mark103

Hi guys, I need your help, I have got a problem with the if statement. When I don't insert the pass function in the url like this: http://www.mysite.com/myscript.php?image=myimagelocation&strings=mystrings&user=test I will get this on my php page: PASSWORD are missing <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbusername'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $image = clean($_GET['image']); $strings = clean($_GET['strings']); $username = clean($_GET['user']); $pass = clean($_GET['pass']); if($username == '' && $pass) { $errmsg_arr[] = 'username are missing'; $errflag = true; }elseif($username && $pass =='') { $errmsg_arr[] = 'PASSWORD are missing'; $errflag = true; } if($username == '' && $pass == '') { $errmsg_arr[] = 'username or password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['image'])) { $insert[] = 'image = \'' . clean($_GET['image']) . '\''; } if(isset($_GET['strings'])) { $insert[] = 'strings = \'' . clean($_GET['strings']) . '\''; } if(isset($_GET['user'])) { $insert[] = 'user = \'' . clean($_GET['user']) .'\''; } if(isset($_GET['pass'])) { $insert[] = 'pass = \'' . clean($_GET['pass']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); $required_fields = array('image', 'strings', 'user'); if($image && $strings && $username) { echo "working 1"; } elseif($username && $pass) { echo "working 2"; } } } ?> Do anyone know how to fix this?

i am not sure what you mean by debug the problem and how do you mean it?

sorry but i am not sure what you mean by debug the problem and how do you mean it?

I am trying to get pass on the if variable that if i do not insert one of those functions with included image, strings or user, i will always keep get this: Username or password are missing. It's just a mess up when I insert the functions in the url and I can't get pass on the if username and password variable. This one is where i am working on it. if($image == '' && $strings == '' && $username == '') { $errmsg_arr[] = 'image, strings or username are missing'; $errflag = true; } elseif($username == '' && $password == ''){ $errmsg_arr[] = 'Username or password are missing.'; $errflag = true; } Please read the first post i have already posted on above of what i have already said it and i am not going to say it twice or over again.

yes i have tried it and i couldn't find the solution. Any idea?

please can someone help?????????

Hi guys, I have got a problem with the if variable statements. When I insert the text of the image location, the name of the strings and when I did not insert the username, all I get this: image, strings or username are missing Username or password are missing. It should not display with the "Username or password are missing.", only the "image, strings or username are missing". The area of the code i am working on: if($image == '' && $strings == '' && $username == '') { $errmsg_arr[] = 'image, strings or username are missing'; $errflag = true; } elseif($username == '' && $password == ''){ $errmsg_arr[] = 'Username or password are missing.'; $errflag = true; } <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'myusername'); define('DB_PASSWORD', 'mypassword'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $image = clean($_GET['image']); $strings = clean($_GET['strings']); $username = clean($_GET['user']); $pass = clean($_GET['pass']); $delete = clean($_GET['delete']); if($image == '' && $strings == '' && $username == '') { $errmsg_arr[] = 'image, strings or username are missing'; $errflag = true; } elseif($username == '' && $password == ''){ $errmsg_arr[] = 'Username or password are missing.'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['image'])) { $insert[] = 'image = \'' . clean($_GET['image']) . '\''; } if(isset($_GET['strings'])) { $insert[] = 'strings = \'' . clean($_GET['strings']) . '\''; } if(isset($_GET['user'])) { $insert[] = 'username = \'' . clean($_GET['user']) .'\''; } if(isset($_GET['pass'])) { $insert[] = 'pass = \'' . clean($_GET['pass']) . '\''; } if(isset($_GET['delete'])) { $insert[] = 'delete = \'' . clean($_GET['delete']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); if(isset($image) && ($strings) && ($username)) { echo "test"; } elseif($username && $delete == 'all') { if ($delete != NULL) { mysql_query("DELETE FROM user_list WHERE username='$username'"); $deleted = mysql_affected_rows(); if($deleted > 0) { echo("The data are now deleted"); } else { echo("The user's data is empty"); } }else{ echo("failed"); } mysql_close($link); } } } ?> Do anyone know how i can get pass on those methods if I enter the images, the name of the strings and the username or the or the username with the password?

