowDialog() = DialogResult.OK Then Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName) Dim fileStream As New FileStream(OpenFileDialog1.FileName, FileMode.Open, FileAccess.Read) Dim fileData As Byte() = New Byte(fileStream.Length - 1) {} fileStream.Read(fileData, 0, CInt(fileStream.Length)) fileStream.Close() Dim command As New OdbcCommand("INSERT INTO your_table (picture_name, picture_data) VALUES (?, ?)", connection) command.Parameters.AddWithValue("pictureName", fileName) command.Parameters.AddWithValue("pictureData", fileData) connection.Open() command.ExecuteNonQuery() connection.Close() ComboBox1.Items.Add(fileName) End If End Sub
Remember to replace<server_name>
,<database_name>
,<username>
, and<password>
with your actual database connection details. Also, adjust the table and column names accordingly.
These steps should allow you to display, insert, update, and upload pictures to a database using VB.NET and ODBC/ODBCCommand.
Certainly! Here's the updated code with try-catch blocks for the update and upload operations:Change the update and upload code with try catch
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Try
If ComboBox1.SelectedIndex <> -1 Then
Dim selectedPictureName As String = ComboBox1.SelectedItem.ToString()
Dim ms As New MemoryStream()
PictureBox1.Image.Save(ms, PictureBox1.Image.RawFormat)
Dim pictureData As Byte() = ms.ToArray()
Dim command As New OdbcCommand("UPDATE your_table SET picture_data = ? WHERE picture_name = ?", connection)
command.Parameters.AddWithValue("pictureData", pictureData)
command.Parameters.AddWithValue("pictureName", selectedPictureName)
connection.Open()
command.ExecuteNonQuery()
connection.Close()
End If
Catch ex As Exception
MessageBox.Show("Error updating picture: " & ex.Message)
Finally
connection.Close()
End Try
End Sub
Private Sub Button2_Click(sender As Object, e As EventArgs) Handles Button2.Click
Try
If OpenFileDialog1.ShowDialog() = DialogResult.OK Then
Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName)
Dim fileStream As New FileStream(OpenFileDialog1.FileName, FileMode.Open, FileAccess.Read)
Dim fileData As Byte() = New Byte(fileStream.Length - 1) {}
fileStream.Read(fileData, 0, CInt(fileStream.Length))
fileStream.Close()
Dim command As New OdbcCommand("INSERT INTO your_table (picture_name, picture_data) VALUES (?, ?)", connection)
command.Parameters.AddWithValue("pictureName", fileName)
command.Parameters.AddWithValue("pictureData", fileData)
connection.Open()
command.ExecuteNonQuery()
connection.Close()
ComboBox1.Items.Add(fileName)
End If
Catch ex As Exception
MessageBox.Show("Error uploading picture: " & ex.Message)
Finally
connection.Close()
End Try
End Sub
connection.Close()
statement is also included in the finally block to ensure that the database connection is closed even if an exception occurs.<server_name>
, <database_name>
, <username>
, and <password>
with your actual database connection details. Also, adjust the table and column names accordingly.include code that when the upload is succesful there will be a message that will pop or show saying that the upload is successfulCertainly! Here's the updated code with try-catch blocks for the update and upload operations:
Code:Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click Try If ComboBox1.SelectedIndex <> -1 Then Dim selectedPictureName As String = ComboBox1.SelectedItem.ToString() Dim ms As New MemoryStream() PictureBox1.Image.Save(ms, PictureBox1.Image.RawFormat) Dim pictureData As Byte() = ms.ToArray() Dim command As New OdbcCommand("UPDATE your_table SET picture_data = ? WHERE picture_name = ?", connection) command.Parameters.AddWithValue("pictureData", pictureData) command.Parameters.AddWithValue("pictureName", selectedPictureName) connection.Open() command.ExecuteNonQuery() connection.Close() End If Catch ex As Exception MessageBox.Show("Error updating picture: " & ex.Message) Finally connection.Close() End Try End Sub Private Sub Button2_Click(sender As Object, e As EventArgs) Handles Button2.Click Try If OpenFileDialog1.ShowDialog() = DialogResult.OK Then Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName) Dim fileStream As New FileStream(OpenFileDialog1.FileName, FileMode.Open, FileAccess.Read) Dim fileData As Byte() = New Byte(fileStream.Length - 1) {} fileStream.Read(fileData, 0, CInt(fileStream.Length)) fileStream.Close() Dim command As New OdbcCommand("INSERT INTO your_table (picture_name, picture_data) VALUES (?, ?)", connection) command.Parameters.AddWithValue("pictureName", fileName) command.Parameters.AddWithValue("pictureData", fileData) connection.Open() command.ExecuteNonQuery() connection.Close() ComboBox1.Items.Add(fileName) End If Catch ex As Exception MessageBox.Show("Error uploading picture: " & ex.Message) Finally connection.Close() End Try End Sub
In these updated code blocks, the try-catch blocks are added to catch any exceptions that may occur during the update and upload operations. If an exception is caught, a message box will display the error message. Theconnection.Close()
statement is also included in the finally block to ensure that the database connection is closed even if an exception occurs.
