在回发上设置viewstate

Question

我按下按钮时尝试设置ViewState变量,但它只在我第二次单击按钮时才有效.这是代码隐藏:

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;

}

有什么我想念的吗？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:样本非常简单,"使用txtName.Text而不是ViewState"不是正确答案,我需要将信息放在ViewState中.

Answer 1

Page_Load火灾前btnSubmit_Click.

如果你想在回发事件被激活后做某事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());
}

基本顺序是:

• 页面init触发(init无法访问ViewState)
• 读取ViewState
• 页面加载火灾
• 任何事件都会发生
• PreRender开火
• 页面渲染