Multiple Pages Problem!!!

[j-dearden]

Hey

im having a little trouble creating a script to display so many results per page from my query. it displays the number of results in my case 2 but when i click next it says cannot execute query can any one provide any suggestiions im running out of ideas, any help would be most appreciated. my code is below

<?php

$conn = @mysql_connect("some_host", "some_user", "some_pwd")
or die("Sorry unable to connect to MySQL");

$rs = @mysql_select_db( "", $conn ) or die ( "Sorry unable to select specified database" );

$category = $POST['category'];
$criteria = $POST['criteria'];

// how many rows to show per page
$rowsPerPage = 2;

// by default we show first page
$pageNum = 1;

// if $GET['page'] defined, use it as page number
if(isset($GET['page']))
{
$pageNum = $_GET['page'];
}

// counting the offset
$offset = ($pageNum - 1) * $rowsPerPage;

$query = "SELECT * FROM customer_table WHERE $category LIKE '%$criteria%' LIMIT $offset, $rowsPerPage";
$result = mysql_query( $query, $conn ) or die ( "Sorry unable to execute query" );

while($row = @mysql_fetch_array($result))
{
echo "<table><tr>
<td width='30' height='30' align='center'><input name='id' type='radio' value='$row[customer_id]'></td>
<td width='438' height='30'><font face='arial' size='2'>&nbsp;&nbsp;$row[first_name] $row[surname]</font></td>
</tr></table>";
}

// how many rows we have in database
$query   = "SELECT COUNT(*) AS numrows FROM customer_table";
$result  = mysql_query($query) or die('Error, query failed');
$row     = mysql_fetch_array($result, MYSQL_ASSOC);
$numrows = $row['numrows']; 

// how many pages we have when using paging?

$maxPage = ceil($numrows/$rowsPerPage);

// print the link to access each page
$self = $_SERVER['PHP_SELF'];
$nav = '';
for($page = 1; $page <= $maxPage; $page++)
{
if ($page == $pageNum)
{
$nav .= " $page "; // no need to create a link to current page
}
else
{
$nav .= " <a href=\"$self?page=$page\">$page</a> ";
}

}

// creating previous and next link
// plus the link to go straight to
// the first and last page

if ($pageNum > 1)
{
$page = $pageNum - 1;
$prev = " <a href=\"$self?page=$page\">[Prev]</a> ";

$first = " <a href=\"$self?page=1\">[First Page]</a> ";

}
else
{
$prev = '&nbsp;'; // we're on page one, don't print previous link
$first = '&nbsp;'; // nor the first page link
}

if ($pageNum < $maxPage)
{
$page = $pageNum + 1;
$next = " <a href=\"$self?page=$page\">[Next]</a> ";

$last = " <a href=\"$self?page=$maxPage\">[Last Page]</a> ";

}
else
{
$next = '&nbsp;'; // we're on the last page, don't print next link
$last = '&nbsp;'; // nor the last page link
}

// print the navigation link
echo $first . $prev . $nav . $next . $last;

[kakki]

for debugging, you can
change this

... mysql_query( "whatever" );

to

// ... mysql_query
echo ( "whatever" );

( just comment the mysql_query command and echo the current statement )
afterwards you can see on the screen what went wrong( or if that wont help ),
execute that in phpmyadmin to see the mysql-error code

[habbardone]

To help the de-bugging:

also using r_print($_GET)

and r_print($_POST)

will help as it shows what is in the arrays.

[bogu]

habbardone wrote:

To help the de-bugging:

also using r_print($_GET)

and r_print($_POST)

will help as it shows what is in the arrays.

Can u add a link for r_print, from php manual?

[kakki]

i think the command is "print_r"

[bogu]

kakki wrote:

i think the command is "print_r"

I know, just trying to make him see the error ... 😃