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Hi,
There is the requirement like this:,
One Login page is there for User Name, Password. Here the LAN user can only Login. How we will validate for a LAN user using C#.
Thanks in Advance.
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Hi,
I need to hide the first row in the datagrid(Not header).
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Here is the code.....hope it will help you...
GridView1.Rows[0].Visible = false;
Regards,
Sandeep Kumar.V
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Thanks sandeep. It works fine.
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I want to show the selected records Details on the next page with in the Popup window.iam displaying all the details in the Datagrid,i add one Link button in the Item template when click on this link button it should show the popup window.am binding the data while page loading.
Javascript Code as follows
function openWindow(url) {
window.open(url,'Calendar',
'width=500,height=400,left=200,top=250');
}
LinkButton PostbackUrl code as follows
PostBackUrl="javascript:openWindow('UserDetails.aspx?id=<%# DataBinder.Eval(Container.DataItem,"ID") %>');">LinkButton
when am running this page am getting this error message
'The server tag is not well formed'.It is showing in the Postbackurl code line.what is wrong in this line.
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Kurian_Kurian wrote: PostBackUrl="javascript:openWindow('UserDetails.aspx?id=<%# DataBinder.Eval(Container.DataItem,"ID") %>');">LinkButton
Replace with below code.......
PostBackUrl="javascript:openWindow('UserDetails.aspx?id=<%# DataBinder.Eval(Container.DataItem,'ID') %>');
Regards,
Sandeep Kumar.V
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Hi Sandeep,
I Tried with your code,it is not showing any error,page is loading,but when i click on the Link Button am not getting the popup window,that time not showing any error message.
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hey post your code completely so that i can help you better
Regards,
Sandeep Kumar.V
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Hi Sandeep am sending my code.
why the code is not showing after posting.what format i should use for showing the complete code.Now it is showing some part only.
Javascript function
function openWindow(url) {
window.open(url,'Calendar',
'width=500,height=400,left=200,top=250');
}
Link Button Code
<asp:templatecolumn headertext="Name">
<itemtemplate>
<asp:linkbutton id="lbName" runat="server" text="<%# DataBinder.Eval(Container, "DataItem.NAME") %>"
="" postbackurl="javascript:openWindow('UserDetails.aspx?id=<%# DataBinder.Eval(Container,'DataItem.ID') %>');">LinkButton
if am just passing 'userdetails.aspx' it will work.i will get popup window.but i want to add each record id also along with that.
-- modified at 11:00 Wednesday 17th October, 2007
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Kurian_Kurian wrote: why the code is not showing after posting.what format i should use for showing the complete code.Now it is showing some part only.
Check the "Ignore HTML" checkbox.
---
"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
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Hi, I am using AJAX to dynamically produce a DataList according to the input in a TextBox. I am using the 'OnTextChanged' event to achieve this and a problem that occurs is that I will lose focus in my TextBox as soon I begin to type anything.
An easy solution for this would be to remove the TextBox from the UpdatePanel but as I am using the UpdateProgress to produce a neat loading gif each time the OnTextChanged event occurs I have to keep the TextBox inside the UpdatePanel right?
I have seen a number of examples on the net which suggest using the LastFocus property but have had no luck getting that to work. Am I right in thinking that my problem cannot be solved with the LastFocus property as I also have an UpdateProgess in my code?
I am very new to ASP.NET and would highly appreciate any advice with this - my code is fairly large so I'll enclose a snippet of it, that will hopefully make my problem a bit clearer:
<asp:updatepanel id="upPnlLabels" runat="server">
<contenttemplate>
asp:TextBox ID="txtSearch" OnTextChanged="txtSearch_OnTextChanged" AutoPostBack="true" runat="server">
<asp:updateprogress id="upMyProgress" runat="server" associatedupdatepanelid="upPnlLabels">
<progresstemplate>
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Sorry I can't figure out how to post html on here, still hope to get some feedback on my problem
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i have on html page on clicking on the button of the page iam moving to the another page which is getting on the same page ie first page itself .now on clicking on the close button of the second page.i want the second page values in the first page .
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hi any one tell me how to upload asp . net project?
karthi
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Use FTP? It sounds like you are trying to upload a asp.net site to a web host or something? Usually they have an ftp site that allows you to copy files.
Or are you asking about how to upload files on a asp.net web site?
Ben
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hi
i have upload the asp.net project in wephost. but i have error in that page..
error:
Details: To enable the details of this specific error message to be viewable on remote machines, please create a <customerrors> tag within a "web.config" configuration file located in the root directory of the current web application. This <customerrors> tag should then have its "mode" attribute set to "Off".
karthi
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go to the web.config file and set debug=true and then upload the site again, the error will appear, see what's the problem and then upload it once more after correcting the error.
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hey i have a pdf document in the web server and i want to print it with a button in my web page
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Hi,
I am doing a web aplication, and in web.config my connection string is there. but while accessing the connection string i am getting that it is "empty"
Any idea for it !!!
Thanks in Advance
Happy Programming
-----
Abhijit
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Can you give some more idea about how u r retrieving connection key from Web.config file?
This time will also pass.....Remember this in bad days of Life.
Keep Smiling
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Hi,
Thanks for resposne !!!
please check the reply that i have post to jintal
Happy Programming
-----
Abhijit
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hi
please give me the code how u set And get The connection string
Jintal Patel
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That is used in web.config File
<connectionStrings>
<add name="mycon" connectionString="Data Source=M53;Initial Catalog=mydb;User ID=sa"
providerName="System.Data.SqlClient" />
</connectionStrings>
Now in in .cs file
SqlDataSource1.ConnectionString = ConfigurationManager.ConnectionStrings["mycon"].ConnectionString;
And in a aspx page there is a gridview and datasource id of gridview is SqlDataSource1.
While Opening that page i am gettig the error " Connection String is empty",
Any idea about that !!!
Thanks in Advance !!!!
Happy Programming
-----
Abhijit
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hi
If u want to open connection then try with
according your code
<connectionStrings>
<add name="mycon" connectionString="Data Source=M53;Initial Catalog=mydb;User ID=sa"
providerName="System.Data.SqlClient" />
</connectionStrings>
SqlConnection con = new SqlConnection(ConfigurationManager.ConnectionStrings["mycon"].ToString());
u will be able to open connection ..
OR also try like this
<appSettings>
<add key="connectionString" value="Data Source==M53;Initial Catalog=mydb;User ID=s<a;Password=sa;"/>
</appSettings>
conn = new SqlConnection(System.Configuration.ConfigurationManager.AppSettings["connectionString"]);
-- modified at 2:36 Thursday 18th October, 2007
-- modified at 2:37 Thursday 18th October, 2007
Jintal Patel
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