表单提交按钮的事件背后的代码未执行。背后ASP.NET提交按钮事件未执行
<%@ Page Language="C#" AutoEventWireup="true" EnableEventValidation="true" Inherits="_Default" %>
<form id="form1" runat="server">
<input id="Submit1" type="submit" value="Update" runat="server" onclick="btnclick_Click" />
<asp:TextBox ID="tbStatus" enableViewState="true" runat="server" TextMode="MultiLine"
Width="296px" Height="67px" ReadOnly="True"></asp:TextBox>
</form>
代码:
protected void Page_Load(object sender, EventArgs e)
{
if (!Page.IsPostBack)
{
tbStatus.Text = "Status: Page loaded";
}
}
protected void btnclick_Click(object sender, EventArgs e)
{
tbStatus.Text += "\nButton clicked. ";
}
的btnclick_Click从未执行。如果我不是改变的Page_Load这样:
protected void Page_Load(object sender, EventArgs e)
{
if (!Page.IsPostBack)
{
tbStatus.Text = "Status: Page loaded";
}
} else {
tbStatus.Text += "\nUpdated. (Page_Load PostBack). ";
}
然后textarea的更新两次,在第三sudmit我跑进错误消息:
The state information is invalid for this page and might be corrupted.
我只想按钮提交事件来执行。提前感谢任何有关我做错事情的信息!
为什么不使用服务器控制? –
哦!我不相信我错过了!这实际上也是我的代码,可怕的错误!谢谢! – Plarsen