在回发时设置viewstate

时间:2020-03-05 18:46:57  来源:igfitidea点击:

我试图在按下按钮时设置ViewState变量,但是它仅在第二次单击按钮时有效。下面是代码:

protected void Page_Load(object sender, EventArgs e)
{
    if (Page.IsPostBack)
    {
        lblInfo.InnerText = String.Format("Hello {0} at {1}!", YourName, DateTime.Now.ToLongTimeString());
    }
}

private string YourName
{
    get { return (string)ViewState["YourName"]; }
    set { ViewState["YourName"] = value; }
}

protected void btnSubmit_Click(object sender, EventArgs e)
{
    YourName = txtName.Text;

}

我有什么想念的吗?这是设计文件的表单部分,就像POC一样非常基本:

<form id="form1" runat="server">
<div>
Enter your name: <asp:TextBox runat="server" ID="txtName"></asp:TextBox>
<asp:Button runat="server" ID="btnSubmit" Text="OK" onclick="btnSubmit_Click" />
<hr />
<label id="lblInfo" runat="server"></label>
</div>
</form>

PS:该示例非常简化,"使用txtName.Text代替ViewState"不是正确的答案,我需要将信息显示在ViewState中。

解决方案

回答

" Page_Load"在" btnSubmit_Click"之前触发。

如果我们想在发回事件触发后执行某些操作,请使用Page_PreRender

//this will work because YourName has now been set by the click event
protected void Page_PreRender(object sender, EventArgs e)
{
    if (Page.IsPostBack)
        lblInfo.InnerText = String.Format("Hello {0} at {1}!", YourName, DateTime.Now.ToLongTimeString());
}

基本顺序为:

  • 页面初始化触发(init无法访问ViewState)
  • 读取ViewState
  • 页面加载触发
  • 发生任何事件
  • 预渲染火灾
  • 页面渲染