Member 11328325 Ответов: 1

Как вставить данные в базу данных сервера mysql онлайн


мой код

Imports System.Data.SqlClient
Imports MySql.Data.MySqlClient
Imports System.IO
Public Class ADD1
    Dim conn As MySqlConnection
    Private imagename As String
    'this code declare variable as Byte

    Private imagge() As Byte
    'this code declare variable as MemoryStream

    Private memor As New MemoryStream
    Dim OpenFile As New OpenFileDialog()
    Private Sub ADD_Load(sender As Object, e As EventArgs) Handles MyBase.Load
        idtxt.Focus()
    End Sub

    Private Sub LinkLabel4_LinkClicked(sender As Object, e As LinkLabelLinkClickedEventArgs) Handles LinkLabel4.LinkClicked
        img.Image = Nothing
    End Sub

    Private Sub PictureBox3_Click(sender As Object, e As EventArgs) Handles PictureBox3.Click
        Try
            conn.ConnectionString = "server=111.111.xxx.xxx;Port=4xxx; Uid=axxusxxx; pwd=xxx; database=xxxrus"
            conn.Open()

            'this code insert these rows into database
            Dim cmd As New MySqlCommand
            cmd.Connection = conn
            cmd.CommandText = "INSERT INTO UOBexam([ID],[NAME],[MIDDLE],[LAST],[TELL],[FACULTY],[YEAR],[SEMESTER],[COURSE1],[COURSE2],[COURSE3],[COURSE4],[COURSE5],[COURSE6],[COURSE7],[GPA],[NOTE],[REPORT],[IMAGE]) VALUES (@ID,@NAME,@MIDDLE,@LAST,@TELL,@FACULTY,@YEAR,@SEMESTER,@COURSE1,@COURSE2,@COURSE3,@COURSE4,@COURSE5,@COURSE6,@COURSE7,@GPA,@NOTE,@REPORT,@IMAGE)"

            With cmd
                .Parameters.AddWithValue("@ID", idtxt.Text)
                .Parameters.AddWithValue("@NAME", name1.Text)
                .Parameters.AddWithValue("@MIDDLE", middle.Text)
                .Parameters.AddWithValue("@LAST", last.Text)
                .Parameters.AddWithValue("@TELL", tell.Text)
                .Parameters.AddWithValue("@FACULTY", ComboBox1.Text)
                .Parameters.AddWithValue("@YEAR", year.Text)
                .Parameters.AddWithValue("@SEMESTER", semester.Text)
                .Parameters.AddWithValue("@COURSE1", course1.Text)
                .Parameters.AddWithValue("@COURSE2", course2.Text)
                .Parameters.AddWithValue("@COURSE3", course3.Text)
                .Parameters.AddWithValue("@COURSE4", course4)
                .Parameters.AddWithValue("@COURSE5", course5.Text)
                .Parameters.AddWithValue("@COURCE6", course6.Text)
                .Parameters.AddWithValue("@COURSE7", course7.Text)
                .Parameters.AddWithValue("@GPA", GPA.Text)
                .Parameters.AddWithValue("@NOTE", NOTE.Text)
                .Parameters.AddWithValue("@REPORT", REPORT.Text)
            End With
            conn.Close()
    
        Catch ex As Exception
            MsgBox(ex.Message)
            conn.Close()
            'clean
        End Try
    End Sub

    Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
        Try
            With OpenFile
                .InitialDirectory = Application.StartupPath
                .FileName = ""
                .Title = "Photo:"
                .Filter = "Image files: (*.jpg)|*.jpg|(*.jpeg)|*.jpeg|(*.png)|*.png|(*.Gif)|*.Gif|(*.bmp)|*.bmp| All Files (*.*)|*.*"
                If .ShowDialog = Windows.Forms.DialogResult.OK Then
                    img.Image = System.Drawing.Image.FromFile(.FileName)

                Else
                    'Wala lang po..
                End If
            End With
        Catch ex As Exception
            MsgBox(ex.Message(), MsgBoxStyle.Critical, "Error...")
        End Try
    End Sub

    Private Sub LinkLabel1_LinkClicked(sender As Object, e As LinkLabelLinkClickedEventArgs) Handles LinkLabel1.LinkClicked
 
    End Sub
End Class


Что я уже пробовал:

соединение в порядке, но проблема в том, что я не могу вставить данные

$*Developer - Vaibhav*$

телеканал CNN.Открыть();
cmd = new SqlCommand(sql, cnn);
УМК.Метод executenonquery();
УМК.Располагать();
телеканал CNN.Рядом();

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$*Developer - Vaibhav*$

C# ADO.NET SqlCommand - ExecuteNonQuery

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