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i found a serious problem that when i click on button then postback happen but i stopped the postback.
my code is follows. please see what is wrong in it.
-----------------------------------------------------
<%@ Page Language="C#" AutoEventWireup="true" CodeFile="Default.aspx.cs" Inherits="_Default" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" >
<head runat="server">
<title>Untitled Page</title>
<script type="text/javascript" language="javascript">
function ShowDiv()
{
var modalWindow = document.createElement('div');
modalWindow.style.position = 'absolute';
modalWindow.style.height = '100px';
modalWindow.style.width = '100px';
modalWindow.style.left = (modalWindow.style.height - getDocHeight())/2;
modalWindow.style.top = (modalWindow.style.width - getDocWidth())/2;
modalWindow.style.backgroundColor = '#C0C0C0';
modalWindow.innerHTML = 'hello...';
document.body.appendChild(modalWindow);
return false;
}
function getDocHeight()
{
return Math.max(document.getElementById('TextBox1').style.height );
}
function getDocWidth()
{
return Math.max(document.getElementById('TextBox1').style.width );
}
</script>
</head>
<body>
<form id="form1" runat="server">
<div>
<asp:Button ID="Button1" runat="server" Style="left: 745px; position: relative;
top: 298px" Text="Button" OnClientClick=" return ShowDiv()" />
<asp:TextBox ID="TextBox1" runat="server" Height="180px" Style="left: 307px; position: relative;
top: 264px" TextMode="MultiLine" Width="432px"></asp:TextBox></div>
</form>
</body>
</html>
here ShowDiv() function return false. another thing i notice that when i comment all the line in function ShowDiv() then postback is not happening. i just do not understand if i generate a div and add to page from javascript then why postback happening.
please tell me the solution to avoid postback when i will generate div from any function. if possible please rectify my code.
thanks
tbhattacharjee
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Tridip Bhattacharjee wrote: another thing i notice that when i comment all the line in function ShowDiv() then postback is not happening.
This is happening because there is an error in your code. Once there is an error, return false; is not getting executed, that's why postback occurs.
The error is in lines
modalWindow.style.left = (modalWindow.style.height - getDocHeight())/2; modalWindow.style.top = (modalWindow.style.width - getDocWidth())/2;
here you are trying to like 180px-100px, so this is throwing an error. Get the height, remove px from the height then try, it will work
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thanks a lot.
tbhattacharjee
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Your welcome 
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Because of Masterpage's heaving occupying space, it is difficult to design content pages.
Is it possible to switch content page w/t or without MasterPage in DesignView.
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I don't think, there is any way.
But another way you can do, create a page without master page, do your designing and and at last, change the page to content page.
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No , its not possible, mybe what you can do , create a Master page and dont do anything to it, just add content pages and start working and you will have a lot of space. When you are done , then link it with the final master page
Vuyiswa Maseko,
Spoted in Daniweb-- Sorry to rant. I hate websites. They are just wierd. They don't behave like normal code.
C#/VB.NET/ASP.NET/SQL7/2000/2005/2008
http://www.vuyiswamaseko.com
vuyiswa@its.co.za
http://www.itsabacus.co.za/itsabacus/
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hi guys. i have strange bug or problem with this two controls. i will simplify my code as much as i can.
i have this markup:
<asp:FormView ID="fv" runat="server" DataSourceID="sds" DataKeyNames="id">
</asp:FormView>
<asp:DetailsView ID="dv" runat="server" DataSourceID="sds" DataKeyNames="id">
</asp:DetailsView>
<asp:SqlDataSource ID="sds" runat="server" ConnectionString="<%$ ConnectionStrings:TestConnectionString %>"
ProviderName="System.Data.OleDb" SelectCommand="SELECT * FROM [tablica]" >
</asp:SqlDataSource>
and lets say i need on first PageLoad get SelectedValue from eather of these:
protected void Page_Load(object sender, EventArgs e)
{
object fvValue = fv.SelectedValue;
object dvValue = dv.SelectedValue;
}
so here is strange situation (looks like evil bug!). fvValue and dvValue have null value, but debuging code using Visual Studio with brakepoints; when i collapse fv or dv and navigate to SelectedValue property it changes from null to "something"?
anyone encounter this situation?
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I have a page which does an upload functionality. Once it is uploaded successfully, I should be able to go to the third step of a Wizard control in another page. Is it possible to do so? Could anybody please help me on this? Any links or tutorials is fine.
Thanks in advance
Success is the good fortune that comes from aspiration, desperation, perspiration and inspiration.
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If you are using the ASP.NET Wizard control then you should be able to set the ActiveStepIndex to the appropriate page.
