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having numbered pages


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#1 kenzo

kenzo
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Posted 14 August 2006 - 12:22 PM

im trying to have numbered pages in my results page, but for some reason i keep getting the mysql_fetch_array(): error message, i dont know what im doing wrong, any ideas, the code is below.....



	<? 
			
				// mumber of pages to be displayed....
				
				$display = 10;
				
				//how many Pages Kenzo?
				
				if (isset($_GET['np']))
				{
				
				$num_pages = $_GET['np'];
				}
			
				else
				
				{
					// count this s.....
				
					$query = " SELECT COUNT (*) FROM tablex ORDER BY id ASC";
					$result = mysql_query ($query);
					$row = mysql_fetch_array ($result, MYSQL_NUM);
					$num_records = $row[0];
					
					
					}
				
				
					// how many pages?
				
					if ($num > $display)
					{
					$num_pages = ceil ($num_records/$display);
					}
					else
					{
					$num_pages = 1;
					}
				
				
				
			
			
				// check 
				
				if (isset($_GET['s']))
				{
				$start = $_GET['s'];
				}
				else
				{
				$start = 0;
				}
			
			
			
				// catch location
				$location =  $_POST['cmp'] ;
				$category =  $_POST['cat'] ;
			
			
			
 				 $query = mysql_query("
   				 SELECT *
    			 FROM
        		 tablex
				 WHERE
				 location = '$location' AND 
				 category = '$category'
				 ORDER BY
				 id
    			ASC
				LIMIT $start, $display
    			 ",$db);
				 }
		
				
				$num = mysql_num_rows($query);
				
				
				
				// number of comapnies check.......
				if ($num > 0) {
				echo "<span class='redno'> $num $category</span> in <span class='redno'> $location </span> displayed........<br><br>";
				}
				else
				{
				echo "Search returned no companies....Sorry.";
				}
  				?>
				


  <br><br>			




		
		<?
		
		

		// print companies and duplicate rows......
		while ($row = mysql_fetch_array($query)) {
		
		
		echo "<div class='rsbar'></div><br><table class='phptable'><tr><td><span class='mheader'>COMPANY NAME :</span></td><td>" . $row['companyname'] . " </td></tr><tr><td><span class='mheader'>LOCATION :</span></td><td>" . $row['location'] . "</td></tr><tr><td><span class='mheader'>NATURE OF BUSINESS :</span></td><td>" . $row['category'] . "</td></tr><tr><td><span class='mheader'></span></td><td> <a href='../login_pass.php'>Contact Details</a></td></tr></table><br>";
		

		
		}
		
		mysql_close();
		
		
		
		
		// Make Links........
		
		if ($num_pages > 1)
		{
		echo "<br>";
		
		$current_page = ($start/$display) + 1;
		
		
					// make previous button
					
					if ($current_page != 1) 
					{
					echo "<a href='catcmpresults.php?s=" . ($start - $display) . "&np=" . $num_pages . "'> Previous </a>";
					
					}
		
		
		
		
		
		// make numbered pages
		
		for ($i = 1; $i <= $num_pages; $i++)
				{
				if ($i != $current_page)
							{
							echo "<a href='catcmpresults.php?s=" . (($display * ($i - 1))) . "&np=" . $num_pages . "'>" . $i . "</a>";
							
							}
							else
							{
							echo $i . " ";
							}
				
				}
		
					
					// make next button
					
					if ($current_page != $num_pages) 
					{
					echo "<a href='catcmpresults.php?s=" . ($start - $display) . "&np=" . $num_pages . "'> Previous </a>";
					
					}
		
		echo "<br>";
		
		}
		
		?>




#2 shocker-z

shocker-z
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Posted 14 August 2006 - 12:25 PM

change $result = mysql_query ($query);

for

$result = mysql_query ($query) or die("Failed with the following error: ".mysql_error());


And paste back the error :)

Regards
Liam
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