求各位帮帮忙。。。windowsform
我现在有两个窗口；一个作为主界面，另一个是一对话框的形式出现；

我的目的：想单击对话框那个界面的button按钮时，在主界面中显示相应的信息；

问题是：每点击一次，就会出现一个一样的主界面；请问怎么解决？？？？
代码：form1中

private void btnEnterInsetStep_Click(object sender, EventArgs e)
{
this.Close();
Form1 f1 = new Form1();
f1.Show();
int index = f1.dgvRecipeEdit.Columns.Add("C", "");
string headerText = "step" + index.ToString();
f1.dgvRecipeEdit.Columns[index].HeaderText = headerText;
f1.dgvRecipeEdit.Columns[index].SortMode = DataGridViewColumnSortMode.NotSortable;
f1.dgvRecipeEdit.Columns[index].DefaultCellStyle.Font = new Font("宋体", 11F);
f1.dgvRecipeEdit.Columns[index].DefaultCellStyle.Alignment = DataGridViewContentAlignment.MiddleCenter;
f1.dgvRecipeEdit.Columns[index].Width = 130;

}
form2中：
private void btnInsertRecipeStep_Click(object sender, EventArgs e)
{
FormInsertStep fis = new FormInsertStep();
fis.ShowDialog();


[解决办法]
传值方法参考http://www.cnblogs.com/cosoft/archive/2011/08/08/2130659.html
[解决办法]
你用下单例模式应该可以解决了
[解决办法]
代码：form1中

private void btnEnterInsetStep_Click(object sender, EventArgs e)
{
this.Close();
Form1 f1 = new Form1();
f1.Show();
int index = f1.dgvRecipeEdit.Columns.Add("C", "");
string headerText = "step" + index.ToString();
f1.dgvRecipeEdit.Columns[index].HeaderText = headerText;
f1.dgvRecipeEdit.Columns[index].SortMode = DataGridViewColumnSortMode.NotSortable;
f1.dgvRecipeEdit.Columns[index].DefaultCellStyle.Font = new Font("宋体", 11F);
f1.dgvRecipeEdit.Columns[index].DefaultCellStyle.Alignment = DataGridViewContentAlignment.MiddleCenter;
f1.dgvRecipeEdit.Columns[index].Width = 130;

}
form2中：
private void btnInsertRecipeStep_Click(object sender, EventArgs e)
{
FormInsertStep fis = new FormInsertStep();
fis.ShowDialog();

在form1，中show form1，那就只能出现form1窗体了
[解决办法]

C# code

  private void btnEnterInsetStep_Click(object sender, EventArgs e)  {  this.Close();  Form1 f1 =Form1.GetInstance();  f1.Show();......................}Form1 窗体中        static   Form1 form = null;        public static   Form1  GetInstance()        {            if (form == null || form.IsDisposed)            {                form = new   Form1 ();            }            return form;        }