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我想在我的 Asp.new c# 表单的网格视图中显示我的数据。我正在使用 Wamp Server 的 MySQL 数据库。我已经厌倦了将数据库与网格视图绑定。请帮忙。

我的代码:

      public partial class Temp : System.Web.UI.Page
{

    string conString = ConfigurationManager.ConnectionStrings["AASProject"].ConnectionString;

    protected void Page_Load(object sender, EventArgs e)
    {
        MySqlConnection con = new MySqlConnection(conString);
        con.Open();

        gvFoodDetail.RowEditing+= new GridViewEditEventHandler(gvFoodDetail_RowEditing);
        gvFoodDetail.RowDeleting += new GridViewDeleteEventHandler(gvFoodDetail_RowDeleting);
        gvFoodDetail.RowUpdating += new GridViewUpdateEventHandler(gvFoodDetail_RowUpdating);
        //gvFoodDetail_RowCancelingEdit += new GridViewCancelEditEventArgs(gv);
       // gvFoodDetail.RowEditing += new GridViewEditEventHandler(gvFoodDetail_RowEditing);
       // gvFoodDetail.RowDeleting += new GridViewDeleteEventHandler(gvFoodDetail_RowDeleting);
       // gvFoodDetail.RowUpdating += new GridViewUpdateEventHandler(gvFoodDetail_RowUpdating);
       // gvFoodDetail.RowCancelingEdit += new GridViewCancelEditEventHandler(gvFoodDetail_RowCancelingEdit);

        gvFoodDetail.PageIndexChanging += new GridViewPageEventHandler(gvFoodDetail_PageIndexChanging);

        if (!IsPostBack)
        {
            gvFoodDetail.Visible = true;
            loadFoodDB();
        }

    }
    public void loadFoodDB()
    {

        string getFoodDetails = "Select Emp_ID,intime,outtime,date From datetime";
        MySqlConnection con = new MySqlConnection(conString);
        con.Open();

        MySqlCommand cmd = new MySqlCommand(getFoodDetails, con);
        DataTable dt = new DataTable();
        MySqlDataAdapter da = new MySqlDataAdapter(cmd);
        da.Fill(dt);
        con.Close();

        gvFoodDetail.DataSource = dt;
        gvFoodDetail.DataBind();
    }
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1 回答 1

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datagridview.Datasource = datasource
datagridview.databind()

这应该设置并显示数据。您必须手动或自动处理列和所有内容。

于 2012-04-09T13:30:09.947 回答