我已经阅读了所有的问题和答案从以前的职位,但不适合我的问题
问题:我想显示从db(mysql)到datagrid的数据,但问题是 MySqlDataReader
无法识别
SelectCommand
Fill
Update
作为 MySqlDataReader
我遵循教程中完全相同的代码编写方法,为什么会出现这个问题??
下面我附上了我的代码。我希望有人能帮我。
Imports MySql.Data.MySqlClient
Public Class AdminAddItem
Dim MysqlConnection As MySqlConnection
Dim FoodCommand As MySqlCommand
Dim DataFoodRead As MySqlDataReader
Dim dbDataSetFood As New DataTable
Private Sub btnLoadFood_Click(sender As Object, e As EventArgs) Handles btnLoadFood.Click
'declare new connection
MysqlConnection = New MySqlConnection
'prepare connection string
MysqlConnection.ConnectionString =
"server = localhost; userid = root; password = 1234 ; database = kedaikopimamba_db"
Dim bSource As New BindingSource
Try
MysqlConnection.Open()
Dim query As String
query = "select * from kedaikopimamba_db.foodtable"
FoodCommand.Connection = MysqlConnection
FoodCommand.CommandText = query
DataFoodRead.SelectCommand = FoodCommand
DataFoodRead.Fill(dbDataSetFood)
bSource.DataSource = dbDataSetFood
dgDisplayItem.DataSource = bSource
DataFoodRead.Update(dbDataSetFood)
MysqlConnection.Close()
Catch ex As MySqlException
MessageBox.Show(ex.Message)
Finally
MysqlConnection.Dispose()
End Try
End Sub
End Class
1条答案
按热度按时间dxpyg8gm1#
代码的哪一部分是错误?请让我们知道,以便于识别问题。