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

Monday, December 3rd, 2012 Programming

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="”&quot;">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.

countries-dropdown-list-values-from-database

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

values-fetched-from-db-on-country-selection

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

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

Friday, August 5th, 2011 PHP, Programming

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:

Creating Link Boxes in CSS

Saturday, September 11th, 2010 Programming

[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>

<body>
<a href="default.asp" target="_blank">This is a link</a>
</body>
</html>

[/html]

Output:

The link box will be created with the Specified Styles.

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

Monday, August 30th, 2010 Programming

[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!!”.

External css

Monday, August 23rd, 2010 Programming

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">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<link rel="stylesheet" type="text/css" href="external.css" />

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />

<title>CSS</title>

</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:

Explode function in php

Monday, August 23rd, 2010 PHP

How to split a name using explode function?
1. Get the name in a variable
2. Specify the position from which you need to split the name.
3. If you need to split the name after an underscore, use “_”.
4. If you need to split the name after @, use “@”.

[php]

<?php

$name  = "john_abraham@gmail.com";
$pieces = explode("_", $name);
echo "firstname = $pieces[0]"."<br />";

$pieces = explode("@", $name);
echo "username = $pieces[0]";

?>

[/php]

Output will be displayed as:
firstname = john
username = john_abraham

TAGS:

IF Statement In PHP

Monday, August 23rd, 2010 PHP

[php]

<html>

<head>
<title>IF statements in PHP.</title>
</head>
<body>

<?php

$name = "Rose";

if ( $name == "Rose" )

echo "Your name is Rose!<br />";

else

echo "Welcome to my homepage!";
?>
</body>
</html>

[/php]

Output:

If the Given input is Rose,the output will be

Your name is Rose!

Else

Welcome to my homepage!

Switch Case in Java Script

Monday, August 16th, 2010 Programming

[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!

Java Script for Mouse Events

Monday, August 9th, 2010 Programming

[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>

<body>
<img border="0"   src="E:\pictures\baby.jpg" id="b1"
onmouseover="mouseOver()" onmouseout="mouseOut()" /></a>
</body>
</html>

[/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”.

To upload a file using php

Thursday, August 5th, 2010 PHP

Steps:

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

2. Create new php file and copy uploader.php code and save it as uploader.php.

3. Run upload.html, then it will display the html data.

4. Click browse button and select any file or image and press upload button.

5. Then it will transfer that specific file or image to the created folder and display the output as file is uploaded.

upload.html:

[php]

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>upload</title>
</head>

<body>
<form enctype="multipart/form-data" method="post" action="uploader.php">
<input type="file" name="file" />
<input type="submit" name="submit" value="upload" />
</form>
</body>
</html>

[/php]

It displays the output as:


uploader.php
[php]</form><form action="uploader.php" method="post" enctype="multipart/form-data"><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head><body>
<?php
if(isset($_REQUEST[‘submit’]))
{
$to="uploaded/".$_FILES[‘file’][‘name’];
move_uploaded_file($_FILES[‘file’][‘tmp_name’],$to);
echo "file is uploaded";

}
?>
</body>
</html>

[/php]

//create a new folder uploaded

TAGS:

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