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Hi Madaxe1,
This file is called index page and text writen in
and
is the place where you are suppose to add your text which will be displayed to user just put your content there.
If you want to format your text I will suggest you use http://www.quackit.com/html/online-html-editor/[^]
this will help you to format your text and do not forget to put a title, short description and some key words in place of
YOUR TITLE HERE,YOUR DESCRIPTION HERE and YOUR KEYWORDS HERE.
Thats all done!! have fun
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Lets say that I want to develop a website where all pages will be dynamically generated and 'printed' into static HTML files. I will then serve these HTML files from a container (such as an Amazon EC2 instance) that has no server side languages or database running in the background.
I will make a script to have the HTML files automatically refreshed at a set period.
My reasons for doing this would be to decouple the link between traffic to my website and server load (in other words, so my website can scale without any need for additional CPU / RAM on the server).
Here are some benefits that i expect from doing it this way
I want a cheap way to host a high traffic website that can accommodate large spikes in traffic (from viral content)
I want it to load as fast as possible -- without any delay from generating content dynamically for each visitor.
My Questions are as follows:
What are the downsides of structuring a website this way?
Is this flat out a bad idea?
What is this type of architecture called?
My Background
I have made a few simple web apps using PHP / mySQL and am familiar with javascript, hmtl, css--- basically a new / amateur web developer.
Thanks for the input!
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Hi,
I don't see any downside of it. I usually do it for many of my small CMS website (< 50-60 pages).
But you need to check from the hosting provider that how much memory/caching you are allowed to use.
For a huge/biger website, you wish to choose to cache only frequently used pages. There are many caching softwares which handles it itself. 
Life is a computer program and everyone is the programmer of his own life.
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In a vB.NET 2010 web form application, I am changing an existing web page. I want the user to be able to select various customers
based upon if the first name, last name, or middle names are selected. The idea is for the user to select one, two, and/or three
of the parameters and hit the search key.
The problem is the results only return data based upon only one parameter. If data for more than one paramter is selected, the selection
criteria is ignored.
When I run the sql in sql server management studio, I get the results I want.
Thus can you tell me what I can do to solve the problem?
The code that is having the error is listed below;
<%@ Page Title="" Language="VB" MasterPageFile="~/MasterPage.master" AutoEventWireup="false" CodeFile="Support.aspx.vb" Inherits="letters_Support" %>
<asp:Content ID="Content1" ContentPlaceHolderID="MPHead" Runat="Server">
</asp:Content>
<asp:Content ID="Content2" ContentPlaceHolderID="MPContent" Runat="Server">
<asp:Panel runat="server" ID="pnlSupport">
<h3 style="text-align: center; border-style:none; ">Support Page</h3>
<table style="width: 100%;">
<tr>
<td>
Last Name:
</td>
<td>
<asp:TextBox ID="lname" runat="server"></asp:TextBox>
</td>
<td>
First Name:
</td>
<td>
<asp:TextBox ID="fname" runat="server"></asp:TextBox>
</td>
<td>
Middle Name:
</td>
<td>
<asp:TextBox ID="mname" runat="server"></asp:TextBox>
</td>
<td>
Search by Name:
</td>
<td>
<asp:Button ID="btnSearch" runat="server" Text="Search" />
</td>
</tr>
</table>
<br />
<asp:GridView ID="GridViewSupport" runat="server" AutoGenerateColumns="False"
PageSize="25"
GridLines="Both"
EnableSortinAndPagingCallbacks="true"
HeaderStyle-CssClass="ui-widget-header"
PagerStyle-CssClass="ui-widget ui-widget-content"
Width="100%"
EmptyDataText="No records were selected."
