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Programming

I am using http://www.erichynds.com/jquery/jquery-ui-multiselect-widget/

Samples here

Initially it seemed tough to get it working as a beginner, but eventually I managed to successfully get it to work.

Step 1:

Download the Jquery Code Or You Can Include it From CDN (Content Delievery Network)

For multiple select drop-down, We need to include the following resource,

Download links are

After including the above files your code will look like:

[php]
<link rel="stylesheet" type="text/css" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1/themes/ui-lightness/jquery-ui.css" />
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.js">
</script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1/jquery-ui.min.js"></script>
<script type="text/javascript" src="../src/jquery.multiselect.js"></script>
[/php]

Step 2:

Use JavaScript Function to Trigger Multi-Select Event

[php]
<script>
$(document).ready(function(){
$("#check_list").multiselect();
});
</script>
[/php]

For customizing multi-select,We have to pass in a object with one or more options like header.height etc. For all available options Click Here

Step 3:

In <select> tag define multiple attribute for selecting multiple options and define name in array format as specified below.

Note that at the end of check_list in the <name>tag [] symbol is used,that denotes the array

[php]
<select name="check_list[]" id="check_list" multiple="multiple">
<option value="1"> Apple </option>
<option value="2"> Orange </option>
<option value="3"> Mango </option>
<option value="4"> Grapes </option>
<option value="5"> Strawberry </option>
</select>
[/php]

Step 4:

When you have embedded this multi-select dropdown in your form,

Output will be Displayed as

Step 5:

Onclicking submit button the value selected in drop-down is posted as an array.Let us assume we are using PHP to get this value.

[php]
if(!empty($_POST[‘check_list’])){ //Checking if check_list control is empty or not
foreach($_POST[‘check_list’] as $check){ //Looping through the array of values in foreach
print_r($check);//return the value of the variable
$test = implode(‘,’,$check);// implode tha array value to pass it in a query
}
}
[/php]
Your complete code will look like this
[html]
<html lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>jQuery MultiSelect Dropdown</title>
<link rel="stylesheet" type="text/css" href="../jquery.multiselect.css" />
<link rel="stylesheet" type="text/css" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1/themes/ui-lightness/jquery-ui.css" />
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.js">
</script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1/jquery-ui.min.js"></script>
<script type="text/javascript" src="../src/jquery.multiselect.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#check_list").multiselect();
});
</script>
</head>
<body id="test">
<form id="multiple" method="post" enctype="multipart/form-data">
<select title="Basic example" multiple="multiple" name="example-basic" size="5">
<option value="1">Apple</option>
<option value="2">Orange</option>
<option value="3">Mango</option>
<option value="4">Grapes</option>
<option value="5">Strawberry</option>
</select>
<input type="submit" name="submit" value="submit" />
</form>
<?php
if(isset($_POST[‘submit’])){
if(!empty($_POST[‘check_list’])){
foreach($_POST[‘check_list’] as $check){
print_r($check);
$test = implode(‘,’,$check);
}
}
$fruits = mysql_query("SELECT * FROM fruits WHERE fruit_id IN(".$test.")");
$fetch_fruits = mysql_fetch_array($fruits);
$name = $fetch_fruits[‘fruit_name’];
$desc = $fetch_fruits[‘desc’];
echo ‘<div>"’.$name.’"</div>’;
echo ‘<div>"’.$desc.’"</div>’;
}
?>
</body>
</html>
[/html]

For working example of Multi-select dropdown checkbox Click Here

Must know before reading this article

PHP Basics
HTML Basics
Jquery

Programming No Comments

Tutorial to get values from database on selection of dropdown in PHP

PHP Fetch Tutorial

Fetching Data From Database on selection of dropdown value

Step 1:

Get the list of countries from database and populate in drop-down.

[php]
$ctry_qry = mysql_query(“SELECT * FROM countries ORDER BY name;”);?>
//Calling Function
<select name="”ctry”"> <option value="”"">Select Country</option></select>
<select name="”ctry”"><option value="”<?PHP">”><!–?PHP echo $ctry[‘name’]; ?–></option></select>
[/php]

Consider that we choose India from the dropdown list.

Step:2

Add a simple javascript function which will run onchange

[javascript]
function reloaddata(ctryid) {
window.location.href="http://localhost/rotatetheglobe/admin/try.php?ctryid=" + ctryid;
}
[/javascript]

Step:3

Reload the page in JS and pass the currently selected country id

Step:4

Get the country id from query string and fetch the required values from the database

[php]
<?PHP</pre>
<div><code>if</code><code>(isset(</code><code>$_REQUEST</code><code>[</code><code>’ctryid'</code><code>])) {</code></div>
<div><code>$ctryinfo_qry</code> <code>= mysql_query(</code><code>"SELECT * FROM country_info where ctry_root_id=’"</code><code>.</code><code>$_REQUEST</code><code>[‘ctryid</code><code>’]."'</code> <code>ORDER BY ctry_info_id ;");</code></div>
<div><code>?></code></div>
<pre>[/php]

Fetch array is used to fetch the respective row from record set . Using fetch array in while loop is used get a new row of MySQL information until the condition is TRUE.When there are no more rows the function will return FALSE causing the while loop to stop!

[php]
<textarea id="msgpost" name="ctry_about" rows="10" cols="50"> <?php echo $ctry_inf[‘ctry_about’];?></textarea>
<!–?php } ?–>

[/php]

CLICK HERE to view the complete output

Possible Errors to occur:

JS Error
Couldn’t get the value from the query string
Page not found

Must Follow

Make Sure the querystring variable and the $_REQUEST variable is same

Make sure your onchange function is same name as that you call in your code.

