tommyda

I need some help building a function that searches a specified page on an external site for a line of code (link), then returns true or false depending on whether it is there or not. I know it is possible because directorys use this function. I am willing to pay $10 though paypal or moneybookers to the first one who can help me Thanks for reading Tommy Arnold

Is it possible to search an external page for a certian line of code. I need it so i can keep track of backlinks so if any one can help me or direct me to a website that can it would be much appreciated!

No errors just a blank screen?? Full source for list.php <?php $payment = $_POST['payment']; $subcat = $_POST['subcat']; include"mysql.php"; // Get all the data from the "example" table $result = mysql_query("SELECT * FROM listings WHERE ('$payment' = 'yes')")or die(mysql_error()); // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { // Print out the contents of each row echo $row['title']; }; ?>

What i need to do is when i call the url www.mysite.com/list.php?site=fakesite that should set the $site var to 'fakesite' then select all from database where 'fakesite' = 'active' But its not pulling any results from the database is it reading '$site' as the previusly posted $POST['fakesite'] ?? <?php $site = $_POST['site']; include"mysql.php"; // Get the data from the table $result = mysql_query("SELECT * FROM listings WHERE ('$site' = 'active)")or die(mysql_error()); include'table.php'; ?>

Thank you

I know its going to be something small but i cant find it please help Error is on line 12 This is the code from lines 12 and 13 line 12 $id = $row['casino']; line 13 if($id == '0'){die'Not found';}; When i remove the code from line 13 it works

Fixed it myself using tizag.com // Get all the data from the "example" table $result = mysql_query("SELECT * FROM cat_gambling")or die(mysql_error()); // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { $id = $row['poker']; $result2 = mysql_query("SELECT * FROM listings WHERE id = $id")or die(mysql_error()); $row2 = mysql_fetch_array($result2) or die(mysql_error()); echo $row2['site']; };

When i try to run this script i get an error (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 '' at line 1) the purpose of the script is to get all id's (int) from table cat_gambling Then use the id's to grab information from the listings table eventually i will display the info in seperate tables using (while) function <?php include'mysql.php'; $result = mysql_query("SELECT * FROM cat_gambling")or die(mysql_error()); $id = $row['id']; while($row = mysql_fetch_array( $result )) { $result2 = mysql_query("SELECT * FROM listings WHERE id = $id")or die(mysql_error()); while($row2 = mysql_fetch_array( $result2 )) { echo $row2['site']; };}; ?> Can anyone help me find a way to fix it please ? EDIT by toplay: Changed topic subject and moved to MySQL forum.

the menu on the left of www.pokedepositoptions.com burns my eyes but im not sure if its because im tired or not so need a few more opinions Thanks in advance

Pretty simple at the end of your code put something like Include"button.php"; OR echo "<a href='nextpage.php'><img src'button.jpg'></a>"; ?>

Problem solved I included displayrating.php before setting $affid swapped them both round now it works!

Thank you i changed it but it still will not display the correct rating on paypal-poker.php