DataSourceID="sqlTEST" AllowPaging="True" AllowSorting="True" >
<Columns>
<asp:BoundField DataField="lastName" HeaderText="Last Name" SortExpression="lastName" ItemStyle-HorizontalAlign="Center" ItemStyle-CssClass="ui-widget-content" ControlStyle-CssClass="text required ui-widget-content ui-corner-all"/>
<asp:BoundField DataField="firstName" HeaderText="First Name" SortExpression="firstName" ItemStyle-HorizontalAlign="Center" ItemStyle-CssClass="ui-widget-content" ControlStyle-CssClass="text required ui-widget-content ui-corner-all"/>
<asp:BoundField DataField="middleName" HeaderText="Middle Name" SortExpression="middleName" ItemStyle-HorizontalAlign="Center" ItemStyle-CssClass="ui-widget-content" ControlStyle-CssClass="text required ui-widget-content ui-corner-all"/>
</Columns>
</asp:GridView>
<asp:SqlDataSource ID="sqlTEST" runat="server"
ConnectionString="<%$ ConnectionStrings:sqlTEST %>" SelectCommand="SELECT [lastName]
,[firstName]
,[middleName]
,[suffix]
,a.[userid]
from [TEST].[dbo].[Identity]
where ([lastName] like '%' + @lname + '%' and
[firstName] like'%' + @fname + '%' and
[middleName] like'%' + @mname + '%')
or ([lastName] like '%' + @lname + '%' and
[firstName] like'%' + @fname + '%' and
@mname is null)
or ([lastName] like '%' + @lname + '%' and
[middleName] like'%' + @mname + '%' and
@fname is null)
or ([firstName] like '%' + @fname + '%' and
[middleName] like'%' + @mname + '%' and
@lname is null)
or ([lastName] like '%' + @lname + '%' and
@fname is null and
@mname is null)
or (@lname is null and
[firstName] like'%' + @fname + '%' and
@mname is null)
or (@lname is null and
@fname is null and
[middleName] like'%' + @mname + '%')
or (@fname is null and @lname is null and @mname is null)
order by [lastName], [firstName], [middleName]"
CancelSelectOnNullParameter="false" >
<SelectParameters>
<asp:ControlParameter ControlID="lname" Name="lname" PropertyName="Text" ConvertEmptyStringToNull="true" />
<asp:ControlParameter ControlID="fname" Name="fname" PropertyName="Text" ConvertEmptyStringToNull="true" />
<asp:ControlParameter ControlID="mname" Name="mname" PropertyName="Text" ConvertEmptyStringToNull="true" />
</SelectParameters>
</asp:SqlDataSource>
</asp:Panel>
</asp:Content>
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Try simplifying your WHERE clause to:
where ([lastName] like '%' + @lname + '%' Or @lname Is Null)
And ([firstName] like '%' + @fname + '%' Or @fname Is Null)
And ([middleName] like '%' + @mname + '%' Or @mname Is Null)
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Hello i made a module where i parse a bunch of names from the DB and i print em in a dropdownlist in the backend. Then i select some names and that names will be posted in the frontend.
I wanted to add an extra element (textbox) where it will print the count of the names i selected.
I wrote the code but i get a NULL value text box and no errors at consol.log . As the script is running at jsfiddle i suspect i have an error at including the library.
Im searching 2 days now with no luck.Could a joomla developer guru enlight me on what im doing wrong ?
my jquery script
jQuery(document).ready(function($){
var count =$("#jform_params_foreignmanuf :selected").length;
$("#jform_params_manucounter").val(count);
});
Multiselect Dropdown list:
{
protected $type = 'Manulist';
protected function getOptions()
{
$db= JFactory::getDbo();
$query = $db->getQuery(true);
$query->select('*');
$query->from('#__virtuemart_manufacturers_en_gb');
$db->setQuery($query);
$items = $db->loadObjectList();
$options = array();
if ($items)
{
foreach($items as $item)
{
$options[] = JHtml::_('select.option', $item->virtuemart_manufacturer_id , $item->mf_name,$item->slug);
}
}
$options = array_merge(parent::getOptions(), $options);
JHtml::_('script', JUri::root() . 'modules/mod_manufacturers/js/js.js');
}
}
Textbox Counter code:
<?php
defined('JPATH_PLATFORM') or die;
JFormHelper::loadFieldClass('Text');
class JFormFieldManulist extends JFormFieldText
{
protected $type = 'Countermanu';
protected function getOptions()
{
$doc = JDocument::getDocument();
$js = '
jQuery(document).ready(function($){
var count = $("#jform_params_foreignmanuf :selected").length;
$("#jform_params_manucounter").val(count);
});
';
$doc->addScriptDeclaration($js);
JHtml::_('script', JUri::root() . 'modules/mod_manufacturers/js/js.js');
}
}
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Hi, suppose I have a Div on a page and inside of this Div I have a Span, a List, and another Div.
