asp.net Button needs to be fired twice

问题

I am generating dynamic html controls. The controls are being generated on Init, after the user has caused a postback by pressing a button. However, he needs to press it twice so that the controls are generated. How can I fix this? Code:

 protected override void OnInit(EventArgs e)
    {
        if (Page.IsPostBack)
        {
            if (Session["id"] != null)
            {
                string id= Session["id"].ToString();
                GenerateDynamicControls(id);

            }
        }
    }

 protected void Page_Load(object sender, EventArgs e)
    {
        Session["id"] = null;
    }

protected void Button1_Click(object sender, EventArgs e)
    {
        string id = TextBox1.Text;
        Session["id"] = id;
    }

回答1:

There is no need to use session for these purposes. Plus your code might fail after subsequent postbacks.

protected override void LoadViewState(object state)
{
    base.LoadViewState(state);
    var id = this.ViewState["DynamicControlGeneration"] as string;
    if (id != null)
        GenerateDynamicControls(id);
}

protected void Button1_Click(object sender, EventArgs e)
{
    string id = TextBox1.Text;
    this.ViewState["DynamicControlGeneration"] = id;
    GenerateDynamicControls(id);
}

回答2:

Session["id"] is set to null on page load. When the page is posted back after button click, the OnInit method is called first and it get the value of Session["id"] as null. After that the button click event is executed and Session["id"] is set. So when you click the button second time the OnInit has the value other than null for Session["id"] and your code is executed on the second click.

回答3:

Call GenerateDynamicControls(id); when button is clicked. That way you will have your controls on first click. And when page reloads they will be recreated in OnInit.

protected void Button1_Click(object sender, EventArgs e)
{
    string id = TextBox1.Text;
    Session["id"] = id;
    GenerateDynamicControls(id);
}

回答4:

Try this,

protected override void OnInit(EventArgs e)
{
     if (Page.IsPostBack)
     {
         string id = Request.Form[TextBox1.ClientID].ToString();
         GenerateDynamicControls(id);
     }
}

来源：https://stackoverflow.com/questions/15496748/asp-net-button-needs-to-be-fired-twice

标签

ASP.NET

init