links working now

my new rating system is working ok here www.pokerdepositoptions.com/display.php?affid=1 For some reason the ratings are not showing properly in paypal-poker.php it is showing 0star.jpg even if it has a rating look here www.pokerdepositoptions.com/paypal-poker-deposit.php // Displayrating.php <? include'connect.php'; $result1 = mysql_query("SELECT * FROM ratings WHERE (affid = '$affid')"); $num1 = mysql_num_rows($result1); if ($num1 > 1){$txt = 'Ratings';} else {$txt='Rating';} $query2 = "SELECT AVG(rating) FROM ratings WHERE (affid = '$affid')"; $result2 = mysql_query($query2) or die(mysql_error()); // Print out result while($row2 = mysql_fetch_array($result2)){ $actualrating = $row2['AVG(rating)']; $roundrating = round($actualrating); $ratecode = "<img border='0' src='images/".$roundrating."star.jpg'> ".$num." ".$txt; } ?> paypal-poker.php <?php include"connect.php"; include"displayrating.php"; $result = mysql_query("SELECT * FROM affiliates WHERE paypal = '<li>paypal</li>' ORDER BY rank DESC")or die(mysql_error()); echo"<blockquote><p class='style10'>Paypal Poker Sites and casinos </p></blockquote>"; echo"<table width='100%' border='1' bordercolor='#EAEAEA' align='center' cellspacing='0' cellpadding='0' bgcolor='white'> "; echo"<tr ><td width='120'><center>Site</center></td>"; echo"<td><center>Link</center></td>"; echo"<td><center>Bonus</center></td>"; echo"<td><center>Bonuscode</center></td>"; echo"<td><center>Usa</center></td>"; echo"<td><center>Rank</center></td>"; echo"<td><center>Review</center></td>"; echo"</tr><tr><td>"; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { $affid = $row['id']; // Print out the contents of each row into a table echo"</tr><tr><td>"; echo "<a href='"; echo $row['url']; echo"'>"; echo"<center><img src='"; echo"$row[i120x120]' width='120' height='120' border='0' />"; echo"</a></center>"; echo"</td><td><center>"; echo"<a href='"; echo"$row"; echo"'>"; echo $row['site']; echo"</a>"; echo"</center></td><td><center>"; echo"$row[bonus]"; echo"</center></td><td><center>"; echo"$row[bonuscode]"; echo"</center></td><td><center>"; echo"$row[usa]"; echo"</center></td><td><center>"; echo $ratecode; echo"</center></td><td><center>"; echo"<a href='http://www.pokerdepositoptions.com/display.php?affid="; echo"$row[id]"; echo" '>Read more</a></center></tr><tr>"; } echo"</td><td></tr></table>"; ?> If anyone can help me it would be much appreciated !

THANK YOU !

How do i check if $_POST['post'] is a number OR is not a number like (IS_NAN)

I am stuck and cannot find any information online about User Enumeration and what i should do to prevent it can anyone help me

i keep getting the error T_CONSTANT_ENCAPSED_STRING when i remove the code below the script works so i know that there is something wrong with it if ($rating > "5"){die "Error: Rating must be between 1 and 5 please contact the support team";}; ????????/

Are there any security holes on my site www.pokerdepositoptions.com

Are there any security issues with my new site www.pokerdepositoptions.com It's not finished by the way

http://www.pokerdepositoptions.com/rate.php?affid=2&rating=5 <<<<RATE http://www.pokerdepositoptions.com/displayrating.php?affid=2 <<<<DISPLAY RATING Works like a dream

It works perfectly and i have learned a new way round handling ifelse's thank you!

There is no error at the moment but i need to put a Else{location: login.php} at the end of the page and its not working is there any other way of putting if if else else in that order

I am trying to build a rating system for my website but i am stuck as i need to check whether a user is logged in and if they ARE then check wether they have already rated and if they HAVE then rate but if they HAVENT say "sorry you have already rated" But if they are not logged direct them to the login page and say "you must be logged in to rate a site <?php include"include/session.php"; include"includes/mysql.php"; //check if logged in if($session->logged_in){ $user = $session->username; $rating=$_GET['rating']; $site=$_GET['affid']; //if logged in check whether this user has rated this site $con = "SELECT * FROM ratings WHERE user='$user' AND affid='$affid';"; $res = mysql_query($con); if (mysql_num_rows($res) > 0) {echo'You have already rated this page';} // yes, pull in the user details } else{ // no, user doesn't exist $insert ="INSERT INTO ratings (user, rating, affid) VALUES ('$user','$rating','$affid')"; $insert2 = mysql_query($insert); if(!$insert2) die(mysql_error()); echo('Thank you for rating');} } else{die'please login';} ?>

Ahhhh thank you its a verchar because originally I had a dollar sign in the values. I changed it to int and its all working perfectly thank you for your quick response!