Re: Trying to make site map

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Thank you very much

On Jan 3, 2010, at 11:42 AM, Simcha Younger wrote:

On Mon, 21 Dec 2009 14:53:58 -0700
Zach Hicken <zach@xxxxxxxxxxxxx> wrote:
You are going to end up with alot of repetitous code if you repeat the process for every level.

I would instead select all page data, and then construct a nested array to show the page structure, and then work off of that.

example:

               $pages = array();
$sql = "SELECT * FROM $Gen WHERE Category = '$Cat' ORDER BY Page_Above desc"; //start from the lower levels, build up $result = mysql_query($sql, $conn) or die(mysql_error());
               //go through each row in the result set
               while ($pageArray = mysql_fetch_array($result)) {
if(isset($pages[$pageArray['id']])){ ksort($pages[$pageArray['id']]); $pageArray['children'] = $pages[$pageArray['id']]; unset( $pages[$pageArray['id']]);
                                                       }
$pages[$pageArray['Page_Above']][$pageArray['id']] = $pageArray;
                               }
$pages = $pages[0]; // remove unnamed top-level (nodes not properly set as children)
sort($pages);
                       function showpages($p, $level=0){
                               $line = "%d\t%s\t%d\r\n";
printf($line, $p['id'], $p['name'], $p['Page_Above']); if(isset($p['children'])) foreach($p['children'] as $child) showpages($child);
                       }
foreach($pages as $page) showpages($page);



I am trying to create a ui for a page management script. During this
step the user chooses which existing page the new page will link
under. Each record has a field called Page_Above, which references the
primary key number (id) of the page above it. Currently I have 4
records in the database:
(id, name, Page_Above)
1, Page1, 0
2, Page2, 1
3, Page3, 2
4, Page4, 1

Here is the pertinent snippet:

		
				include "config.php";
				$conn = mysql_connect($server, $DBusername, $DBpassword);
				mysql_select_db($database,$conn);
				$sql = "SELECT * FROM $Gen WHERE Category = '$Cat' AND Page_Above
= 0";
				$result = mysql_query($sql, $conn) or die(mysql_error());
				//go through each row in the result set and display data
				while ($pageArray = mysql_fetch_array($result)) {
					
					$prime_id  = $pageArray['id'];
					$Name = $pageArray['Name'];
					print ("<tr><td bgcolor=#ffffff>$Name</td><td bgcolor=#ffffff
align=left valign=top>");
					
				
							$sql = "SELECT * FROM $Gen WHERE Page_Above = $prime_id";
							$result = mysql_query($sql, $conn) or die(mysql_error());
							//go through each row in the result set and display data
							while ($secpageArray = mysql_fetch_array($result)) {
								// give a name to the fields
								$second_id  = $secpageArray['id'];
								$Name = $secpageArray['Name'];
								print ("$Name<br>");
								
								
							$sql = "SELECT * FROM $Gen WHERE Page_Above = $second_id";
							$result = mysql_query($sql, $conn) or die(mysql_error());
							//go through each row in the result set and display data
							while ($thpageArray = mysql_fetch_array($result)) {
								// give a name to the fields
								$third_id  = $thpageArray['id'];
								$Name = $thpageArray['Name'];
								print ("</td><td>");
								print ("$Name<br>");
								
								
							}
						
						}
					
					}

The results I am getting are incomplete, it only pulls one page per
level instead of all the pages per level, Like this:
"Page1, Page 2, Page3"
it skips Page4.

When I remove the request for the third level, then I get:
"Page1,Page2,Page4"  Which is correct up to that point. It breaks
apart when I try to go on the third level.

Any ideas how I can get this to work? In the end there will be 5 levels.
Thanks


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Simcha Younger <simcha@xxxxxxxxxxxx>

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