Develop Reference

how to add hide event to the button in webform?

im trying to hide div by clicking the button, but it does hide it for a second and return the div (looks like it refreshing page).here is my html code :
<div id="page">
<p>First Paragraph</p>
<p>Second Paragraph</p>
<p>Yet one more Paragraph</p>
<asp:Button ID="Button1" runat="server" Text="Button" />
</div>
and here is my jquery code:
$(document).ready(function () {
$("#Button1").click(function () {
$("#page").hide();
});
});

What Rick said in his answer AND you need to add Type="Button" to your button.
<Button ID="Button1" type="button">Button</button>
The default button type is submit, which will cause the page refresh. AND, as Rick mentioned, ASP buttons will also submit. So, you need to use an HTML button and set the type="button" attribute.
From MDN
The type of the button. Possible values are:
submit: The button submits the form data to the server. This is the default if the attribute is not specified, or if the attribute
is dynamically changed to an empty or invalid value.
reset: The button resets all the controls to their initial values.
button: The button has no default behavior. It can have client-side scripts associated with the element's events, which are
triggered when the events occur.

An ASP Button with runat="server" causes a server post-back. If you aren't going to be acting on that click from the server, just use a client-side button like:
<button id="Button1" type="button">Button</button>

Set the client id mode to static on the button, then it'll work. It's not working because that hasn't been specified so the name is changing.
<asp:Button ID="Button1" ClientIDMode ="Static" runat="server" Text="Button" />
If you wanted to disable the postback to simply hide something without processing anything, then just using a client side button is the best option.

Server Controls, Client Controls, Panels, Default Buttons and Client Events

Ok so I would like to be able to be in the part number textbox press enter and it do a 'Quick Search'. However when press enter it activates the 'Search' Button instead. The 'Search' Button as the diagram shows is a default button of the panel it is in. But the 'Quick Search' is not in that same panel so I am kinda stumped on how to change this action so it calls click on the button on the 'Quick Search' and not the 'Search'. If this doesn't make since ask more questions and I will update the diagram and question.
Thanks in advance!
New Facts
In the browser Render ... these are in the same form
I want the Quick Search Button to be a client side button which makes it hard to use a panel and a default button

Try to place Part Number textbox and Quick Search button in separate Panel with DefaultButton="bnQuickSearck" attribute like follows:
<asp:Panel runat="server" DefaultButton="bnQuickSearck">
<asp:TextBox ID="tbPartNumber" runat="server"></asp:TextBox>
<asp:Button ID="bnQuickSearck" runat="server" Text="Quick Search" />
</asp:Panel>
EDIT If you have client-based button you can use the following code on javascript:
<div id="divId">
<input id="txtId" type="text" />
<input id="btnId" type="button" value="Quick Search" onclick="alert('test')"; />
</div>
<script>
$("#divId").bind("keypress", function (e) {
if (e.keyCode == 13) {
$("#btnId").click();
return false;
}
});
</script>
Make sure jQuery installed in your web application in this case.

It sounds like you have one big form. If this is the case, when you hit enter in the Part Number field it activates the default action, in this case it would seem to be the "Search" button. If you can break up your forms, they'll each have their own default actions. This will eliminate any need for crazy redirects, AJAX, server-side foo, etc.

add numbers to textbox

<input type=text id="txtNum"/>
ex. of button
<asp:Button ID="btnNum1" runat="server" CssClass="btnNumbers" Text="1" />
<asp:Button ID="btnNum5" runat="server" CssClass="btnNumbers" Text="5" />
Basically I have a huge numberpad on my page using Buttons label from 0-9 and backspage and clear. This is going to be for a touchscreen device
I am not sure how to go about when a user touches button1 to place a '1' in my textbox. Then if they hit button5 my textbox value would append the 5 to the 1.
I would like to use javascript to perform this task so it does not do a postback for every button click please help.

The easiest way to prevent the postback is to not use the asp.net button control. Just use standard html button
<button type="button" class="btnNumbers">1</button>
Then just attach to the click event for the button and append it's value to the textbox.

Add following method
$(document).ready(function() { $(".btnNumbers").click(function() { $("#txtNum").val($("#txtNum").val() +$(this).val()); return false; });
});

How to show a html button for asp:button click

I have an asp.net web page with asp:button control. I need to show two (normal) HTML buttons for the click event of the asp:button control.
My requirement is when the page load for the first time there will be only asp:button there visible. After I click on that asp:button other two HTML buttons should be visible. But they should be visible for all the other postbacks. I mean if there would be any post backs, that HTML buttons should be visible constantly. How could I do that? Please help me. I tried to implement that using jquery hide and show.

You can use <asp:Button> controls for the other two buttons, and can set Visible="false" initially, then setting Visible="true" when you need to show them. This way, the server can do everything, and retain viewstate too so you don't have to reshow the buttons everytime the page posts back.
ASP buttons are standard buttons but can trigger the functionality you want through the OnClientClick property:
<asp:Button .. UseSubmitBehavior="false" Text="Move up" OnClientClick="moveSlider(1);return false;" />
<asp:Button .. UseSubmitBehavior="false" Text="Move down" OnClientClick="moveSlider(-1);return false;" />
Using return false; makes sure that the button doesn't postback, and UseSubmitbehavior="false" renders an <input type='button' /> instead.
HTH.

aspx submit form error

Im working with asp.net framework 4.0 and I have this code:
form id="form1" runat="server" method="get" action="Profile.aspx"
// some code
asp:Button runat="server" ID="SubmitButton" Text="Submit"
Each time i click the submit button i get this error:
Validation of viewstate MAC failed. If this application is hosted by a Web Farm or cluster, ensure that configuration specifies the same validationKey and validation algorithm. AutoGenerate cannot be used in a cluster.
any idea how to fix it ???

This is caused by cross-page POST (i.e. you are submitting the ViewState of the first page to the second). You can add PostBackUrl to the button like this:
<asp:Button runat="server" ID="SubmitButton" Text="Submit" PostBackUrl="~/WebForm2.aspx" />
Alternatively you can handle the click event of the button in the first page, move some of the logic in this handler and do a Response.Redirect (i.e. GET request) to the second page. The right solution depends on your particular case.

If you want to send post back to your current page, remove in form tag method="get" action="Profile.aspx" attributes. And handle in codebehind post data from your page.
If your want to send data you another page like Profile.aspx, use PostBackUrl attribute of the button control, like Stilgar wrote for you. And then in Profile.aspx codebehind to get access to control from your current page use somethink like this:
If(Page.PreviousPage != null)
{
var textBox = Page.PreviousPage.FindControl("ControlID") as TextBox;
if(textBox != null)
{
//Use your logic here
}
}
Hope it will be helpful for you!
Best regards, Dima.