|
PHP - Can You Put A Youtube Link In Script To Display In A Page
This is my script I want to display but it does not show in the page just blank.
<td><object width="220" height="148"><param name="movie" value="http://www.youtube.com/user/THEWORLDOFTRAVEL#p/search/2/uX9nd3xM_MY=1&h1=enUS&color1=0x234900&color3=34900&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/user/THEWORLDOFTRAVEL#p/search/2/uX9nd3xM_MY=1&h1=enUS&color1=0x234900&color3=34900&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="220" height="148"></embed></object></td> Similar Tutorialshi guys, i am trying to create a button that when u press if in the text are will appear <video>...</video> i know this is js do don't worry i will manage to do it, my problem is in php, i did a function that recognizes the link between <video>...</video> and then it embades it and shows me a youtube video instead, the problem i got is that if i have in the text area more <video>...</video>, like: bla bla bal<video>youtube link</video> and the other vid <video>youtube link</video> i dont know how to make it embade both, i dont know how to check if there is more, and the second problem i got is that all the text that was not betwen <video>...</video> will disapear after the vide is embadded, here is the code i done until now Code: [Select] <form action="test1.php" method = "POST"> <textarea name="body" id="text" cols="30" rows="10"></textarea> <input type="submit" value = "submit"> </form> <?php function video($string){ preg_match_all("#<video\b[^>]*>(.*?)</video>#", $string, $output); preg_match_all("#(?<=v=)[a-zA-Z0-9-]+(?=&)|(?<=v\/)[^&\n]+(?=\?)|(?<=v=)[^&\n]+|(?<=youtu.be/)[^&\n]+#", $output[0][0], $match); return "<iframe width='560' height='315' src='http://www.youtube.com/embed/".$match[0][0]."' frameborder='0' allowfullscreen></iframe>"; } if (isset($_POST['body'])&&!empty($_POST['body'])) { $body = $_POST['body']; print_r(video("$body")); } ?> Anyone have access to a youtube video/audio downloader script? I'm sure there are one of these laying around somewhere... Looking for 1. Hello.
I have a bit of a problem. When I fetch the link field from the database.i don't see an actual link on the page.
One more thing, what type of field should I use to store the link in the database? Probably there is where I went wrong.
All help is
How could I from the varying URLs parse only the YouTube video ID? http://www.youtube.com/watch?v=SwrawcORlp0&feature=player_embedded http://www.youtube.com/watch?v=SwrawcORlp0&feature=popular How can I from the URLs above only extract the id into a variable $url SwrawcORlp0 Thanks, df I wanted to store a link in the db and display the same link on the front end which will take me to the site. i tried with href but i get some string error. Also i wanted to add a youtube embed tag to display it again on query to the db. in short is possible to store the url in the db and display the same url. thanks My code: Code: [Select] <?php $username = $_SESSION['loggedin']; $extract_user_rank = mysql_query("SELECT `rank` FROM users WHERE username='$username'"); if($extract_user_rank['rank'] > 0) { echo "<a href='control_panel.php'>Admin Control Panel</a>"; } else { //return nothing } ?> I have my rank in the database set to 1. Why doesn't it output the ACP link? Hi, Is it possible to build a PHP Template page that selects and publishes a row of data from a MySQL Database when a linked is clicked? I would design: Template.php Text links (perhaps on homepage of navigation bar): ProductA, ProductB and ProductC If you click link ProductA Template.php would display data for ProductA and likewise for ProductB and ProductC. I would also like search engines for find ProductA, ProductB and ProductC PHP pages. (Not just my single Template.php) Any ideas as to how this could be done without designing individual PHP pages for each product? Kind regards, Matthew. I have a "Members" page that displays my organizations members info via My SQL. Currently, the database displays "State" quick links at the top and has the members organized by State down the page. If you click on one of the State links at the top, it will navigate to the section of the page with that state and associated members. I want the members associated with a specific state to be displayed only once I click the associated state link -- instead of all of the information showing at once like it is now. The page I am referring to can be seen at this link: http://homesforhorse...rs.com/members/
<?php
