XML parsing

[iNFERiON]

Hi all,

I'm pretty new to XML itself, so let alone XML and PHP. Yet, my employer wants me to use a XML file with the data to be used on the site.

I have the current XML file.

<?xml version="1.0" encoding="ISO-8859-1"?>
<projects>
	<project>
		<code>p00001</code>
		<name>Test Project</name>
	</project>
</projects>

What I'm trying to get done is to have PHP output the info like this:

Project Code: p00001
Project Name: Test Project

How to do this ?
I can't seem to get it done.

Thanks in advance.

[Weedpacket]

What have you tried? What were the results?

[iNFERiON]

Well, first I am trying to create a navigation menu that will show the info I want on the right side. That does not go well. This is the menu source:

<?php
  $projects_menu = array();
  $new_project = $_GET["new_project"];

  $sax = xml_parser_create();
  xml_parser_set_option($sax, XML_OPTION_CASE_FOLDING, false);
  xml_parser_set_option($sax, XML_OPTION_SKIP_WHITE,true);
  xml_set_element_handler($sax, 'sax_start','sax_end');
  xml_set_character_data_handler($sax, 'sax_cdata');
  xml_parse($sax, file_get_contents('projects/projects.xml'),true);
  xml_parser_free($sax);

function sax_start($sax, $tag, $attr) {
  global $projects_menu;
  if ($tag == 'projects') {
    //echo '<ul>';
  } elseif ($tag == 'project') {
  	$projectcode = htmlspecialchars($attr['code']);
    $projectname = htmlspecialchars($attr['name']);
    //echo '<li>' . $projectname . '';
    $projects_menu [] = array('code' => htmlspecialchars($attr['code']));
  }
}

function sax_end($sax, $tag) {
  if ($tag == 'projects') {
    //echo '</ul>';
  } elseif ($tag == 'project') {
    //echo '</li>';
  }
}

function sax_cdata($sax, $data) {
  echo htmlspecialchars($data);
}

if ($new_project != ""){
    ?>Submitted<?
    $projects_menu [] = array('code' => $new_project);
}

function update_menu($menu_data){

  $your_data = "<projects>\n";

  foreach( $menu_data as $menu_item)
  {
  $your_data = $your_data."<project>\n\t<code>".$menu_item['code']."</code>\n</project>\n";
  }

  $your_data = $your_data."</projects>";

// Open the file and erase the contents if any
$fp = fopen("projects/projects.xml", "w");

// Write the data to the file
fwrite($fp, $your_data);

// Close the file
fclose($fp);
}

?>
    <table border="0" cellpadding="0" cellspacing="0" width="100%" height="91%">
      <tr>
        <td width="100%" height="100%" bgcolor="#E7EDF6" class="menu" valign="top">
          <table cellpadding="1" cellspacing="0" border="0">
            <?
            foreach( $projects_menu as $menu_item)
            {
            ?><tr><td width="18"><img src="images/icon_requests.gif"></td><td><a href="editproject.php?pid=<?=$menu_item['code']?>" target="mainFrame"><?=$menu_item['code']?></a></td></tr><?
            }
            ?>
          </table>
        </td>
      </tr>
</table>
  </body>
</html>

This code shows the project code and name, though it is not presented as the link I want, so I must be missing something.