Remember to replace<server_name>
,<database_name>
,<username>
, and<password>
with your actual database connection details. Also, adjust the table and column names accordingly.
These updated code blocks should provide better error handling for the update and upload operations.
Certainly! Here's the updated code with a message box to display a success message when the upload is successful:include code that when the upload is succesful there will be a message that will pop or show saying that the upload is successful
Private Sub Button2_Click(sender As Object, e As EventArgs) Handles Button2.Click
Try
If OpenFileDialog1.ShowDialog() = DialogResult.OK Then
Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName)
Dim fileStream As New FileStream(OpenFileDialog1.FileName, FileMode.Open, FileAccess.Read)
Dim fileData As Byte() = New Byte(fileStream.Length - 1) {}
fileStream.Read(fileData, 0, CInt(fileStream.Length))
fileStream.Close()
Dim command As New OdbcCommand("INSERT INTO your_table (picture_name, picture_data) VALUES (?, ?)", connection)
command.Parameters.AddWithValue("pictureName", fileName)
command.Parameters.AddWithValue("pictureData", fileData)
connection.Open()
command.ExecuteNonQuery()
connection.Close()
ComboBox1.Items.Add(fileName)
MessageBox.Show("Upload successful!")
End If
Catch ex As Exception
MessageBox.Show("Error uploading picture: " & ex.Message)
Finally
connection.Close()
End Try
End Sub
MessageBox.Show()
method with the success message "Upload successful!".<server_name>
, <database_name>
, <username>
, and `<passwordwhen inserting a new picture I want to see the selected photo in a picture box before I click the upload button. and I can do this many times before I upload it and update the databaseCertainly! Here's the updated code with a message box to display a success message when the upload is successful:
Code:Private Sub Button2_Click(sender As Object, e As EventArgs) Handles Button2.Click Try If OpenFileDialog1.ShowDialog() = DialogResult.OK Then Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName) Dim fileStream As New FileStream(OpenFileDialog1.FileName, FileMode.Open, FileAccess.Read) Dim fileData As Byte() = New Byte(fileStream.Length - 1) {} fileStream.Read(fileData, 0, CInt(fileStream.Length)) fileStream.Close() Dim command As New OdbcCommand("INSERT INTO your_table (picture_name, picture_data) VALUES (?, ?)", connection) command.Parameters.AddWithValue("pictureName", fileName) command.Parameters.AddWithValue("pictureData", fileData) connection.Open() command.ExecuteNonQuery() connection.Close() ComboBox1.Items.Add(fileName) MessageBox.Show("Upload successful!") End If Catch ex As Exception MessageBox.Show("Error uploading picture: " & ex.Message) Finally connection.Close() End Try End Sub
In this updated code, after the upload is successful, a message box is displayed using theMessageBox.Show()
method with the success message "Upload successful!".
Remember to replace<server_name>
,<database_name>
,<username>
, and `<password
Add this code that when inserting a new picture I want to see the selected photo in a picture box before I click the upload button. and I can do this many times before I upload it and update the databaseCertainly! Here's the updated code with a message box to display a success message when the upload is successful:
Code:Private Sub Button2_Click(sender As Object, e As EventArgs) Handles Button2.Click Try If OpenFileDialog1.ShowDialog() = DialogResult.OK Then Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName) Dim fileStream As New FileStream(OpenFileDialog1.FileName, FileMode.Open, FileAccess.Read) Dim fileData As Byte() = New Byte(fileStream.Length - 1) {} fileStream.Read(fileData, 0, CInt(fileStream.Length)) fileStream.Close() Dim command As New OdbcCommand("INSERT INTO your_table (picture_name, picture_data) VALUES (?, ?)", connection) command.Parameters.AddWithValue("pictureName", fileName) command.Parameters.AddWithValue("pictureData", fileData) connection.Open() command.ExecuteNonQuery() connection.Close() ComboBox1.Items.Add(fileName) MessageBox.Show("Upload successful!") End If Catch ex As Exception MessageBox.Show("Error uploading picture: " & ex.Message) Finally connection.Close() End Try End Sub
In this updated code, after the upload is successful, a message box is displayed using theMessageBox.Show()
method with the success message "Upload successful!".
Remember to replace<server_name>
,<database_name>
,<username>
, and `<password
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