别忘了
autopostback = true<asp:CheckBox ID="CheckBox1" runat="server" oncheckedchanged="CheckBox1_CheckedChanged" AutoPostBack="true" />
<asp:Panel runat="server" ID="panel1"></asp:Panel>
-
protected void CheckBox1_CheckedChanged(object sender, EventArgs e)
{
panel1.Controls.Add(new Label { Text = "Text goes here" });
}
这允许您添加任何您想要的控件。
只需添加TextBox、Label等控件并使其不可见即可。 然后在
CheckBox1_CheckedChangedVisible<asp:CheckBox ID="CheckBox1" runat="server" oncheckedchanged="CheckBox1_CheckedChanged" />
<asp:TextBox ID="textBox" runat="server" Visible=false />
和
protected void CheckBox1_CheckedChanged(object sender, EventArgs e)
{
textBox.Visible = true;
}
在复选框标签下添加新的文字控件，
<asp:Literal id="lblMsg" runat="server"/>
然后在您的 CheckBox1_CheckedChanged 事件中将此设置为：
lblMsg.Text = "Checked";
并且是的，将复选框的
AutoPostBacktrue我建议使用 javascript 来完成此操作。您的用户不必“回发”，应用程序的感觉会更好，并且您将减少服务器费用。
使用 jQuery，你可以使用这样的东西：
<script language="javascript" type="text/javascript">
$(document).ready(function(){
$("#chk").onclick(function() {
$("#content").toggle();
});
});
</script>
<input type="Checkbox" id="chk"/>
<div id="content" style="display:none">
<asp:TextBox runat="Server" id="oneOfYourControls" />
</div>
jQuery 不是强制性的...您可以使用标准的简单
getElementById()唯一的缺点是你无法动态构建内容，但在大多数情况下这实际上并不是问题
在 Checked Changed Event 中编写如下代码：
protected void CheckBox1_CheckedChanged(object sender, EventArgs e)
{
Label1.Text = "Text";
}
489checkbox
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5
I added a checkbox to my form.
When the checkbox is checked, some information (TextBox, Label etc.) should appear right under the checkbox.
How can I do this?
Code:
<asp:CheckBox ID="CheckBox1" runat="server"
oncheckedchanged="CheckBox1_CheckedChanged" />
C#:
protected void CheckBox1_CheckedChanged(object sender, EventArgs e)
{
}
c#asp.netcheckbox
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edited Aug 6, 2021 at 8:40
Didier Aupest's user avatar
Didier Aupest
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asked Aug 2, 2011 at 7:35
hicay's user avatar
hicay
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Don't forget autopostback = true
<asp:CheckBox ID="CheckBox1" runat="server" oncheckedchanged="CheckBox1_CheckedChanged" AutoPostBack="true" />
<asp:Panel runat="server" ID="panel1"></asp:Panel>
-
protected void CheckBox1_CheckedChanged(object sender, EventArgs e)
{
panel1.Controls.Add(new Label { Text = "Text goes here" });
}
This allows you to add any control you want.
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answered Aug 2, 2011 at 7:44
nthpixel's user avatar
nthpixel
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Just add TextBox, Label etc. controls and make them invisible. Then in the CheckBox1_CheckedChanged function make them visible. This is done by setting the bool Visible property
<asp:CheckBox ID="CheckBox1" runat="server" oncheckedchanged="CheckBox1_CheckedChanged" />
<asp:TextBox ID="textBox" runat="server" Visible=false />
and
protected void CheckBox1_CheckedChanged(object sender, EventArgs e)
{
textBox.Visible = true;
}
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answered Aug 2, 2011 at 7:38
Petar Ivanov's user avatar
Petar Ivanov
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you should wrap all your controls in a panel. It will be more simple to change the visibility of a single panel control than a lot of other controls. Moreover, the structural html tags will also be shown/hidden along the server controls –
Steve B
Aug 2, 2011 at 8:12
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1
Add a new literal control under your checkbox tags,
<asp:Literal id="lblMsg" runat="server"/>
then in your CheckBox1_CheckedChanged event to this:
lblMsg.Text = "Checked";
and yes set the AutoPostBack property of your checkbox to true
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answered Aug 2, 2011 at 7:40
Waqas's user avatar
Waqas
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1
I would suggest doing this using javascript. Your users won't have to "postback" and the feeling on the application will be better, and you will reduce the server charge.
Using jQuery, you can use something like this:
<script language="javascript" type="text/javascript">
$(document).ready(function(){
$("#chk").onclick(function() {
$("#content").toggle();
});
});
</script>
<input type="Checkbox" id="chk"/>
<div id="content" style="display:none">
<asp:TextBox runat="Server" id="oneOfYourControls" />
</div>
jQuery is not mandatory... you can use standard simple getElementById().
The only drawback, is that you cannot dynamically build the content, but in most case it's a not actually a matter
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answered Aug 2, 2011 at 8:01
Steve B's user avatar
Steve B
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In Checked Changed Event write your code like this:
protected void CheckBox1_CheckedChanged(object sender, EventArgs e)
{
Label1.Text = "Text";
}
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answered Aug 2, 2011 at 7:42
thevan's user avatar
thevan
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