PHP - Query With Joins

I have this code which should get all the results from forum_forums and list them in their respective parents using forum_forums.parent_id. It does work sort of but it repeats one of them and doesnt include all of them.

Can anyone see where i am going wrong?

$parent_info_query = $db->query("SELECT 
                    parent_id,
                    parent_name
                    FROM ".DB_PREFIX."parents")
                    or die(mysql_error());

while ($parent_info = mysql_fetch_object($parent_info_query)) { 
 
// Add parent_id into variable for later query
 
 $parent_id = $parent_info->parent_id;

echo '<table class="forum_table" onclick="expandCollapseTable(this)">
<tr id="tr1">
<th class="forum_left_corner"></th>
<th class="forum_parent_name">'.$parent_info->parent_name.'</th>
<th class="empty"></th>
<th class="empty"></th>
<th class="forum_last_post_header">'.LAST_POST.'</th>
</tr>';

// Get Forum information from DB to show all forums
// including who the last post was posted by

$forum_info_query = $db->query("SELECT 

                    ".DB_PREFIX."forums.forum_id,
                    ".DB_PREFIX."forums.forum_name,
                    ".DB_PREFIX."forums.forum_description,
                    ".DB_PREFIX."forums.forum_topics,
                    ".DB_PREFIX."forums.forum_posts,
                    ".DB_PREFIX."forums.forum_last_poster,
                    ".DB_PREFIX."forums.forum_last_post_time,
                    ".DB_PREFIX."forums.forum_last_post,
                    
                    ".DB_PREFIX."members.user_id,
                    ".DB_PREFIX."members.user_username,
                    ".DB_PREFIX."members.user_group,
                    
                    ".DB_PREFIX."topics.topic_id,
                    ".DB_PREFIX."topics.topic_name
                    
                    FROM ".DB_PREFIX."forums
                    
                    JOIN ".DB_PREFIX."members
                    
                    ON ".DB_PREFIX."forums.forum_last_poster
                    = ".DB_PREFIX."members.user_id
                    
                    JOIN ".DB_PREFIX."topics
                    
                    ON ".DB_PREFIX."forums.forum_id
                    = ".DB_PREFIX."topics.forum_id
                    
                    WHERE ".DB_PREFIX."forums.parent_id 
                    = $parent_id") 
                    
                    or trigger_error("SQL", E_USER_ERROR);
                    
while ($forum_info = mysql_fetch_object($forum_info_query)) {

im trying to improve my code by using table joins but when i use the following code it just returns everything in the database rather than the results that equal $forum_id which is the get value 2. There should only be 2 results.

 $query = $db->query("SELECT 
                    ".DB_PREFIX."topics.topic_id,
                    ".DB_PREFIX."topics.topic_name,
                    ".DB_PREFIX."topics.topic_poster,
                    ".DB_PREFIX."topics.topic_time_posted,
                    ".DB_PREFIX."topics.topic_views,
                    ".DB_PREFIX."topics.topic_replies,
                    ".DB_PREFIX."topics.topic_last_poster,
                    ".DB_PREFIX."topics.topic_last_post_time,
                    ".DB_PREFIX."topics.topic_locked,
                    ".DB_PREFIX."topics.topic_sticky,
                    
                    ".DB_PREFIX."parents.parent_id,
                    ".DB_PREFIX."parents.parent_name,
                    
                    ".DB_PREFIX."forums.forum_name,
                    
                    ".DB_PREFIX."members.user_username,
                    ".DB_PREFIX."members.user_group
                    
                    FROM ".DB_PREFIX."topics
                    
                     JOIN ".DB_PREFIX."members
                     JOIN ".DB_PREFIX."parents
                     JOIN ".DB_PREFIX."forums
                    
                    ON ".DB_PREFIX."topics.topic_poster 
                    = ".DB_PREFIX."members.user_username
                    
                    WHERE ".DB_PREFIX."forums.forum_id
                    = ".$forum_id."
                    