I know the language. I've read a book. - _Madmatt
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Thank you for the reply.
I am using ASP.Net Wizard Control only. But I need to come frome another page to the Wizard page. For eg. I have two pages, a.apx and b.aspx. Assume that Wizard controlis in b.aspx and Upload functionality in a.aspx. Once the upload becomes successful in a.aspx,I should be able to navigate to the 2nd or 3rd step of Wizard Control in b.aspx. Is this possible in Wizard Control?
Is it possible to write something like FindControl in a.aspx by referring the b.aspx page and setting the activestepindex in a.aspx itself so that it directly come to the 2nd or 3rd step of Wizard control in b.aspx?
Thanks
Success is the good fortune that comes from aspiration, desperation, perspiration and inspiration.
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I think you need to stop and think a bit, and learn/understand the asp.net model.
You can't find a control on page b from page because it doesn't exist until the page is served.
What you can do is pass the data to set the wizard from page a to page b via query string or session variable. Then in your page load event set the wizard property appropriately. This is basic ASP.NET development.
I know the language. I've read a book. - _Madmatt
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What are industry standards on customizing a product to fit your needs, such as aspdotnetstorefront? Changes may include source code change, database structure changes, etc. When there is any upgrade to the product, it should be smooth process w/o loosing the customizations.
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how to show busy indicator in the center of multiline textbox control when user click on button. when i am using MS-Ajax and multiline textbox & button in update panel then i need when user write something in textbox and click button for submitting it then a busy indicator will show just at the center of that multi line textbox. i have seen this type of effect in MSDN forum. can anyone help me with sample code.
thanks
tbhattacharjee
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You can't add a control to the textbox, what is most likely happening is a div containing the progress indicator is floated over the textbox. You can use javaScript to asertain the positions and widths to position it correctly.
I know the language. I've read a book. - _Madmatt
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thanks but it will be better if u give me that kind of code to simulate the situation.
tbhattacharjee
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My 1 Vote.
Actually it would be far better and you would learn a great deal more by researching what has been suggested and properly understanding it. It may be easier for you if someone does your work for you,but its certainly not better...
Rhys
"With no power comes no responsibility"
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I have the following classes:
namespace MyDept.MyTeam<br />
public class AnElement1 : ConfigurationElement<br />
{
namespace MyDept.MyTeam<br />
public abstract class ConfigurationElement : System.Configuration.ConfigurationElement, ICustomTypeDescriptor<br />
{
I am trying to serialize AnElement1 to an xml string format, so I make the following call to create a serializer:
XmlSerializer xmlSerializer = new<br />
XmlSerializer(typeof(MyDept.MyTeam.AnElement1));
But I get the following inner exception:
{"Types 'System.Configuration.ConfigurationElement' and MyDept.MyTeam.ConfigurationElement' both use the XML type name, 'ConfigurationElement', from namespace ''. Use XML attributes to specify a unique XML name and/or namespace for the type."}
Has anyone encountered this? I've googled a bit but couldn't find a solution that doesn't require changing either class names. I can't change the abstract class name as it is being used somewhere else and I need the behaviour of this class in my configuration elements.
----------------------------------------------------------
Lorem ipsum dolor sit amet.
modified on Thursday, November 11, 2010 6:16 PM
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I want to know how can I use the image control with Browse button so it's just a simple button which will invoke OpenDialog and when selecting the file then it will be shown in the image control?
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It sin't that difficult and there are many examples available. Upload the file, assign the path to the image control.
I know the language. I've read a book. - _Madmatt
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There is no any event of browse control which can be handled by server side. you should call the onchange event and implement javascript function and assign selected image in image control.
It's better take html control
Farogh Haider
Web developer
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I am retrieving a formatted html from my sql server text field into asp:Literal... I want ti know how to set the wordwrap for the asp:Literal?
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Enclose your literal in a div tag and set the width of that div.
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No, you don't need to wrap it in a div. The literal control will be rendered as a span which you should apply appropriate CSS styles to.
From all of the responses you have given you have much to learn before giving solid advice to others.
I know the language. I've read a book. - _Madmatt
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Can you please try the following code at your system?
<asp:Literal ID="myLiteral" runat="server" Text="Hello World"></asp:Literal>
Mark Nischalke wrote: The literal control will be rendered as a span
I looked at the html source generated from browser. There is no span at all. Please let me know if I am wrong.
He is talking about Literal Control not the Label Control.
Mark Nischalke wrote: From all of the responses you have given you have much to learn before giving solid advice to others.
Well, no one is perfect. Nither me nor you. We all are learning. I am just trying my best to help others by sharing my opinion/experience.
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