I don't like that idea and this isn't what i want, i still want to use the code like this: echo '<p id="images"> <a href="images.php?id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?id='.$row['id'].'">Delete</a> </td> | <p> <p id="test">'.$row['Public'].'</td>''; I can't find out how to use the code to allow to use in each paragragh. You can see what my data is showing on my php at the moment. my data 1 | Images | Link | Delete | Enabled my data 2 | Images | Link | Delete | Disabled I want something to be shown on my php like this: my data 1 | Images | Link | Delete | Enabled my data 2 | Images | Link | Delete | Disabled Any idea how i can do this?

i don't want a new paragraph in the same line as the other tags. Do you know how to put that tags of <p id="test">'.$row['Public'].' in the same line as the other tags? if so plz can you post an example code.

Do anyone know how i can put the html tags in the same line as the other html tags? Here's an example: Images | Link | Delete | Enabled On my code, it break the tags to the new line without put on the same line as the other tags. Here's the code: [code]<?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'myusername'); define('DB_PASSWORD', 'mypass'); define('DB_DATABASE', 'mydbtable'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); if($username == '') { $errmsg_arr[] = 'username ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'PASSWORD ID missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result) > 0) { $qrytable1="SELECT images, id, public FROM members WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo '<p id="images"> <a href="images.php?id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?id='.$row['id'].'">Delete</a> </td> | <p> <p id="test">'.$row['Public'].'</td>''; } } } ?> [/code]

Hi guys, I have added a column in a database, I know how to read the information from the database but I don't know how to check with the column name called "public" whether if the cell has text called enabled or disabled, then add the href link in enabled or disabled text in my php. I need it to allow the clients to control on their own account. <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'myusername'); define('DB_PASSWORD', 'mypass'); define('DB_DATABASE', 'mydbtable'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); if($username == '') { $errmsg_arr[] = 'username ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'PASSWORD ID missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result) > 0) { $qrytable1="SELECT images, id, public FROM members WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo '<p id="images"> <a href="images.php?id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?id='.$row['id'].'">Delete</a> </td> | Enabled | Disabled'; } } } ?> What I am trying to achieve is to check the column name called "public" whether if it has text that stored in the database (it could be enabled or disabled) then add the hyperlink in that text of enabled or disabled in the php page so it would allow the user to click one of the link to allow the hyperlink to switch over from enabled to disabled. E.G: A user is going to click the enabled text with the hyperlink to change the text in the database from enabled to disabled and remove the hyperlink in the php page from enabled to add the hyperlink in the disabled text. Do anyone know how to do this? Any advice would be much appreciated. Thanks in advance.

Thanks for your quick replied and thanks for your help MadTechie. I can see the problem is fixed, however i would like to print out on my php page when i enter the function name as favorites. When I enter the function name as favorites, it did not print out when I use echo. Code: [code]<?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $favorites = clean($_GET['favorites']); $id = clean($_GET['id']); if($favorites && $id == ''){ // both are empty $errmsg_arr[] = 'favorites id are missing.'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); } $insert = array(); if(isset($_GET['id'])) { $insert[] = 'id = \'' . clean($_GET['id']) .'\''; } if(isset($_GET['favorites'])) { $insert[] = 'favorites = \'' . clean($_GET['favorites']) . '\''; } if($favorites && $id) { echo "its working!"; } ?> Any idea?

Hi guys, I have a trouble with my php snippet, when I insert the var function in the url bar something is like: http://www.mysite.com/delete.php?favorites&id=0 or http://www.mysite.com/delete.php?whateveritis&id=0 It doesn't get pass the favorites function to delete the id. It is the same things that it goes for each different function. Here's the current code: <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbtablename'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $favorites = clean($_GET['favorites']); $id = clean($_GET['id']); if($favorites && $id == ''){ // both are empty $errmsg_arr[] = 'favorites id are missing.'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['id'])) { $insert[] = 'id = \'' . clean($_GET['id']) .'\''; } if(isset($_GET['favorites'])) { $insert[] = 'favorites = \'' . clean($_GET['favorites']) . '\''; } if($favorites && $id) { mysql_query("DELETE FROM favorites WHERE id='$id'"); $deleted = mysql_affected_rows(); if($deleted > 0) { echo "favorites channels is deleted"; } else { echo("favorites is already deleted"); } } } ?> If you do know how to get pass the favorites function, then please say so as i need your help. Any advice would be much appreicated.

