public void FillGrid()
{
string connString = "Server =localhost;Port = 3306; Database =lbmtcs; UID =serverUser; Password =";
MySqlConnection Pathway = new MySqlConnection(connString);
string gridView = "SELECT PIN, Status, Cert_Title_No, Owner_Name, Co_Owner, Street, Barangay FROM landrecord"
+"WHERE PIN = @pin, Status = @Stat AND Cert_Title_No = @Cert AND Owner_Name = @Owner AND Co_Owner = @CO AND Street = @St AND Barangay = @Bar";
MySqlCommand grid = new MySqlCommand(gridView, Pathway);
MySqlDataAdapter MyDA = new MySqlDataAdapter();
MySqlCommandBuilder cmd = new MySqlCommandBuilder(MyDA);
MyDA.SelectCommand = grid;
DataTable table = new DataTable();
MyDA.Fill(table);
BindingSource bSource = new BindingSource();
bSource.DataSource = table;
dGrid.DataSource = bSource;
}
个MySqlException是未处理的:命令执行过程中遇到数据库异常时数据网格
致命错误。
代码有什么问题?
我正在选择一些列以显示在我的数据网格中。
对于初学者我会改变GridView的变量是这样的STRSQL第二你在哪里创建参数..? 路径改变,以sqlConn清理你的代码,并使用变量,只是有意义的FYI – MethodMan
你只是想显示数据,或者你打算保存和更新等?让我知道,因为它改变了答案。 – dash
谢谢破折号..跟着什么u贴..这是新的错误.. 似乎无法找到正确的类型..我在我的数据库中使用int .. DbType ..格式不正确所有Int创建此错误 –