pagination in menu

**lakshmiyb** · 08:01 AM

hi,
this is my menu code
when pagination is clicked, the pagination page will be displayed in div class of this page.
but when link in pagination is clicked it is redirecting to pagination page,

but i want it to be displayed in menu page itself.
please help,

i am waiting for the result.
thanking u in advance.

menu.php

PHP Code:


			
<ul id="nav" class="dropdown dropdown-linear">
 
<li>
<a href="#" id="pagination">pagination</a>
</li>
 
</ul>
 
<table width="100%" border="1" cellpadding="0" cellspacing="0">
<tr valign="top">
<td>
<div id="result" class="functions">
 
</div>
</td>
</tr>
</table>
<script type="text/javascript" src="jquery-1.2.6.min.js"></script>
<script type="text/javascript">
$.ajaxSetup ({
cache: false
});
var ajax_load = "<img src='images/load.gif' class='loading' alt='loading...' />";
//load() functions
 
$("#pagination").click(function(){
$("#result")
.html(ajax_load).load("pagination.php");
});
 
</script>
 
 
 
and
paginationpage.php
 
$page_name="pagination.php"; 
$start=$_GET['start'];
if(strlen($start) > 0 and !is_numeric($start)){
echo "Data Error";
exit;
}
$eu = ($start - 0); 
$limit = 10; $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 "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;".$row['languages'] ."";
}
echo "</table>";
if($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>";
}
?>
</div>

**lizciz** · 10-19-2009, 07:38 AM

I think you should change this

$("#result").html(ajax_load).load("pagination.php" );

into this

$("#result").load("pagination.php");

**lakshmiyb** · 10-19-2009, 07:42 AM

no ,it is not displaying what i want

**lizciz** · 10-19-2009, 08:49 AM

So what is it displaying then?

**lakshmiyb** · 10-19-2009, 08:57 AM

ajax_load is an image of loading, when there is no page it will show the load image, so there will never be any change with r with out (ajax_load)

**lizciz** · 10-19-2009, 11:06 AM

I don't know if this will make any difference, but the only things I can come up with is:

1) Move your script to the <head> section, instead of in the <body>. All javascript should normally be there.

2) Surround all your jQuery code with the ready() function. When I used jQuery I learned that all code must be in there, to make sure the page is ready before calling any jQuery functions.

Code:

$(document).ready(function(){
   // All your code goes here
});

**lakshmiyb** · 10-20-2009, 03:28 AM

no change in the result

i think there must be change in
print "<a href='$page_name?start=$back'>PREV</font></a>";

because when prev or next is clicked it is moving to $page_name i.e, pagination page but i want it to redirect it to menu.php result in div class

**lizciz** · 10-20-2009, 07:40 AM

Aha, I think I just understood the problem :P
When you click "pagination" the pagination page opens correctly, but when you click a link in the pagination page it reloads the entire page? I thought it was the first step that didn't work.

Try adding some jQuery to the pagination page, that makes all the links open in a div. Something like

Code:

$(document).ready(function(){
   $(".link").click(){
      $("#your_div").load($(this).attr("href"));
      return false;
   };
});

And add the class "link" to all your links in pagination.php, as in

PHP Code:


			
echo '<a href="some_page.php" class="link">Some link</a>';

I'm not really that good with jQuery, but I think this will make all the links on the page load the address given by the links' href attribute, into your div with the id "your_div".

**lakshmiyb** · 10-21-2009, 03:03 AM

NO, there is no change in my output
the link is in pagination page but not in to menu page.

**lizciz** · 10-21-2009, 07:30 AM

I just tried this at my computer and got it to work. I'll post my example code and you can try to work something out from it.

Open page1.htm. The first link will open page2 in a div, then the links in page2 will open page3 (and 4, 5) in the seccond div.

page1.htm

HTML Code:

<html>

<head>
<title>Page1</title>
<script type="text/javascript" src="jquery-1.3.2.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
   $("#link1").click(function(){
      $("#div1").load("page2.htm");
   });
});
</script>
</head>

<body>
<a href="#" id="link1">Click this to bring up page2</a>
<br />
<div id="div1" style="width:600px; height:200px; border:1px solid #000;"></div>
<br /><br />

Page 3, 4 and 5 will open here
<div id="div2" style="width:600px; height:300px; border:1px solid #000;"></div>
</body>

</html>

page2.htm

HTML Code:

<script type="text/javascript">
$(document).ready(function(){
   $(".link2").click(function(){
      $("#div2").load($(this).attr("href"));
      return false;
   });
});
</script>
<br />
<a href="page3.htm" class="link2">Click to open page 3</a>
<a href="page4.htm" class="link2">Click to open page 4</a>
<a href="page5.htm" class="link2">Click to open page 5</a>
<br />

And then I have page3.htm, page4.htm and page5.htm with only

HTML Code:

<h1>page 3/4/5</h1>

And also, of course, don't forget the jquery file.

**lakshmiyb** · 10-21-2009, 09:04 AM

thank u very much,

I got it working in same div class with out taking another div class

i am greatfull to u

thanku once again for a great help

**lakshmiyb** · 10-21-2009, 09:07 AM

hi,
i need sql query for searching the all the tables of database and displaying the result,
if u can help me thanku.

**lizciz** · 10-21-2009, 07:34 PM

You'll get much results if you search Google for "php mysql search" or similar, but basically what you want to do is

SELECT * FROM your_table WHERE field LIKE '$search'

where field is the database table field name you want to search in and $search is the word that the user entered to search for.

**lakshmiyb** · 10-22-2009, 03:27 AM

thank u for the reply,

i have the query for the table search,
i have a doubt about the query
how to search the whole database of the site, the search which will be placed at the main page of the site

**lizciz** · 10-22-2009, 04:59 AM

Well it all depends on how your database is structured and how many tables/fields you want to search in. For searching in several fields you can just expand it into

SELECT * FROM your_table WHERE field1 LIKE '$search' OR field2 LIKE '$search' OR field3 LIKE '$search' ...

As I siad, you should be able to find alot of info of this by searching Google.

**lakshmiyb** · 10-22-2009, 05:25 AM

ok, thanku very much