Hi guys, I need a bit of your help. I have successfully deleted a row in a database after I have inserted the value name with a var function that match with a username in the url bar. Now I would like to delete more than one row, but I don't know how to do this? What I am trying to achieve is to enter the url something like this: www.mysite.com/delete.php?favorites=the value1&use=test It will search for a value in the database and then delete it. But if I add the &favorites=the value2 or more than 2, then search for the values in each row while looking for a username before delete the rows Here's the current code: <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $favorites = clean($_GET['favorites']); $username = clean($_GET['user']); if($favorites == '' && $username == ''){ // both are empty $errmsg_arr[] = 'Both name and email are missing. You must enter one or the other.'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['favorites'])) { $insert[] = 'favorites = \'' . clean($_GET['favorites']) . '\''; } if(($favorites) && isset($username)) { mysql_query("DELETE FROM favorites WHERE username='$username' AND favorites='$favorites'"); $deleted = mysql_affected_rows(); if($deleted > 0) { echo "favorites rows is deleted"; } else { echo("favorites are already deleted"); } } } ?> Any advice would be much appreciated. Thanks, Mark

Hi guys, I am recent working on my php to delete the information in a database. When I insert the value in the url bar something is like: www.mysite.com/delete.php?myfavorites=what ever it is&user=test. When I inserted the value in the url bar, it did not search for the value in a database to delete it while match with the username in the url bar. <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $favorites = clean($_GET['favorites']); $username = clean($_GET['user']); if($favorites == '' && $username == '') { // both are empty $errmsg_arr[] = 'Favorites are missing.'; $errflag = true; if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['favorites'])) { $insert[] = 'favorites = \'' . clean($_GET['favorites']) . '\''; } if(isset($_GET['username'])) { $insert[] = 'username = \'' . clean($_GET['username']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); if($favorites && $username) { mysql_query("DELETE valuedata FROM favorites WHERE username='$username'"); $deleted = mysql_affected_rows(); if($deleted > 0) { echo("The value in favorites are deleted"); } else { echo("failed"); } }else{ echo("failed"); } } } mysql_close($link); ?> If you know how to delete the value in a database for each row while it match with the username in the url, please say so as i would like to know how to do it. Any advise would be much appreciated. Thanks, Mark

Look! you don't understand what i am trying to do. I said I want to set the Enabled after the "protected:". Like this: Link | Delete | Protected: Enabled On my site, it's only show like this: Link | Delete | Protected: Enabled Any idea?

Thanks scootstah, I can see it have fixed the problem. However, I want to know how i can set the data next to the protected like "protected: enabled". My site is showing like this: Link | Delete | Protected: Enabled Do you know how i can do that?

Hi guys, I have got a problem with the parse error. Error: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/mysite/public_html/Protected.php on line 64 The parse error are highlighing in line 64. I found out that the problem are coming from this line: Protected: '<p id='Protected_Required'>' . $row["Protected_Required"] . '</p>'';[/Code] here's the current code: [code]<?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $pass = clean($_GET['pass']); if($username == '' && $pass == ''){ // both are empty $errmsg_arr[] = 'Username and password are missing. You must enter both or the other one.'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['user'])) { $insert[] = 'username = \'' . clean($_GET['user']) .'\''; } if(isset($_GET['pass'])) { $insert[] = 'pass = \'' . clean($_GET['pass']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); if($username && $pass) { $query="SELECT id, images, Protected_Required FROM mydatalist WHERE username='$username'"; $result1=mysql_query($query) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo '<a href="http://' . $row["images"] . '">Link</a> </td> | <a href="delete.php?id='.$row['id'].'">Delete</a></td> | Protected: '<p id='Protected_Required'>' . $row["Protected_Required"] . '</p>''; } } } } ?> Do you know how to fix the error I get? Any advice would be much appreciated. Thanks in advance.

thank you very much, problem solved!

It looks like this: whateveristoredinthedb. It just the data that I stored in the db cells. I hope you get this

it extract the data that I stored in the database and it have output the data into the hyperlink to turn into this http://mysite.freesite.com/mydatalink.com. It's doing it wrong which I want the data to turn to the output hyperlink like this http://whateveristoredinthedb. Do you know how to do this and if i need to make the change?

