|
Mircea Grelus wrote: what world are you living in? Nobody's going to send you the solution by email. So you can stop hoping for that.
Exactly.
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
|
|
|
|
|
I am using ajax.net. I am using the update panal. I have already put a script manager control. When run this form then display run time message below this.
"Only one instance of a ScriptManager can be added to the page."
|
|
|
|
|
Mr Gagan just not ask repeated question again and again. Do yourself ok. And try to find out why error is coming. Not like Sonia Gupta to get all the solve code.
Ok ........
Devjit Das.
|
|
|
|
|
Not reapeat same question. My previous error:
"The control with ID 'UpdatePanel1' requires a ScriptManager on the page. The ScriptManager must appear before any controls that need it."
Now error: Only one instance of script manager control.
I am using the webuserconrol when webusercontrol under the use ajax.net
I am using the webusercontrol my aspx page under update panel. then create this problem
Only one instance of script manager control.
when webusercontrol put outside the update panel. and not use scriptmanager control. run project and not display error. but project requirment that under updatepanel.
|
|
|
|
|
Have you tried these[^] solutions?
Cheers,
Mircea
"Pay people peanuts and you get monkeys" - David Ogilvy
|
|
|
|
|
if u install ajax.net setup
and u found
Only one instance of a ScriptManager can be added to the page.?????
then u can try to create another website under chose ajaxenablewebsite
in give template
and put hole file and folder in ajax enable directory
no knowledge in .net
|
|
|
|
|
That's usually because you have already defined the ScriptManager in... for instance the MasterPage.
If you need to access the script manager from a child page, or from a user control use ScriptManagerProxy. Here you will find a selection of articles useful to start with AJAX ASP .net:
http://www.tipsdotnet.com/ArticleBlog.aspx?KWID=2&Area=AJAX&PageIndex=0[^]
HTH
/// -------------------------
Braulio Díez
tipsdotnet.com
/// -------------------------
|
|
|
|
|
how should i call my javascript function when the page is of master type.i mean the javascript code is under the content tag.
<asp:content id="Content1" contentplaceholderid="ContentPlaceHolder1" runat="Server">
function abc()
{
if(confirm('ok'))
{
}
}
i mean there is no body tag(because of master type page).hpow should i call the abc() function on browser closing
imran gave me the code
string attributeKey = "onunload";
string attributeValue = "abc();";
HtmlGenericControl body = (HtmlGenericControl)this.Master.FindControl("bid");
body.Attributes.Add(attributeKey, attributeValue);
which i wrote at form load , on the page having master page properties.
in the master page , i defined the function as below
function abc()
{
alert('hi');
}
function is not arising after clicking the form closing button
Soniagupta1@yahoo.co.in
Yahoo Messenger Id = soniagupta1
|
|
|
|
|
what is this yet u have not solve this problem.
Just use like this.
HtmlGenericControl body = (HtmlGenericControl)this.Master.FindControl("bid");
body.Attributes.Add("onbeforeunload", "alert('Close');");
Hope to solve ur problem.
Devjit Das.
|
|
|
|
|
Let me make u understand again.i want this message to invoke , only on the form closing button , it is emerging on all controls click (button click) of form.How should i deprieve it off , from apppearing on clicking all controls .
Soniagupta1@yahoo.co.in
Yahoo Messenger Id = soniagupta1
|
|
|
|
|
There is no way to know when the user closes the browser. No event is thrown. What you were presented here were methods commonly used to support this functionality. Like the method states, the event fires when the page is unloaded, but the page is unloaded as well when it roundtrips to the server.
Seriously though, you should learn to use Google. While trying to find solutions for your problems you will be forced to read arcticles, comments, forums and you will learn a great deal from that. Instead of posting the same questions here again and again and complain the code doesn't work.
Cheers,
Mircea
"Pay people peanuts and you get monkeys" - David Ogilvy
|
|
|
|
|
Mircea Grelus wrote: Seriously though, you should learn to use Google. While trying to find solutions for your problems you will be forced to read arcticles, comments, forums and you will learn a great deal from that.Instead of posting the same questions here again and again and complain the code doesn't work.
wise truth, Great statement...~5~
Regards
- J O H N -
|
|
|
|
|
u can try
<script language="javascript" type="text/javascript">
function abc()
{
alert('hi');
}
</script>
this code in separet .js file and then using
no knowledge in .net
|
|
|
|
|
Seriously, please tell me that no-one is paying you to ask the same basic question over and over, ignore the answers, and show no signs of having any ability to do research, write code, or even read ?
Christian Graus - Microsoft MVP - C++
"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )
|
|
|
|
|
Hi,
I have a Nested Gridview and each row of my nested Gridview has a radio button. I want the selected row from my nested gridview.
But here is the problem...
I want to access the selected rowid on a Button click, which is outside my Nested Gridview.
Can I save the Rowid for the radio button checked with out postback?
HELP NEEDED !!
Regards
Munawar
|
|
|
|
|
Navigate through the items of the GridView and find the radiobutton control and check whether it's selected or not. you should be able to access the ID too using the Datakey's property of the GridView. something like this...
For Each gv As GridViewRow In GridView1.Rows
If CType(gv.FindControl("rbtn"), RadioButton).Checked Then
RowID = CInt(GridView1.DataKeys(gv.RowIndex).Value)
End If
Next
Regards
- J O H N -
|
|
|
|
|
Hi dears,
I want to test my website on my workgroup network . Then I created a virtual directory in IIS and could brows website from another pc correctly. but it has a big problem . the website is not enable to connect to SQL database.
How can I fix it ?
thx a lot,
www.behzadi.net
|
|
|
|
|
furza_milan wrote: How can I fix it ?
Did you add grant permision for user asp.net
|
|
|
|
|
We have no idea. If IIS is serving the page, then the only thing that should matter is that the server can see SQL Server, the computer you use to browse the site should be irrelevant. I suspect you need to provide more details of what's happening, errors you see, etc.
If you see an error telling you to change your settings so you can see detailed errors, please just do that first, don't post it.
Christian Graus - Microsoft MVP - C++
"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )
|
|
|
|
|
if I catch the error no message occured but when I do not catch the code I just see "the page can not be displayed...!" message.
www.behzadi.net
|
|
|
|
|
did u run the aspnet_regiis -i in the vs 2005 command prompt
Soniagupta1@yahoo.co.in
Yahoo Messenger Id = soniagupta1
|
|
|
|
|
of course I did run aspnet_regiis.exe
I can brows website and can see my page layout but no data from database shown in webpage...
www.behzadi.net
|
|
|
|
|
how it is possible that u r giving the solution to other. Do u think so sonia. Not do again.
Devjit Das.
|
|
|
|
|
Sir, just trying to do it.It's all ur knowledge , which i share with others
Soniagupta1@yahoo.co.in
Yahoo Messenger Id = soniagupta1
|
|
|
|
|
Disable "show friendly http error messages" in the settings in your browser, so that you see the real error message that the server sends.
---
"Anything that is in the world when you're born is normal and ordinary and is just a natural part of the way the world works. Anything that's invented between when you're fifteen and thirty-five is new and exciting and revolutionary and you can probably get a career in it. Anything invented after you're thirty-five is against the natural order of things."
-- Douglas Adams
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