                    ORDER BY ".DB_PREFIX."topics.topic_time_posted $max")
                    or trigger_error("SQL", E_USER_ERROR);

I've got a region-type select form where a user chooses his/her continent, and depending on the selection, a second select form is filled with country values relating to that continent. From there, depending on the country selection, a third select form is dynamically filled with regions relating to that country and continent. See the code attached:

<form action="" method="POST">
    <select id="continent" name="continent" style="min-width: 170px;">
        <option value="">Select Continent</option>
        <?php
        $q = "SELECT * FROM table_areaContinent";
        $r = mysql_query($q) or die(mysql_error());
        
        while($row = mysql_fetch_array($r)){
            $continent_id = $row['cont_id'];
            $continent = $row['continent'];
            
            print '<option value="' . $continent_id . '">' . $continent . '</option>' . "\n";
        }
        ?>
    </select>
    
    <br /><br />
    
    <select id="country" name="country" style="min-width: 170px;">
        <option value="">Select Country</option>
        <?php
        $q = "SELECT * FROM table_areaCountry JOIN table_areaContinent USING (cont_id)";
        $r = mysql_query($q) or die(mysql_error());
        
        while($row = mysql_fetch_array($r)){
            $country_id = $row['country_id'];
            $country = $row['country'];
            $continent_id = $row['cont_id'];
            
            print '<option value="' . $country_id . '" class="' . $continent_id . '">' . $country . '</option>' . "\n";
        }
        ?>
    </select>
    
    <br /><br />

    <select id="region" name="region" style="min-width: 170px;">
        <option value="">Select Region</option>
        <?php
        $q = "SELECT * FROM table_areaRegion JOIN table_areaCountry USING (country_id)";
        $r = mysql_query($q) or die(mysql_error());
        
        while($row = mysql_fetch_array($r)){
            $region_id = $row['region_id'];
            $region = $row['region'];
            $country_id = $row['country_id'];
            
            print '<option value="' . $region_id . '" class="' . $country_id . '">' . $region . '</option>' . "\n";
        }
        ?>    
    </select>
    
    <br /><br />
    
    <input type="submit" value="Submit" name="submit" class="submit" />
</form>

Now, I simply just want to echo out the Continent, Country and Region that the user selected after the form is submitted. I currently have this :

<?php
if(!empty($_POST['submit'])){
    $continent = $_POST['continent'];
    $country = $_POST['country'];
    $region = $_POST['region'];
    
    print '
    <b>Continent: ' . $continent . '<br />
    <b>Country: ' . $country . '<br />
    <b>Region: ' . $region . '<br />
    ';
}
?>

Which echoes out the ID numbers for each variable. What would be the best direction to take when pulling data from multiple tables to echo out the actual Names that the IDs are assigned to?

I have created a file with this basic structure in the referals table:
refREFER = the referer, person who sent the referal link out
refREFED = the refered person, the one who signed up with the link
refREFERIP = the referer's IP
refREFEDIP = the refered person's IP

I'm trying to create a code that will list any user that has signed up on the same IP with a referal link. Everything works except the Refered column, which I can't seem to figure out how to do. Basically I have made the refREFER = the userid in the users table, so it can be output as a username. However I want to do the same with the refREFED but don't know how. Here's my code so far:


<?php
include "globals.php";
if($ir['user_level'] != 2 && $ir['user_level'] != 3) //User level check
{
    die("You can't access this page");
}

$q=mysql_query("SELECT u.userid, u.username, u.laston, r.* 
FROM users u 
LEFT JOIN referals r ON r.refREFER=u.userid 
WHERE r.refREFERIP=r.refREFEDIP 
ORDER BY userid ASC",$c); //Selecting everything that's needed

print "<center><br /><b><font color=white><h1>Referral Multis</h1></b><br />

<table width=75% border=1>
<tr style='background:black'>
<th><font color=grey>Referer</font></th>
<th><font color=grey>Refered</font></th>
<th><font color=grey>IP</font></th> 
<th><font color=grey>Time</font></th>
</tr>";
while(($r=mysql_fetch_array($q) or die(mysql_error()))) //Making all of the select stuff into $r
{
$reftime = date('F j, Y g:i:s a', $r['refTIME']);
if($r['laston'] >= time()-60*60) { $on="<font color=green><b>Online</b></font>"; } else { $on="<font color=red><b>Offline</b></font>"; }
print "
<tr> 
<td><a href='viewuser.php?u={$r['userid']}'><font color=brown><b>{$r['username']}</b></font></a> [{$r['refREFER']}]</td>
<td> {$r['username']} - {$r['refREFED']}</td>
<td>{$r['refREFERIP']}</td>
<td>$reftime</td>
</tr>";
}
print "</table>";

$h->endpage();

?>


At the moment under the Refered column it shows the wrong username, but the correct ID. Any ideas?