update_option('image_default_link_type','none');
include("/home/cingen/config_admin.php");
function listMembers() {
$sql = mysql_query("SELECT c.*, s.*
FROM (".TABLE_MEMBERS." c LEFT JOIN ".TABLE_STATE." s on c.state = s.state_abbr)
WHERE c. status = '1'
ORDER BY c.country, c.state, c.organization ASC");
while ($row = mysql_fetch_array($sql)) {
$display_members = false;
$organization = stripslashes($row['organization']);
$website = stripslashes($row['website']);
if ($website) {
$link = "<a href='http://".$website."' target='_blank'>";
$endlink = "</a>";
} else {
$link = "";
$endlink = "";
}
$display_members .= $link.$organization.$endlink."<br />";
if ($row['address']) $display_members .= stripslashes($row['address'])." ".stripslashes($row['address2'])."<br />";
if ($row['city']) $display_members .= stripslashes($row['city']).", ";
if ($row['state']) $display_members .= stripslashes($row['state'])."";
if ($row['zip']) $display_members .= " ".$row['zip'];
$display_members .= "<br />";
if ($row['contact_name']) $display_members .= "Contact: ".stripslashes($row['contact_name']);
if ($row['contact_title']) $display_members .= ", ".stripslashes($row['contact_title']);
if ($row['phone']) $display_members .= "<br />Tel: ".stripslashes($row['phone']);
if ($row['email']) $display_members .= "<br />".$row['email'];
if ($row['website']) $display_members .= "<br /><a href='http://".$row['website']."' target='_blank'>".$row['website']."</a><br/>";
if ($row['year_est']) $display_members .= "Founded in ".$row['year_est'].".";
if ($row['org501c3'] == "1") $display_members .= " A 501(c)3 non-profit.";
if ($row['gfas'] == "1") $display_members .= "<br />GFAS: Accredited Sanctuary.";
if ($row['gfas'] == "2") $display_members .= "<br />GFAS: Verified Sanctuary.";
if ($row['member_category']) $display_members .= "<br />".$row['member_category'];
$display_members .= "<br /><br />";
$entries[$row['country']][$row['state_name']][] = $display_members;
}
$countrylinks = false;
$statelinks = false;
$display = false;
if(is_array($entries)){
$display .= '
<div class="memberlist">';
foreach($entries as $country=>$state_members){
$countrylinks .= '<a href="#'.$country.'">'.$country.'</a> ';
$display .= '
<h2 id="'.$country.'">'.strtoupper($country).'</h2>
<div class="country">';
if(($state_members)){
foreach($state_members as $state=>$members){
$statelinks .= '<a href="#'.$state.'">'.$state.'</a> ';
$display .= '
<h3 id="'.$state.'">'.strtoupper($state).'</h3>
<div class="state">';
if(is_array($members)){
foreach($members as $key=>$member){
$display .= '
<div class="member">
'.$member.'
</div>';
}
}
$display .= '
</div>';
}
}
$display .= '
</div>';
}
$display .= '
</div>';
}
$statelinks1 = '
<h2>Members List</h2>
<strong>Quick Links</strong><br /><br />
'.$statelinks.'<br /><br />'
.$display;
return $statelinks1;
}
add_shortcode('memberlist', 'listMembers');
function listRescueStandards() {
$display_members = '';
$sql = mysql_query("SELECT vc.*, s.*, m.*
FROM ".TABLE_COMPLIANCE." vc, ".TABLE_STATE." s, ".TABLE_MEMBERS." m
WHERE vc.member_id = m.cid
AND m.status = '1'
AND m.state = s.state_abbr
ORDER BY m.state, m.organization ASC");
while ($row = mysql_fetch_array($sql)) {
$organization = stripslashes($row['organization']);
if ($row['website']) {
$link = "<a href='http://".$row['website']."' target='_blank'>";
$endlink = "</a>";
} else {
$link = "";
$endlink = "";
}
if($x!=$row['state_name']){
$display_members .= "<br /><strong>".strtoupper($row['state_name'])."</strong><br />";
$x = $row['state_name'];
}
$display_members .= $link.$organization.$endlink."<br />
".stripslashes($row['address'])." ".stripslashes($row['address2'])."<br />
".stripslashes($row['city']).", ".stripslashes($row['state'])." ".$row['zip']."<br />";
if ($row['contact_name']) $display_members .= "Contact: ".stripslashes($row['contact_name']);
if ($row['contact_title']) $display_members .= ", ".stripslashes($row['contact_title']);
if ($row['phone']) $display_members .= "<br />Tel: ".stripslashes($row['phone']);
if ($row['fax']) $display_members .= "<br />Fax: ".stripslashes($row['fax']);
if ($row['email']) $display_members .= "<br />".$row['email'];
if ($row['website']) $display_members .= "<br /><a href='http://".$row['website']."' target='_blank'>".$row['website']."</a>";
if ($row['year_est']) $display_members .= "<br />Founded in ".$row['year_est'].".";
if ($row['org501c3'] == "1") $display_members .= "<br />This organization IS registered with the IRS as a 501(c)3.";