If the page is displayed in an Iframe and if none of the elements has an ID, how would I go about adding a temporary ID when they are clicked using jQuery?
Each time a new element is clicked if it doesn't have an ID it will be assigned the same temporary ID given to the element that was clicked last. So the ID has to be removed from the previous element and then given to the new element.
Obviously I have to first call the load method to load the Iframe then inside the load method I will get the contents of the Iframe and find the body element when the page is clicked. Thus far the code below is what I have inside the load method but I need help in finishing it.
$(this).contents().find("body").on('click', function (e) {
var ClickedElement = $(e.target);
}
Thanks in advance for your help.
modified 4-Dec-14 2:34am.
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I am trying to learn how design a better web applications,
from the design and architecture, to the bits and bytes of the code.
i am currently using jersey servlet container, for my REST service.
and i'm using a node.js server as a "front controller" that communicates with my REST API and does all the view rendering.
I am aware that some of the things i did are wrong or just ugly, and i'm trying to learn how to do things right.
so i was thinking of learning spring mvc, but i find it overwhelming.
so my question is: is there a way to gradually integrate spring features to my web application? so that instead of being overwhelmed by all of springs component, i can tackle each component seperatly.
Also, could you recommand tutorials\books about web application design? most of the meterial i found was orianted towards specific technologies, rather then generally how to design a web aplication and how to get different technologies to work together.
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Hi all, I created a folder called Everyone in my project folder and moved some html pages from the project folder to this new folder. I left the css, images, and scripts folders in their original locations.
Then I modified the pages that were moved to the Everyone folder to use relative paths to access the styles, images and scripts. None of the pages in the Everyone folder is displaying correctly.
The following is my folder structure
Project Folder
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-----------------------------------
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Everyone CSS images scripts
I have used the following relative paths on a page in the Everyone folder to get access to a css file and an image from the images folder but they are not working.
<link rel="stylesheet" href="/css/mediaCSS.css" type="text/css" media="screen" />
<img src = "/images/myimage.jpg" />
Please point out why they're not working, thanks in advance
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Everyone is at a directory below your project root where some of your html pages live.
So is the other directories.
To access a directory below your project root from another directory that is below your project root then you first must move upwards to the project root, in a similar way that you navigate the folder structure in a dos prompt.
So, to navigate to the CSS directory from the Everyone directory use ../css/file-name.css the same would be true for the images and scripts directories. (note the double dot (..) to move up a single level in the directory structure)
modified 1-Aug-19 21:02pm.
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Hi Richard, thanks for responding. I have already tried what you've suggested even before posting this question and it did not work. I tried it again just to be certain and sure enough it did not work.
The project folder I mentioned above is my project folder in Visual Studio. Basically I am using form authentication and I wanted to allow everyone to have access to certain files. So I created the Everyone folder in my project folder and configured the Web.Config file for this new folder so that all the files put in it can be accessed by everyone.
modified 25-Nov-14 1:25am.
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I solved this by adding the location element to the webconfig and allowed everyone to access my file
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i just started a class on web development and i think i like what i see.. my problem is that i just cant picture the perfect way of picking colours for design...
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This task would generally be undertaken by a graphic designer. But if you don't have that skill then the best way is to look at lots of different websites and see which colours give a pleasing look and which ones just look wrong. You can also look at Microsoft recommendations[^].
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Do not pick colors - pick a designer...
I'm not questioning your powers of observation; I'm merely remarking upon the paradox of asking a masked man who he is. (V)
תפסיק לספר לה' כמה הצרות שלך גדולות, תספר לצרות שלך כמה ה' גדול!
Skipper: We'll fix it.
Alex: Fix it? How you gonna fix this?
Skipper: Grit, spit and a whole lotta duct tape.
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If you're interested in the graphic design portion, I would suggest taking some Graphic design classes; even the analog design classes can give you a good grasp on the use of complementary palettes and white space.
Some schools even offer UX courses, but that's often in the CIS or even Art catalog rather than the CS one.
