pagination inside the div class with menu

lakshmiyb

hi,
this is my menu code

menu.php
<html>
<body>
<table width="100%" border="0" cellpadding="0" cellspacing="0">
<tr valign="top">
<td align="center" valign="top">
<ul id="nav" class="dropdown dropdown-linear">

<li>
<a href="#" id="pagination">pagination</a>

</li>

</ul>
</td>
</tr>
</table>
<p> </p>
<table width="100%" border="1" cellpadding="0" cellspacing="0">
<tr valign="top">
<td>
<div id="result" class="functions">

</div>
</td>
</tr>
</table>

//load() functions

$("#pagination").click(function(){
$("#result").load("pagination.php");
});

</script>

</body>
</html>

i used css to display the result

and
pagination.php

<?php
//connection to db will goes here
$page_name="zpagination.php";
$start=$_GET['start'];
if(strlen($start) > 0 and !is_numeric($start)){
echo "Data Error";
exit;
}

$eu = ($start - 0);
$limit = 10; // No of records to be shown per page.
$this1 = $eu + $limit;
$back = $eu - $limit;
$next = $eu + $limit;

$query2=" SELECT * FROM footer_languages";
$result2=mysql_query($query2);
echo mysql_error();
$nume=mysql_num_rows($result2);

$query=" SELECT * FROM footer_languages limit $eu, $limit ";
$result=mysql_query($query);
echo mysql_error();

while($row = mysql_fetch_array($result))
{
print "     ".$row['languages'] ."";
}
echo "</table>";
if($nume > $limit ){

echo "<table align = 'center' width='50%'><tr><td align='left' width='30%'>";
if($back >=0) {
print "<a href='$page_name?start=$back'>PREV</font></a>";
}
echo "</td><td align=center width='30%'>";
$i=0;
$l=1;
for($i=0;$i < $nume;$i=$i+$limit){
if($i <> $eu){
echo " <a href='$page_name?start=$i'>$l</font></a> ";
}
else { echo "$l</font>";}
$l=$l+1;
}

echo "</td><td >";

if($this1 < $nume) {
print "<a href='$page_name?start=$next'>NEXT</font></a>";}
echo "</td></tr></table>";

}
?>

when pagination is clicked in menu the out put from pagination page will be displayed inside div class

but when next or prev is clicked it is moving to pagination page
i need all the results to be displayed inside div given in menu.php

can any one heplp me in this , i am waiting for replay

Thanking u in advance

big_nerd

lakshmiyb,

I would recommend you paste an example output, between the original and once you click on the next button.

Also, it will help us who have limited time if you are able to edit your post and put the code within the tags, it keeps proper formatting and helps us to see if there is an error somewhere, without having to copy & paste into one of our own editors/notepad and space it properly.

For future, please keep PHP code in between the [ php] [ /php] bbcode tags (without spaces).

lakshmiyb

menu.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=utf-8" />
<title>MENU</title>
<link href="css/dropdown.linear.css" media="all" rel="stylesheet" type="text/css" />
<link href="css/default/default.ultimate.linear.css" media="all" rel="stylesheet" type="text/css" />
</head>

		<li>
			<a href="#" id="pagination">pagination</a>

		</li>


	</ul>
</td>

</tr>
</table>
<p> </p>
<table width="100%" border="1" cellpadding="0" cellspacing="0">
<tr valign="top">
<td>
<div id="result" class="functions">
OUT PUT WILL BE DISPLAYED HERE
</div>
</td>
</tr>
</table>

//load() functions

$("#pagination").click(function(){
$("#result").load("pagination.php");
});

</script>
</body>
</html>

Pagination.php

<!doctype html public "-//w3c//dtd html 3.2//en">
<html>
<head>
<title>paging</title>


$ebits = ini_get('error_reporting');
error_reporting($ebits ^ E_NOTICE);
//require "config.php";           // All database details will be included here 
$servername='localhost';     // Your MySql Server Name or IP address here
$dbusername='root';                // Login user id here
$dbpassword='';                // Login password here
$dbname='reelload';     // Your database name here

connecttodb($servername,$dbname,$dbusername,$dbpassword);
function connecttodb($servername,$dbname,$dbuser,$dbpassword)
{
global $link;
$link=mysql_connect ("localhost","root","");
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("reelload",$link) or die ("could not open db".mysql_error());
}



$page_name="pagination.php"; //  If you use this code with a different page ( or file ) name then change this 
$start=$_GET['start'];
if(strlen($start) > 0 and !is_numeric($start)){
echo "Data Error";
exit;
}

$eu = ($start - 0); 
$limit = 10;                                 // No of records to be shown per page.
$this1 = $eu + $limit; 
$back = $eu - $limit; 
$next = $eu + $limit; 


/////////////// WE have to find out the number of records in our table. We will use this to break the pages///////
$query2=" SELECT * FROM movie_languages";
$result2=mysql_query($query2);
echo mysql_error();
$nume=mysql_num_rows($result2);
/////// The variable nume above will store the total number of records in the table////

/////////// Now let us print the table headers ////////////////


////////////// Now let us start executing the query with variables $eu and $limit  set at the top of the page///////////
$query=" SELECT * FROM movie_languages limit $eu, $limit ";
$result=mysql_query($query);
echo mysql_error();

//////////////// Now we will display the returned records in side the rows of the table/////////
while($row = mysql_fetch_array($result))
{
print "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;".$row['languages'] ."";
}
echo "</table>";
////////////////////////////// End of displaying the table with records ////////////////////////

/////////////////////////////// 
if($nume > $limit ){ // Let us display bottom links if sufficient records are there for paging

/////////////// Start the bottom links with Prev and next link with page numbers /////////////////
echo "<table align = 'center' width='50%'><tr><td  align='left' width='30%'>";
//// if our variable $back is equal to 0 or more then only we will display the link to move back ////////
if($back >=0) { 
print "<a href='$page_name?start=$back' >PREV</font></a>"; 
} 
//////////////// Let us display the page links at  center. We will not display the current page as a link ///////////
echo "</td><td align=center width='30%'>";
$i=0;
$l=1;
for($i=0;$i < $nume;$i=$i+$limit){
if($i <> $eu){
echo " <a href='$page_name?start=$i' >$l</font></a> ";
}
else { echo "$l</font>";}        /// Current page is not displayed as link and given font color red
$l=$l+1;
}

echo "</td><td >";
///////////// If we are not in the last page then Next link will be displayed. Here we check that /////
if($this1 < $nume) { 
print "<a href='$page_name?start=$next'>NEXT</font></a>";} 
echo "</td></tr></table>";

}
// end of if checking sufficient records are there to display bottom navigational link.

</script>
</body>
</html>

this is what i am using now,
when next is clicked in the output of the menu it is moving to pagination page
but i want the output of the next link in menu page itself
thanku for ur response

lakshmiyb

the problem is solved