Must know before reading this article

PHP Basics
HTML Basics
DB Handling

Coming Up:

How to configure a site to work in your local host when you are running windows

Programming No Comments

Code to get Title and Link in WordPress RSS Feed using PHP

You would often want to display the RSS feed of your blog in your personalized home page or your professional page. This can be achieved, if you have basic knowledge on PHP and HTML.

Just copy and paste the code below in the section where you want the feed to appear:

[php]
<?PHP

$con = file_get_contents(‘http://www.xxxxxx.com/blog/feed/’);/* Feed Url */
$title = getlinks("<title>", "</title>", $con); // gets Title and stored in array
$url = getlinks("<link>", "</link>", $con); // gets Url and stored in array

echo "<ul>";
for($i=0; $i < 6; $i++) {
echo ‘<li><a href="’.$url[$i].’" title="’.$title[$i].’" >’.urldecode($title[$i]).'</a></li>’;
}
echo "</ul>";

function getlinks($s1,$s2,$s) {
$myarray=array();
$s1 = strtolower($s1);
$s2 = strtolower($s2);
$L1 = strlen($s1);
$L2 = strlen($s2);
$scheck=strtolower($s);

do {
$pos1 = strpos($scheck,$s1);
if($pos1!==false) {
$pos2 = strpos(substr($scheck,$pos1+$L1),$s2);
if($pos2!==false) {
$myarray[]=substr($s,$pos1+$L1,$pos2);
$s=substr($s,$pos1+$L1+$pos2+$L2);
$scheck=strtolower($s);
}
}
}
while (($pos1!==false)and($pos2!==false));
return $myarray;
}

[/php]

Tags: PHP code for Rss Feed

PHP, Programming No Comments

Creating Link Boxes in CSS

[html]

<html>
<head>
<style type="text/css">
a:link,a:visited
//Styles to be applied in link Box

{
display:block;
font-weight:bold;
color:#FFFFFF;
background-color:#98bf21;
width:120px;
text-align:center;
padding:4px;
text-decoration:none;
}
a:hover,a:active
{
background-color:#7A991A;
}
</style>
</head>

[/html]

Output:

The link box will be created with the Specified Styles.

Programming 1 Comment

To set an background Image and repeat the image Horizontally using CSS

[html]

<html>
<head>
<style type="text/css">
body
{
background-image:urlC:\Users\systemuser\Pictures\flower.jpg’);
background-repeat:repeat-x;
}
</style>
</head>
<body>
<h1>Hai Friends!!</h1>
</body>
</html>

[/html]

Output:

The image in the given URL will be set as the Background with the Text “Hai Friends!!”.

Programming No Comments

External css

Steps:

1:Create new html file and copy link.html code and save it as link.html.

2.Create new css file and copy external.css code and save it as external.css.

3.link.html will connect css externally.

4.Run link.html to get output.

link.html:

[php]
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

</head>

<body>
<h3 align="center">External css</h3>

<p>This page uses external CSS.</p>

<p>p.first class code</p>

</body>

</html>

[/php]

external.css:

[php]

body{ background-color:#00CCCC;}

p { color:#FFFFFF;

}

h3{ color:#000000;

text-decoration:underline;

text-transform:uppercase;

}

p.first{

color:#33FFCC;

}

[/php]

output:

External css

This page uses external CSS.

p.first class code

Tags: css

Programming No Comments

Switch Case in Java Script

[php]

<html>
<body>
<head>
<title>Switch Case </title>
</head>
<body>
<script type="text/javascript">
var d = new Date();
theDay=d.getDay();
switch (theDay)
{
case 5:
document.write("<b>Finally Friday</b>");
break;
case 6:
document.write("<b>Super Saturday</b>");
break;
case 0:
document.write("<b>Sleepy Sunday</b>");
break;
default:
document.write("<b>I’m really looking forward to this weekend!</b>");
}
</script>
</body>
</html>

[/php]

/*This JavaScript will generate a different greeting based on what day it is. Note that Sunday=0, Monday=1, Tuesday=2, etc.*/

Output:

I’m really looking forward to this weekend!

Programming No Comments

Java Script for Mouse Events

[php]

<html>
<head>
<script type="text/javascript">
function mouseOver()
{
document.getElementById("b1").src ="b_blue.gif";
}
function mouseOut()
{
document.getElementById("b1").src ="b_pink.gif";
}
</script>
</head>

[/php]

output:

The function mouseOver() causes the image to shift to “b_blue.gif”.

The function mouseOut() causes the image to shift to “b_pink.gif”.

Programming 1 Comment

Program to insert Background Image

[php]

<html>
<head>
<style type="text/css">
body {background-image:url(‘systeuser/desktop/bc.jpg’);}/*inserting background image*/
</style>
</head>
<body>
<h1>Hello!</h1>
</body>
</html>

[/php]

output:

Hello!

(If u run this code in Internet Explorer,the background image will be displayed with “Hello!” Message).

Programming No Comments

Programming

Validation Using Javascript

How To Use Jquery Mutliselect Dropdown

Tutorial to get values from database on selection of dropdown in PHP

PHP Fetch Tutorial

Fetching Data From Database on selection of dropdown value

CLICK HERE to view the complete output

Code to get Title and Link in WordPress RSS Feed using PHP

Creating Link Boxes in CSS

To set an background Image and repeat the image Horizontally using CSS

External css

Switch Case in Java Script

Java Script for Mouse Events

Program to insert Background Image

Blogroll

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