Check out my code below and you will see what I am trying to do.

I cannot figure out what is going wrong.
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mskordus/public_html/index.php on line 25

Code: [Select]
<?php
$result = mysql_query("SELECT * FROM students,teachers,specialEd,course,sched,notes WHERE
   students.studentId=notes.studentId AND
   students.studentId=sched.studentId AND
   students.teacherId=teachers.teacherId AND
   students.specialId=specialEd.specialId AND
   students.courseId=course.courseId AND
   ");
echo "<table>";
while ($row = mysql_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['students.fName'] . "</td>";
echo "<td>" . $row['teachers.fName'] . "</td>";
echo "</tr>";
}
echo "</table>";
?>

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=315743.0

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=313836.0

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=320409.0

Hello all,   Based on the suggestion of you wonderful folks here, I went away for a few days (to learn about PDO and Prepared Statements) in order to replace the MySQLi commands in my code. That's gone pretty well thus far...with me having learnt and successfully replaced most of my "bad" code with elegant, SQL-Injection-proof code (or so I hope).   The one-and-only problem I'm having (for now at least) is that I'm having trouble understanding how to execute an UPDATE query within the resultset of a SELECT query (using PDO and prepared statements, of course).   Let me explain (my scenario), and since a picture speaks a thousand words I've also inlcuded a screenshot to show you guys my setup:   In my table I have two columns (which are essentially flags i.e. Y/N), one for "items alreay purchased" and the other for "items to be purchased later". The first flag, if/when set ON (Y) will highlight row(s) in red...and the second flag will highlight row(s) in blue (when set ON).   I initially had four buttons, two each for setting the flags/columns to "Y", and another two to reverse the columns/flags to "N". That was when I had my delete functionality as a separate operation on a separate tab/list item, and that was fine.   Now that I've realized I can include both operations (update and delete) on just the one tab, I've also figured it would be better to pare down those four buttons (into just two), and set them up as a toggle feature i.e. if the value is currently "Y" then the button will set it to "N", and vice versa.   So, looking at my attached picture, if a person selects (using the checkboxes) the first four rows and clicks the first button (labeled "Toggle selected items as Purchased/Not Purchased") then the following must happen:   1. The purchased_flag for rows # 2 and 4 must be switched OFF (set to N)...so they will no longer be highlighted in red. 2. The purchased_flag for row # 3 must be switched ON (set to Y)...so that row will now be highlighted in red. 3. Nothing must be done to rows # 1 and 5 since: a) row 5 was not selected/checked to begin with, and b) row # 1 has its purchase_later_flag set ON (to Y), so it must be skipped over.   Looking at my code below, I'm guessing (and here's where I need the help) that there's something wrong in the code within the section that says "/*** loop through the results/collection of checked items ***/". I've probably made it more complex than it should be, and that's due to the fact that I have no idea what I'm doing (or rather, how I should be doing it), and this has driven me insane for the last 2 days...which prompted me to "throw in the towel" and seek the help of you very helpful and intellegent folks. BTW, I am a newbie at this, so if I could be provided the exact code, that would be most wonderful, and much highly appreciated.   Thanks to you folks, I'm feeling real good (with a great sense of achievement) after having come here and got the great advice to learn PDO and prepared statements.   Just this one nasty little hurdle is stopping me from getting to "end-of-job" on my very first WebApp. BTW, sorry about the long post...this is the best/only way I could clearly explaing my situation.   Cheers guys!

case "update-delete":
	if(isset($_POST['highlight-purchased'])) 
		{
			// ****** Setup customized query to obtain only items that are checked ******
			$sql = "SELECT * FROM shoplist WHERE";
					for($i=0; $i < count($_POST['checkboxes']); $i++)
						{
							$sql=$sql . " idnumber=" . $_POST['checkboxes'][$i] . " or";
						}
					$sql= rtrim($sql, "or");
											
