通过一个按钮的click事件在datagrid中显示数据
例如 这样一个按钮事件
private void btnSearch_Click(object sender, EventArgs e)
{
string source = frmLogin.DBConString;
string select = sqlSelect;
SqlConnection conn = new SqlConnection(source);
SqlDataAdapter da = new SqlDataAdapter(select, conn);
DataSet ds = new DataSet();
da.Fill(ds, "vw_selectDrugInfo");
dataGrid1.SetDataBinding(ds, "vw_selectDrugInfo");
}
这样在datagrid的表头中显示的就是数据库的字段名,那么该怎样自定义表头的内容呢?
例如 这样一个按钮事件
private void btnSearch_Click(object sender, EventArgs e)
{
string source = frmLogin.DBConString;
string select = sqlSelect;
SqlConnection conn = new SqlConnection(source);
SqlDataAdapter da = new SqlDataAdapter(select, conn);
DataSet ds = new DataSet();
da.Fill(ds, "vw_selectDrugInfo");
dataGrid1.SetDataBinding(ds, "vw_selectDrugInfo");
}
这样在datagrid的表头中显示的就是数据库的字段名,那么该怎样自定义表头的内容呢?
2.设置Columns
{
DataBind();
}
private void DataBind()
{
//初始化DataGridView
DataGridViewTextBoxColumn column = new DataGridViewTextBoxColumn();
column.Name = "sno";
column.HeaderText = "学号";
//设置绑定数据源的sno字段
column.DataPropertyName = "sno";
this.dataGridView1.Columns.Add(column); column = new DataGridViewTextBoxColumn();
column.Name = "sname";
column.HeaderText = "姓名";
//设置绑定数据源的sname字段
column.DataPropertyName = "sname";
this.dataGridView1.Columns.Add(column); column = new DataGridViewTextBoxColumn();
column.Name = "sage";
column.HeaderText = "年龄";
//设置绑定数据源的"sage"字段
column.DataPropertyName = "sage";
this.dataGridView1.Columns.Add(column); //设置不自动产生列
this.dataGridView1.AutoGenerateColumns = false;
this.dataGridView1.AllowUserToAddRows = false; //以下进行数据绑定
SqlConnection con = new SqlConnection("server=.;database=student;uid=sa;pwd=0421");
SqlDataAdapter sda = new SqlDataAdapter("select * from studentDetails", con);
DataSet ds = new DataSet();
sda.Fill(ds, "student");
this.dataGridView1.DataSource = ds.Tables["student"];
}
{
DataBind();
}
private void DataBind()
{
//以下进行数据绑定
SqlConnection con = new SqlConnection("server=.;database=student;uid=sa;pwd=0421");
SqlDataAdapter sda = new SqlDataAdapter("select sno as 学号,sname as 姓名,sage as 年龄 from studentDetails", con);
DataSet ds = new DataSet();
sda.Fill(ds, "student");
this.dataGridView1.DataSource = ds.Tables["student"];
}