Yes i got the element since she passed it as an argument and it is the id of the image
Code:
<img src="set1.jpg" width="250" height="188" id="[B]prm[/B]" onClick="animateMe('prm');" alt="Click to start/stop animation">
vee10
Last Activity: Mar 19 '09, 08:09 AM
Joined: Oct 17 '06
Location:
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hi,
Yes i got the element since she passed it as an argument and it is the id of the image
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Hi
In my project i faced the same situation not only for & then i don't know this soln so i have written my own code for displaying & so by seeing this soln now i understood that i have to dig more in to the concepts...Leave a comment:
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Thanks for the solutionLeave a comment:
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Hi,
Its working fine for me in the IE also .
Whats the error u r getting while executing in the IE .Leave a comment:
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Hi,
in the script function u are comparing the disabledFldName s as an instance of array or not but while calling the fuction u r sending it as an array so always it will be false so the previous one which is enabled will not disabled
forexample u first select the radio button 2 so permant address one will be enabled now u selected the radio button 3 and now the disabledFldName s will be only the currentAddress not the...Leave a comment:
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Hi,
U can do this by using two methods
Method 1:
Using Div tag
...Code:java script : function selectBgImg(whichImage){ var color; if (document.body){ document.body.background = bgImg[whichImage]; switch(whichImage) { case 0: color = "images/curvepink.jpg" break; case 1: color = "images/curvegreen.jpg"Leave a comment:
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Hi,
This link may help u
http://www.designerwiz .com/JavaScripts/link_box_descri bes_content.htmLeave a comment:
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hi,
This is just as example i am giving since i have used that in my application but according to ur requirement u should change
...Code:public string GetKeyValue() { string key = "hkey_local_machine", subKey = @"System\ControlSet001\Control\COM Name Arbiter", valueName = "ComDB"; return( GetRegistryKeyValue(key, subKey, valueName)); }Leave a comment:
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Hi,
this is how u can open and close and execute queries
...Code:Partial Class Default4 Inherits System.Web.UI.Page Private ConnectionString As String = "Data Source=PRAVEENAPC;Initial Catalog=testing;Integrated Security=True" Private reader As Data.SqlClient.SqlDataReader = Nothing Private conn As Data.SqlClient.SqlConnection = Nothing PrivateLeave a comment:
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Item Template is nothing but adding any linkbutton or textbox etc ie ading any single item control
EditItem Template is adding some controls when we click the Edit button in the datagrid only these controls will be available after clicking edit in datagrid
but in itemTemplate when the datagrid is loaded this will be available ie we can see the controls under <ITEMTEMPLATE > once datagrid is loaded
but for <EDITITEMTEMPLA TE>...Leave a comment:
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hi..
...Code:SqlConnection conn = new SqlConnection("Data Source=PRAVEENAPC;Initial Catalog=testing;Integrated Security=True"); conn.Open(); SqlCommand cmd = new SqlCommand(); cmd.CommandText = "insert into emp values("; cmd.Connection = conn; for (int i = 0; i < GridView1.Columns.Count; i++) {Leave a comment:
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Hi,
I think on checking the checkbox in the datagrid any of the events is going to fire so u should place a button and do the necessay thing u want to do
if u get any solution plz share that...Leave a comment:
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Hi,
U can check this link
http://www.aspdotnetco des.com/GridView_Insert _Edit_Update_De lete.aspx
if any doubt u can post ur doubt...Leave a comment:
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Hi,
u should initialize the connection property of the sqlcommand variable
...Code:cmd .Connection = conn;
Leave a comment:
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u can check out this link
http://www.codeproject .com/KB/webforms/datagridcheckbo x.aspx...Leave a comment:
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hi...
sorry for posting wrong which is not ur requirement
...Code:<asp:GridView ID="GridView1" runat="server" > <Columns> <asp:TemplateField HeaderText="Check"> <EditItemTemplate> <asp:CheckBox ID="CheckBox1" runat="server" /> </EditItemTemplate>Leave a comment:
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Using Template field u can do that
...Code:<asp:GridView ID="GridView1" runat="server" > <Columns> <asp:TemplateField HeaderText="Check"> <EditItemTemplate> <asp:CheckBox ID="CheckBox1" runat="server" /> </EditItemTemplate>Leave a comment:
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hi,
should not use the && for queries
da = "select * from login where username ='"+textbox1.te xt+"' and password ='"+textbox2.te xt + "'";...Leave a comment:
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hi ..
few days back i was also looking into the solution i got the solution
http://www.aspdotnetco des.com/GridView_Insert _Edit_Update_De lete.aspx
i am also just posting my code but its not using the database
...Code:<%@ Page Language="C#" AutoEventWireup="true" CodeFile="Default8.aspx.cs" Inherits="Default8" %> <!DOCTYPE html
Leave a comment:
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hello...
break should be placed for every case
example
at which position u r getting the object not supported...Code:switch() { case "Frank": break; case "Anthony": ---- break; }Leave a comment:
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