			$statement = $conn->prepare($sql);
			$statement->execute();
																						
			// *** fetch results for all checked items (1st query) *** //
			$result = $statement->fetchAll();

			$statement->closeCursor();
											
			// Setup query that will change the purchased flag to "N", if it's currently set to "Y"
			$sqlSetToN = "UPDATE shoplist SET purchased = 'N' WHERE purchased = 'Y'";
											
			// Setup query that will change the purchased flag to "Y", if it's currently set to "N", "", or NULL
			$sqlSetToY = "UPDATE shoplist SET purchased = 'Y' WHERE purchased = 'N' OR purchased = '' OR purchased IS NULL";

			$statementSetToN = $conn->prepare($sqlSetToN);
			$statementSetToY = $conn->prepare($sqlSetToY);
											
			/*** loop through the results/collection of checked items ***/
			foreach($result as $row)
				{
					if ($row["purchased"] != "Y")
						{
							// *** fetch one row at a time pertaining to the 2nd query *** //
							$resultSetToY = $statementSetToY->fetch();
															
							foreach($resultSetToY as $row)
								{															
									$statementSetToY->execute();																
								}
						}
					else
						{
							// *** fetch one row at a time pertaining to the 2nd query *** //
							$resultSetToN = $statementSetToN->fetch();
															
							foreach($resultSetToN as $row)
								{															
									$statementSetToN->execute();																
								}
						}
				}
				break;
		}

 CRUD Queston.png   20.68KB   0 downloads    

Here is my code:
// Start MySQL Query for Records
$query = "SELECT codes_update_no_join_1b" .
"SET orig_code_1 = new_code_1,
       orig_code_2 = new_code_2" .
"WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)";
$results = mysql_query($query) or die(mysql_error());
// End MySQL Query for Records


This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this???

Code: [Select]
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1

If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

  Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");
 
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>
 
<div id="wrap">
 
<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>
 
<? include_once("includes/navigation.php"); ?>
 
<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php
 
if(!$_SESSION['user'])
{
 
        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);
 
        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>
       
        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>
       
        </select></td></tr>
       
        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");
               
                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }
       
}
else
{
        echo "You are currently logged in!";
}
 
?>
</div>
</div>
 
<div style="clear: both;"> </div>
</div>
 
<div id="footer">
<a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop
</div>
</div>
 
</body>
</html>

I was just wondering if it's possible to run a query on data that has been returned from a previous query?  For example, if I do

Code: [Select]
$sql = 'My query';
$rs = mysql_query($sql, $mysql_conn);

Is it then possible to run a second query on this data such as

Code: [Select]
$sql = 'My query';
$secondrs = mysql_query($sql, $rs, $mysql_conn);

Thanks for any help

I'm trying to update every record where one field in a row is less than the other. The code gets each row i'm looking for and sets up the query right, I hope I combined the entire query into one string each query seperated by a ; so it's like
UPDATE `table` SET field2= '1' WHERE field1= '1';UPDATE `table` SET field2= '1' WHERE field1= '2';UPDATE `table` SET field2= '1' WHERE field1= '3';UPDATE `table` SET field2= '1' WHERE field1= '4';UPDATE `table` SET field2= '1' WHERE field1= '5';
this executes properly if i run the query in phpMyAdmin, however when I run the query in PHP, it does nothing... Any advice?

Say I have this query:

site.com?var=1

..I have a form with 'var2' field which submits via get.

Is there a way to produce:

site.com?var=1&var2=formdata

I was hoping there would be a quick way to affix, but can't find any info. Also, the query could sometimes be:

site.com?var2=formdata&var=1

I would have to produce:

site.com?var2=updatedformdata&var=1

Is my only option to further parse the query?

What would be the correct way to close a mysql query?

At current the second query below returns results from the 1st query AND the 2nd query
The 3rd query returns results from the 1st, 2nd and 3rd query.
 etc etc.

At the moment I get somthing returned along the lines of...