if ($row['org501c3'] != "1") $display_members .= "<br />This organization is NOT registered with the IRS as a 501(c)3.";
$display_members .= "<br /><br />";
}
return "<div class='memberlist'>" . $display_members . "</div>";
}
add_shortcode('standardslist', 'listRescueStandards');
Greetings! I have a website www.lanceronlinejobs.com/our_franchises.php I have a franchise images. I want to display each franchise record from database whenever a user click on franchise link. Here is my code. ourfranchises.php code: Code: [Select] <?php include('fconnection.php'); $sql = mysql_query("SELECT * FROM tbl_franchise ORDER BY id DESC") or die(mysql_error()); $row3 = mysql_fetch_array($sql); ?> <a href="franchiseDetails.php?city=<?php echo $row3['city'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>---------------------------------------------------------------------- franchiseDetails.php code: Code: [Select] <?php $id = $_GET['id']; $query = "SELECT * FROM tbl_franchise WHERE id = '$id' ORDER BY id DESC LIMIT 1"; $result = mysql_query($query); if (!$result) { echo "NO RECORD FOUND"; } else { while($row3 = mysql_fetch_array($result)): ?> Manager Name: <?php echo $row3['manager_name'];?> Please help me. Any help would be appreciated. Thanks. Hello guys, I want to make an xml feed into my webpage, but the xml is generated from an .asp file on another website, so it's not actually an xml file, but xml output. Now, I want to retrieve this xml output on my webpage, but I'm having difficulties while doing this. What I've tried so far: - With PHP: Code: (php) [Select] <?php echo file_get_contents("http://link-on-other-website.asp?XML=1"); ?> - With javascript: Code: (javascript) [Select] <script type="text/javascript"> if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.open("GET","http://link-on-other-website.asp?XML=1", false); xmlhttp.send(); xmlDoc=xmlhttp.responseXML; document.write("<table border='1'>"); var x=xmlDoc.getElementsByTagName("RECORD"); for (i=0;i<x.length;i++) { document.write("<tr><td>"); document.write(x[i].getElementsByTagName("NAME")[0].childNodes[0].nodeValue); document.write("</td><td>"); document.write(x[i].getElementsByTagName("DATE")[0].childNodes[0].nodeValue); document.write("</td></tr>"); } document.write("</table>"); </script> When using the PHP code, I can retrieve it obviously, but I get an <?xml ... > declaration within my body, which is not allowed. When using the Javascript code, I can't even retrieve anything. And the goal is to style this xml output to match my website. That's the whole purpose of this xml link, because I couldn't style the iframe I had earlier. Can someone help me on this, on how to properly embed this asp generated xml output, within my HTML5 document? ps: I've asked for the xml-feed link from that other website, so using the above PHP code is legit in my case. Trying to figure out how to make it so name is a link to the profile when its echo anyone know how to do this? im at a huge stand still Code: [Select] <?php $sql = "SELECT name FROM users WHERE DATE_SUB(NOW(),INTERVAL 5 MINUTE) <= lastactive ORDER BY id ASC"; $query = mysql_query($sql) or die(mysql_error()); $count = mysql_num_rows($query); $i = 1; while($row = mysql_fetch_object($query)) { $online_name = htmlspecialchars($row->name); echo '<a href="Inbox.php">"'[$goauld]'</a>"'; ?> how can i do this pls? i just want to display a link "Free Chat" when i click on it, it must display a dropdown with options "yes" and "no" the link must not go anywhere yet, only when yes is selected it must go to a page i have the code: Code: [Select] <a href="" id="link" name="link" onClick="yesnolist(1)">Free Chat</a> then i have javascript: Code: [Select] <script type="text/JavaScript"> function yesnolist() { var e = document.getElementById("link"); var strUser1 = e.options[e.selectedIndex].value; if (strUser1 == "1") { window.location.href = "http://........."; } return strUser1; } </script> the above is not correct, can some-one help me please? thanks Well the subject line is pretty explicit. I found this script that uploads a picture onto a folder on the server called images, then inserts the the path of the image on the images folder onto a VACHAR field in a database table. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?