Hi all, I'm recent working on my php to extract some data from mysql database. Is it possible to output the database data into <a href> turn to the hyperlink instead of using http://mysite.freesite.com/mydatalink.com to this http://mydatalink.com? The script I have made so far is doing something wrong and I can't be able to find the solution. here's the current code: <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var) { return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />', $errmsg_arr); } else { $insert = array(); if(isset($_GET['user'])) { $insert[] = 'username = \'' . clean($_GET['user']) .'\''; } if (count($insert)>0) { $names = implode(',',$insert); if($username) { $qrytable1="SELECT id, links FROM mydatalist WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo '<p id="images"> <a href="images.php?id='.$row['id'].'">Images</a></td> | <a href="' . $row["links"] . '">Link</a> </td>'; } } } } ?> I don't want to adding an <a href> into the first echo statement, because I am stored the links in the database which you can see that I am using ['links']. I need to ouput the data from the db to turn into hyperlink. If anyone could assist with this I'd be very grateful! Thanks, Mark

Hi guys, I'm recent working on my php to add the row in mysql database. I need a bit of your help as I would like to know how to check the data in mysql database if it have added in mysql and then print it out without adding another same row? here's the current code: <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $pass = clean($_GET['pass']); $login = clean($_GET['login']); $all = clean($_GET['all']); if($username == '' && $pass == ''){ // both are empty $errmsg_arr[] = 'Username and password are missing. You must enter both or the other one.'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['user'])) { $insert[] = 'username = \'' . clean($_GET['user']) .'\''; } if(isset($_GET['pass'])) { $insert[] = 'pass = \'' . clean($_GET['pass']) . '\''; } if(isset($_GET['login'])) { $insert[] = 'LoggedUser = \'' . clean($_GET['login']) . '\''; } if(isset($_GET['all'])) { $insert[] = 'all = \'' . clean($_GET['all']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); if($username) { $qrytable1="SELECT username, LoggedUser FROM Online_Users WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo "<p id='LoggedUser'>"; echo $row['LoggedUser'] . "</p>"; } } if($username && $login == '1') { $sql="INSERT INTO Online_Users (username, LoggedUser) VALUES ('$_GET[user]','$_GET[login]')"; $result = mysql_query($sql); if($result) { $qrytable1="SELECT username, LoggedUser FROM Online_Users WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo "<p id='LoggedUser'>"; echo $row['LoggedUser'] . "</p>"; } } } elseif($username == 'all') { $names = implode(',',$insert); $result = mysql_query("SELECT * FROM Online_Users") or die('Error: ' . mysql_error()); $num_rows = mysql_num_rows($result); echo "There are $num_rows Users Online right now"; } } } ?>

Hi guys, I need your help with my php script. On my script, I'm currently working with image where I can hot-link them from another website while it is on protected. when you click right-mouse button on firefox, you could see something like "view page source" which is disabled. when you click on "save page as", you can save the image as "image.php". when you open them, you would not be able to read due to the image but you can find the real image link in image.php where i want to protect them. here's the currently code: <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'myusername'); define('DB_PASSWORD', 'mypassword'); define('DB_DATABASE', 'mydbname'); $id = (int)$_GET['id']; $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var) { return mysql_real_escape_string(strip_tags($var)); } $qrytable1="SELECT images FROM image_list WHERE id=$id"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { $image = $row['images']; $details = getimagesize($image); header ('Content-Type: ' . image_type_to_mime_type($details[2])); echo readfile($image), "<p id='images'>", $row['images'] . "</p>"; } ?> I guess that there must be a way to protected the url in the php script where i can hide them. It should be easy to modify but I am not sure what line I need to adjust to make it protected. Can you please help me in what line I need to modify in order to protected the hot-linking in my php?

i think i have found the solution. I have changed the id define instead of using the username define. every image have changes when i input each different id. however, this is the last thing i need before this will get solve. Can you please help me how to set the html tag for this line? echo "<p id='images'>", readfile($image); I am getting the url output on my php page. So once we set the html tag, this will be solve.