QUERY 1 RESULTS
Accommodation 1
Accommodation 2
Accommodation 3

QUERY 2 RESULTS
Restaurant 1
Restaurant 2
Restaurant 3
Accommodation 1
Accommodation 2
Accommodation 3

QUERY 3 RESULTS
Takeaways 1
Takeaways 2
Takeaways 3
Restaurant 1
Restaurant 2
Restaurant 3
Accommodation 1
Accommodation 2
Accommodation 3

Code: [Select]

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<?php 
  include($_SERVER['DOCUMENT_ROOT'].'/include/db.php');
?>
<title>Untitled Document</title>
<style type="text/css">
<!--
-->
</style>
<link href="a.css" rel="stylesheet" type="text/css" />
</head><body>
<div id="listhold">

<!------------------------------------------------------------------------------------------------------------------------------------------------------>

<div class="list"><a href="Placestostay.html">Places To Stay</a><br />
<?php
$title ="TITLE GOES HERE";
$query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Accommodation' AND confirmed ='Yes' ORDER BY name");
echo mysql_error();
while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0];
$i = -1;
foreach($nt as $value)
{$i++;
$FileName = str_replace(' ','_',$nt[$i]) . ".php";
$FileUsed = str_replace('_',' ',$nt[$i]);
echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>"  . $nt[$i] . "</a>" . "<br/>";
$FileHandle = fopen($FileName, 'w') or die("cant open file");
$pageContents = file_get_contents("header.php");
fwrite($FileHandle,"$pageContents");}
fclose($FileHandle);
?>
</div>

<!------------------------------------------------------------------------------------------------------------------------------------------------------>

<div class="list"><a href="Eatingout.html">Eating Out</a><br />
<?php
$title ="TITLE GOES HERE";
$query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Restaurant' AND confirmed ='Yes' ORDER BY name");
echo mysql_error();
while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0];
$i = -1;
foreach($nt as $value)
{$i++;
$FileName = str_replace(' ','_',$nt[$i]) . ".php";
$FileUsed = str_replace('_',' ',$nt[$i]);
echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>"  . $nt[$i] . "</a>" . "<br/>";
$FileHandle = fopen($FileName, 'w') or die("cant open file");
$pageContents = file_get_contents("header.php");
fwrite($FileHandle,"$pageContents");}
fclose($FileHandle);
?>
</div>

<!------------------------------------------------------------------------------------------------------------------------------------------------------>

<div class="list"><a href="Eatingin.html">Eating In</a><br />
<?php
$title ="TITLE GOES HERE";
$query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Takeaways' AND confirmed ='Yes' ORDER BY name");
echo mysql_error();
while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0];
$i = -1;
foreach($nt as $value)
{$i++;
$FileName = str_replace(' ','_',$nt[$i]) . ".php";
$FileUsed = str_replace('_',' ',$nt[$i]);
echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>"  . $nt[$i] . "</a>" . "<br/>";
$FileHandle = fopen($FileName, 'w') or die("cant open file");
$pageContents = file_get_contents("header.php");
fwrite($FileHandle,"$pageContents");}
fclose($FileHandle);
?>
</div>

<!------------------------------------------------------------------------------SKILLED TRADES BELOW--------------------------------------------------->

<div class="list"><a href="Skilledtrades.html">Skilled Trades</a><br/>
<?php
$title ="TITLE GOES HERE";
$query = mysql_query("SELECT DISTINCT subtype FROM business WHERE type ='Skilled Trades' AND confirmed ='Yes' ORDER BY name");
echo mysql_error();
while($ntx=mysql_fetch_row($query)) $nt[] = $ntx[0];
$i = -1;
foreach($nt as $value)
{$i++;
$FileName = str_replace(' ','_',$nt[$i]) . ".php";
$FileUsed = str_replace('_',' ',$nt[$i]);
echo "<a href='" . str_replace(' ','_',$nt[$i]) . ".php?title=$title&subtype=$FileUsed'>"  . $nt[$i] . "</a>" . "<br/>";
$FileHandle = fopen($FileName, 'w') or die("cant open file");
$pageContents = file_get_contents("header.php");
fwrite($FileHandle,"$pageContents");}
fclose($FileHandle);
?>
</div>