> The script seems to work fine because I managed to upload a picture which was successfully inserted into my images folder and into the database. Now the problem is, I can't figure out exactly how to write the script that displays the image on an html page. I used the following script which didn't work. Code: [Select] //authenticate user //Start session session_start(); //Connect to database require ('config.php'); $sql = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' "); $row = mysql_fetch_assoc($sql); $imagebytes = $row['image']; header("Content-type: image/jpeg"); print $imagebytes; Seems to me like I need to alter some variables to match the variables used in the insert script, just can't figure out which. Can anyone help?? Hello all, I am new to php and was wondering if i could get some guidance here. I am using phpAdmin 2.6.0 and running Mysql 4.1.21. here is my situation.... I have a script that allows us to upload a new product name, product code, category and a PDF file to the data base. There is also a folder on the server that has the PDF files in it. I think I deleted the code on the page (library.php) that displays the files for the client to download. My goal here is after I upload everything I want it to then be displayed on the page with a link to the PDF file. Here is the page that has the links on it. I hope that I explained this correctly. I am not a programmer but do have some idea and have been reading up on php to try and figure this out. I am looking to create the script that would display the links on the library.php page. Any help would be great. The other link is to the script that allows us to upload. www.pennstateind.com/library.php www.pennstateind.com/lib-admin.php I would like it to list a bunch of music in a folder but the user dosent have accesses to except me. So the page he can view will list all the files in the other folder and has a downloadable link to it so he can download the files Thanks Dan I came up with the following script for displaying an uploaded picture on a webpage but for some reason I can't pinpoint, the picture won't be displayed. I checked my database from php myadmin and sure enough, the picture was successfully uploaded but somehow, the picture won't be displayed. So here is the display script. I named it display_pic.php Code: [Select] <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //include the config file require('config.php'); $image = stripslashes($_REQUEST[imname]); $rs = mysql_query("SELECT* FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main' "); $row = mysql_fetch_assoc($rs); $imagebytes = $row[image]; header("Content-type: image/jpeg"); print $imagebytes; ?> The tag on the html page that's supposed to display the picture reads something like this <img src="display_pic.php" width="140" height="140"> And just in case this might help, I will include the image upload script below, which I think worked just fine because my values were successfully inserted into the database. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Make sure the user actually // selected and uploaded a file if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) { // Temporary file name stored on the server $tmpName = $_FILES['image']['tmp_name']; // Read the file $fp = fopen($tmpName, 'r'); $data = fread($fp, filesize($tmpName)); $data = addslashes($data); fclose($fp); // Create the query and insert // into our database. $query = "INSERT INTO images (member_id, image_cartegory, image_date, image) VALUES ('{$_SESSION['id']}', 'main', NOW(), '$data')"; $results = mysql_query($query); // Print results print "Thank you, your file has been uploaded."; } else { print "No image selected/uploaded"; } // Close our MySQL Link mysql_close(); } //End of if statmemnt. ?> So any insights as to why the script fails to display the image? Any help is appreciated. Hello I have an array that I would like to display as a family tree... and am looking for an example script that will allow me to do this. I want to be able to create X number of generations and fill the boxes with data from my array.... So, if I want 5 generations I want to generate a tree with 63 elements or boxes Any help would be appreciated Thanks I had posted a similar post a few days back, about an display script which is supposed to retrieve a stored image from my database and displays it on an html page, but fails to do so. Someone suggested that I upload my image onto a folder on the server and then save the image path name in my database. I found this script(below) which does just that. It uploads the image into a folder on the server called images and stores the image path into a table in my database called images and I know this script works because i saw the file saved in the images