Or, as everyone else suggests, if you prefer to be closer to the metal you should pass that work on to someone else.
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hey.. What's up??
I am developer and work on fiverr & oDesk.. If you have any problem in your website or you need creative website then you can feel free to contact me..
click here for visit my profile  skype: shobuj_bd2020
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There is NO perfect way to pick colors. My advice is to make friends with some artists, join some art forums like Behance or deviant art and make use of some of these talented people, because color sense is something you have, not something you can get.
I usually ask my wife who is a Virgo and she can look at a strip of cloth or sample, go shopping and come back with an exact match. It's something they are born to do naturally.
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The Report Viewer Web Control HTTP Handler has not been registered in the application's web.config file. Add <add verb="*" path="Reserved.ReportViewerWebControl.axd" type="Microsoft.Reporting.WebForms.HttpHandler, Microsoft.ReportViewer.WebForms, Version=11.0.0.0, Culture=neutral, PublicKeyToken=89845dcd8080cc91"> to the system.web/httpHandlers section of the web.config file, or add <add name="ReportViewerWebControlHandler" precondition="integratedMode" verb="*" path="Reserved.ReportViewerWebControl.axd" type="Microsoft.Reporting.WebForms.HttpHandler, Microsoft.ReportViewer.WebForms, Version=11.0.0.0, Culture=neutral, PublicKeyToken=89845dcd8080cc91"> to the system.webServer/handlers section for Internet Information Services 7 or later.
I am getting the above error while I try to run RDLC report.
The following things I have added to the solution:
1. ReportViewer control on the .aspx page.
<rsweb:ReportViewer ID="ReportViewer2" runat="server" Font-Names="Verdana" Font-Size="8pt" WaitMessageFont-Names="Verdana" WaitMessageFont-Size="14pt">
<LocalReport ReportPath="reports\Report1.rdlc">
<DataSources>
<rsweb:ReportDataSource DataSourceId="SqlDataSource1" Name="DataSet1" />
</DataSources>
</LocalReport>
</rsweb:ReportViewer>
2. Registered assembly on .aspx page
<%@ Register Assembly="Microsoft.ReportViewer.WebForms, Version=11.0.0.0, Culture=neutral, PublicKeyToken=89845DCD8080CC91" Namespace="Microsoft.Reporting.WebForms" TagPrefix="rsweb" %>
3. Added reference Microsoft.ReportViewer.WebForms to the project
4. In web.config tags added are:
<compilation debug="true" targetFramework="4.5">
<assemblies>
<add assembly="Microsoft.Build.Framework, Version=4.0.0.0, Culture=neutral, PublicKeyToken=B03F5F7F11D50A3A"/>
<add assembly="Microsoft.ReportViewer.WebForms, Version=11.0.0.0, Culture=neutral, PublicKeyToken=89845DCD8080CC91"/>
<add assembly="Microsoft.ReportViewer.Common, Version=11.0.0.0, Culture=neutral, PublicKeyToken=89845DCD8080CC91"/>
</assemblies>
<buildProviders>
<add extension=".rdlc" type="Microsoft.Reporting.RdlBuildProvider, Microsoft.ReportViewer.WebForms, Version=11.0.0.0, Culture=neutral, PublicKeyToken=89845dcd8080cc91"/>
</buildProviders>
</compilation>
Kindly provide the solution to resolve this error of RDLC report.
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Why cant you do it yourself. Its telling you clearly what to do.
The Report Viewer Web Control HTTP Handler has not been registered in the application's web.config file.
Add to the system.web/httpHandlers section of the web.config file, or add to the system.webServer/handlers section for Internet Information Services 7 or later.
Regards,
CodingK
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add below line in system.web in web.config file
<httphandlers>
<add path="Reserved.ReportViewerWebControl.axd" verb=" * " type="Microsoft.Reporting.WebForms.HttpHandler,
Microsoft.ReportViewer.WebForms, Version=10.0.0.0, Culture=neutral,
PublicKeyToken=b03f5f7f11d50a3a">
<add path="Reserved.ReportViewerWebControl.axd" verb="*" type="Microsoft.Reporting.WebForms.HttpHandler, Microsoft.ReportViewer.WebForms, Version=10.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a"
="" validate="false">
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