Pagination Error

[yash87]

Hi, Im trying to do pages for my search result. However, when I click on the page number. This error appears (below) :

 

Notice: Undefined index: search in C:\wamp\www\I-Document\new.php on line 8

ERROR: Select from dropdown

 

This message only should appear when there is no input in dropdown and no search input.

Im not sure how to correct this. Please help! Thank u.

 

new.php

<?php

//connecting to the database
include 'config.php';

if (!isset($_POST['submit'])) {
        
    $search = mysql_escape_string($_POST['search']);
$dropdown = empty($_POST['dropdown'])? die ("ERROR: Select from dropdown") : mysql_escape_string($_POST['dropdown']); 

	//max displayed per page
	$per_page = 10;

	//get start variable
	$start = $_GET['start'];

	//count records
	$record_count = mysql_num_rows(mysql_query("SELECT * FROM document"));

	//count max pages
	$max_pages = $record_count / $per_page; //may come out as decimal

	if (!$start)
	   $start = 0;

	//display data
	$query = mysql_query("SELECT * FROM document WHERE $dropdown LIKE '%$search%' LIMIT $start, $per_page");


	echo "<b><center>Search Result</center></b><br>";
     
	$num=mysql_num_rows($query);


	if ($num==0)
			echo "No results found";
		 else
		 {
			echo "$num results found!<p>"; 
		  }
		  
	echo "You searched for <b>$search</b><br /><br /><hr size='1'>";
  
	  echo "<table border='1' width='600'>
			<th>File Reference No.</th>
			<th>File Name</th>
			<th>Owner</th>
			<th>Borrow</th>
			 </tr>";

		while ($rows = mysql_fetch_assoc($query))  
		{
		  echo "<tr>";
		  echo "<td>". $rows['file_ref']  ."</td>";
		  echo "<td>". $rows['file_name'] ."</td>";
		  echo "<td>". $rows['owner'] ."</td>";
		  echo "<td><a href=add_borrower.php?id=" . $rows['id'] . ">Borrow</a></td>";
		  echo "</tr>";
		}
		  echo "</table>"; 
   


	//setup prev and next variables
	$prev = $start - $per_page;
	$next = $start + $per_page;

	//show prev button
	if (!($start<=0))
		   echo "<a href='new.php?start=$prev'>Prev</a> ";

	//show page numbers

	//set variable for first page
	$i=1;

	for ($x=0;$x<$record_count;$x=$x+$per_page)
	{
	 if ($start!=$x)
		echo " <a href='new.php?start=$x'>$i</a> ";
	 else
		echo " <a href='new.php?start=$x'><b>$i</b></a> ";
	 $i++;
	}
}
	//show next button
	if (!($start>=$record_count-$per_page))
		   echo " <a href='new.php?start=$next'>Next</a>";


?>

[mike12255]

ok im  going to take a shot at this but its late and im tired so hope I can still help.

 

this line:

 

  $search = mysql_escape_string($_POST['search'])

 

is pointless post is used to collect values sent using post or send values however a little above that line you typed this:

 

if (!isset($_POST['submit'])) {

 

so your saying that the submit button has not been pushed yet meaning a POST value has not been sent. Or have you redirected yourself to this page using a form on another page?

[yash87]

Nup, the search codes are fine.Its just the pagination part which is not working. When i click page '2', instead of showing the result.. it shows the error above.

[mike12255]

here is my new theory

 

when going from your first page to your second you lose:

($_POST['search']);

 

try setting it to a variable and passing it through the url using.