folder and the path name inserted into the images table. Code: [Select] <?php //This file inserts the main image into the images table. //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user //Start session session_start(); //Connect to database require ('config.php'); //Check whether the session variable id is present or not. If not, deny access. if(!isset($_SESSION['id']) || (trim($_SESSION['id']) == '')) { header("location: access_denied.php"); exit(); } else{ // Check to see if the type of file uploaded is a valid image type function is_valid_type($file) { // This is an array that holds all the valid image MIME types $valid_types = array("image/jpg", "image/jpeg", "image/bmp", "image/gif"); if (in_array($file['type'], $valid_types)) return 1; return 0; } // Just a short function that prints out the contents of an array in a manner that's easy to read // I used this function during debugging but it serves no purpose at run time for this example function showContents($array) { echo "<pre>"; print_r($array); echo "</pre>"; } // Set some constants // This variable is the path to the image folder where all the images are going to be stored // Note that there is a trailing forward slash $TARGET_PATH = "images/"; // Get our POSTed variable $image = $_FILES['image']; // Sanitize our input $image['name'] = mysql_real_escape_string($image['name']); // Build our target path full string. This is where the file will be moved to // i.e. images/picture.jpg $TARGET_PATH .= $image['name']; // Make sure all the fields from the form have inputs if ( $image['name'] == "" ) { $_SESSION['error'] = "All fields are required"; header("Location: member.php"); exit; } // Check to make sure that our file is actually an image // You check the file type instead of the extension because the extension can easily be faked if (!is_valid_type($image)) { $_SESSION['error'] = "You must upload a jpeg, gif, or bmp"; header("Location: member.php"); exit; } // Here we check to see if a file with that name already exists // You could get past filename problems by appending a timestamp to the filename and then continuing if (file_exists($TARGET_PATH)) { $_SESSION['error'] = "A file with that name already exists"; header("Location: member.php"); exit; } // Lets attempt to move the file from its temporary directory to its new home if (move_uploaded_file($image['tmp_name'], $TARGET_PATH)) { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); header("Location: images.php"); echo "File uploaded"; exit; } else { // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to // Make sure you chmod the directory to be writeable $_SESSION['error'] = "Could not upload file. Check read/write persmissions on the directory"; header("Location: member.php"); exit; } } //End of if session variable id is not present. ?> Now the display image script accompanying this insert image script is shown below. Code: [Select] <?php // Get our database connector require("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>Dream in code tutorial - List of Images</title> </head> <body> <div> <?php // Grab the data from our people table $sql = "SELECT * from images WHERE image_cartegory = 'main'"; $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo "<div class=\"picture\">"; echo "<p>"; // Note that we are building our src string using the filename from the database echo "<img src=\"images/ " . $row['filename'] . " \" alt=\"\" />"; echo "</p>"; echo "</div>"; } ?> </div> </body> </html> Well, needless mentioning, the picture isn't displayed. Just a tiny jpeg icon is displayed at the top left hand corner of the page. Figuring out that the problem might be the lack of a header that declares the image type, I add this line header("Content-type: image/jpeg"); but all it does is display a blank white page, without the tiny icon this time. What am I missing? Any clues?? Hello all, I have followed the stop on this page to what I thought was a tee: http://www.inkplant.com/code/pull-tw...-your-site.php but I guess I was wrong because I am getting errors. I have setup the database and table properly and created all files, but when I do step 4 and place the code on my site I am getting this error Code: [Select] Fatal error: Call to undefined function dbQuery() The api pull went fine and it stored the tweets in the database, but it can't display them. Is there a missing step about connecting to the database somewhere? Thanks for any help in need of an external link counter script! please help cheers matt |
|||||||