[SOLVED] Limiting search page numbers

[86Stang]

I'm wanting to limit the number of page links that result from a search.  So, instead of seeing "Prev |1|2|3|4

|5|6|7|8|9|Next" etc., you would like to show only 3 pages in either direction of the current page.  I.E. if you're on page 5 you'd see "2|3|4|5|6|7|8".

 

Here's what I have and it displays ALL pages:

For($i = 1 ; $i <= $TotalNumberOfPages ; $i++)	
{	
	If($i == $CurrentPage) 
		{
			$Nav .= "<B>$i</B>";
		}
	Else
		{
			$Nav .= "<A HREF=\"search.php?page=" . $i . "&keywords=" .urlencode($keywords) . "\">$i </A>";
		}
}

 

teng84, I promise not to edit the code snippet this time!

[tinker]

$pos = 55;

$start = $pos - 3;
if($start<0){ $start = 0; }
$end = $pos + 3;
if($end>$max){ $end = $max; }

For($i = $start ; $i <= $end ; $i++)
{
...
}

[86Stang]

No results.  Why $pos = 55?  What's significant about the number 55?  Would that be causing the blank results?

[sasa]

TRY

For($i = $CurrentPage > 3 ? $CurrentPage - 3 : 1; $i <= ($CurrentPage + 3 < $TotalNumberOfPages ? $CurrentPage + 3 : $TotalNumberOfPages) ; $i++)

[86Stang]

Works perfect.  You're a rockstar sasa!!

[86Stang]

This is solved but I have a question -- what do the question marks in that last code snippet do exactly?

[sasa]

$i = $CurrentPage > 3 ? $CurrentPage - 3 : 1;

means

if($CurrentPage > 3) $i = $CurrentPage - 3; else $i = 1;