i get this error!!!

i have this code, i tried to put it with my code but it get some error.

<?

	include 'connect.php';

	$tbl_name="news";

	$adjacents = 3;
	
	$query = "SELECT COUNT(*) as num FROM $tbl_name";
	$total_pages = mysql_fetch_array(mysql_query($query));
	$total_pages = $total_pages[num];
	
	$targetpage = "news.php"; 
	$limit = 2;
	$page = $_GET['page'];

	if($page) 
		$start = ($page - 1) * $limit;
	else
		$start = 0;
	
	$sql = "SELECT * FROM $tbl_name order by a_id asc LIMIT $start, $limit";
	$result = mysql_query($sql);

	if ($page == 0) $page = 1;
	$prev = $page - 1;
	$next = $page + 1;
	$lastpage = ceil($total_pages/$limit);
	$lpm1 = $lastpage - 1;
	
$pagination = "";
	if($lastpage > 1)
	{	
		$pagination .= "<div class=\"pagination\">";
		//previous button
		if ($page > 1) 
			$pagination.= "<a href=\"$targetpage?page=$prev\">« previous</a>";
		else
			$pagination.= "<span class=\"disabled\">« previous</span>";	
		
		//pages	
		if ($lastpage < 7 + ($adjacents * 2))	//not enough pages to bother breaking it up
		{	
			for ($counter = 1; $counter <= $lastpage; $counter++)
			{
				if ($counter == $page)
					$pagination.= "<span class=\"current\">$counter</span>";
				else
					$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";					
			}
		}
		elseif($lastpage > 5 + ($adjacents * 2))	//enough pages to hide some
		{
			//close to beginning; only hide later pages
			if($page < 1 + ($adjacents * 2))		
			{
				for ($counter = 1; $counter < 4 + ($adjacents * 2); $counter++)
				{
					if ($counter == $page)
						$pagination.= "<span class=\"current\">$counter</span>";
					else
						$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";					
				}
				$pagination.= "...";
				$pagination.= "<a href=\"$targetpage?page=$lpm1\">$lpm1</a>";
				$pagination.= "<a href=\"$targetpage?page=$lastpage\">$lastpage</a>";		
			}
			//in middle; hide some front and some back
			elseif($lastpage - ($adjacents * 2) > $page && $page > ($adjacents * 2))
			{
				$pagination.= "<a href=\"$targetpage?page=1\">1</a>";
				$pagination.= "<a href=\"$targetpage?page=2\">2</a>";
				$pagination.= "...";
				for ($counter = $page - $adjacents; $counter <= $page + $adjacents; $counter++)
				{
					if ($counter == $page)
						$pagination.= "<span class=\"current\">$counter</span>";
					else
						$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";					
				}
				$pagination.= "...";
				$pagination.= "<a href=\"$targetpage?page=$lpm1\">$lpm1</a>";
				$pagination.= "<a href=\"$targetpage?page=$lastpage\">$lastpage</a>";		
			}
			//close to end; only hide early pages
			else
			{
				$pagination.= "<a href=\"$targetpage?page=1\">1</a>";
				$pagination.= "<a href=\"$targetpage?page=2\">2</a>";
				$pagination.= "...";
				for ($counter = $lastpage - (2 + ($adjacents * 2)); $counter <= $lastpage; $counter++)
				{
					if ($counter == $page)
						$pagination.= "<span class=\"current\">$counter</span>";
					else
						$pagination.= "<a href=\"$targetpage?page=$counter\">$counter</a>";					
				}
			}
		}
		
		//next button
		if ($page < $counter - 1) 
			$pagination.= "<a href=\"$targetpage?page=$next\">next »</a>";
		else
			$pagination.= "<span class=\"disabled\">next »</span>";
		$pagination.= "</div>\n";		
	}
?>

<?
while($row = mysql_fetch_array($result))
{
	
$id = $row["a_id"];
$title = ucwords($row["title"]); 
$news = $row["short"];
$date = $row["a_date"];

echo $short;
echo "<br>";
}
	
?>

<?=$pagination?>

Code (markup):

the ERROR is :

Notice: Use of undefined constant num - assumed 'num'

$total_pages = $total_pages[num];

i dont know what is that mean??

 

Means you're using dumb standards.

$total_pages = $total_pages['num'];

I'm not gonna go into details, the error notice sums it up anyway.

 

The error doens't tell what is wrong, and why not giving the user what he needs, i think it's better to tell what he's doing wrong then kick his arse... next time give him the sollution like i do!

$query = "SELECT COUNT(*) as num FROM $tbl_name";
$total_pages = mysql_fetch_array(mysql_query($query));
$total_pages = $total_pages[num];

as you can see its not returning an array but an result, so try to use mysql_result instead of mysql_fetch_array

 

Actually now I'm gonna kick both your arses because what he is actually asking for help with is another issue:

"Notice: Use of undefined constant num - assumed 'num'"

This means he's used a string which is not encapsulated with either single or double quotation marks, this means PHP will automatically look for a constant (as it should do), if the user has not previously used a function like define('num', 'blah'); then it will default to use that unencapsulated string as it's base value. Hence 'num' will be returned.

It's roughly 4 times slower and returns a warning, before defaulting to $total_pages = $total_pages['num']; anyway. It's vital unless you want the definition feature that you put it in quotes, preferably single unless parsing is required - and even then single is preferable.

It also has the downside if someone had done a operation like define('num', 'blah'); then it would instead look for the index 'blah' (i.e. $total_pages['blah']) because the constant had been changed to 'blah' as it should have been. Therefore the result of this is unlikely to be what you want.

Dan

 

Thanks Danltn, thats worked

and thanks also to EricBruggema

You guys rock man

 

It isn't an error, just a "Notice"

It's really $total_pages['num'] (Note, "num" is a String)
In php, you can definite constants
Variables are prefixed with the $, but constants have no prefix.
Fortunately, php is smart enough to realize that in $total_pages[num], since you have no constant named num, you must mean the string 'num'.

 

Now how can i display this in the above script:

Displaying 1 - 10 of 50 news [Total 5 Pages]

waiting for your help guys...