supermerc Posted April 6, 2007 Share Posted April 6, 2007 Hey, im trying to make this code work, its for a rating system, but im getting this error: 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 'WHERE id='20'' at line 1 I dont know whats wrong ??? This is my code <?php $rating_posted=$_GET['vote'];//pased variable by the the stars value $id=(INT)$_GET['member_id']; $query=mysql_query("SELECT total_votes, total_value, used_ips FROM $tableName WHERE id='".$id."' ")or die(" Error: ".mysql_error()); $numbers=mysql_fetch_assoc($query); $checkIP=unserialize($numbers['used_ips']); $count=$numbers['total_votes'];//how many votes total $current_rating=$numbers['total_value'];//total number of rating added together and stored $sum=$rating_posted+$current_rating;// add together the current vote value and the total vote value $tense=($count==1) ? "vote" : "votes";//plural form votes/vote $voted=@mysql_fetch_assoc(@mysql_query("SELECT title FROM $tableName WHERE used_ips LIKE '%".$_SERVER['REMOTE_ADDR']."%' AND id='$id' ")); //Pattern match ip:suggested by Bramus! http://www.bram.us/ - this variable searches through the previous ip address that have voted and returns true or false if($voted){ echo "<div class=\"rating\">". "<ul class=\"star-rating\">". "<li class=\"current-rating\" style=\"width:". @number_format($current_rating/$count,2)*20 ."px;\">Current rating.</li>". "<li class=\"one-star\">1</li>". "<li class=\"two-stars\" >2</li>". "<li class=\"three-stars\">3</li>". "<li class=\"four-stars\">4</li>". "<li class=\"five-stars\">5</li>". "</ul>". "<p>Rating: <strong>".@number_format($current_rating/$count,2)."</strong> {".$count." ".$tense." cast} <br />You have previously voted.</p></div>";//show the current value of the vote with the current numbers }else{ if(isset($_GET['vote'])){ if($sum==0){ $added=0;//checking to see if the first vote has been tallied }else{ $added=$count+1;//increment the current number of votes } if(is_array($checkIP)){ array_push($checkIP,$_SERVER['REMOTE_ADDR']);//if it is an array i.e. already has entries the push in another value }else{ $checkIP=array($_SERVER['REMOTE_ADDR']);//for the first entry } $insert=serialize($checkIP); mysql_query("UPDATE $tableName SET total_votes='".$added."', total_value='".$sum."', used_ips='".$insert."' WHERE id='".$_GET['id']."'"); echo "<div class=\"rating\"><p>Rating: <strong>".@number_format($sum/$added,2)."</strong> {".$added." ".$tense." cast} <span>Thank you for your vote!</span></p></div>";//show the updated value of the vote }else{ ?> <div class="rating"> <p>How clear was this tutorial?</p> <ul class="star-rating"> <li class="current-rating" style="width:<?php echo @number_format($current_rating/$count,2)*20 ?>px;">Current rating</li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 1 star out of 5" class="one-star">1</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 2 stars out of 5" class="two-stars" >2</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 3 stars out of 5" class="three-stars" >3</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 4 stars out of 5" class="four-stars" >4</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 5 stars out of 5" class="five-stars" >5</a></li> </ul> <?php echo "<p>Rating: <strong>".@number_format($sum/$count,2)."</strong> {".$count." ".$tense." cast}</p></div>";//show the current updated value of the vote } // end isset get vote } //end voted true, false ?> Please help me! Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/ Share on other sites More sharing options...
AndyB Posted April 6, 2007 Share Posted April 6, 2007 If that's all your code ... you need to connect to the database and select a database and define the tablename before you execute a query. Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222673 Share on other sites More sharing options...
supermerc Posted April 6, 2007 Author Share Posted April 6, 2007 it does connect I got external files connectdb closedb etc. Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222674 Share on other sites More sharing options...
fert Posted April 6, 2007 Share Posted April 6, 2007 where is $tablename defined Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222675 Share on other sites More sharing options...
trq Posted April 6, 2007 Share Posted April 6, 2007 Firstly. Calling functions within functions like this.... $voted=@mysql_fetch_assoc(@mysql_query Is just begging for trouble. You have absolutely no error handling in place. As for which query is failing. I'll assume its the fist one. The syntax for running a SELECT should be something like... <?php $sql = "SELECT total_votes, total_value, used_ips FROM $tableName WHERE id = $id"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { $row = mysql_fetch_assoc($result); // display data. } else { echo "No results found"; } } else { echo "Query failed<br />$sql<br />". mysql_error(); } ?> This will print the actual query if there is a problem. Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222676 Share on other sites More sharing options...
supermerc Posted April 6, 2007 Author Share Posted April 6, 2007 Ok thx a lot thorpe it did change something although im getting a new error now, Query failed SELECT total_votes, total_value, used_ips FROM WHERE id = 19 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 'WHERE id = 19' at line 1 Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/randomy/public_html/star_rating.php on line 15 Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222678 Share on other sites More sharing options...
trq Posted April 6, 2007 Share Posted April 6, 2007 Look at the outputted query. SELECT total_votes, total_value, used_ips FROM WHERE id = 19 As others have posted, you have failed to define $tableName. Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222680 Share on other sites More sharing options...
supermerc Posted April 6, 2007 Author Share Posted April 6, 2007 its in another file, ill try defining it in that page Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222681 Share on other sites More sharing options...
trq Posted April 6, 2007 Share Posted April 6, 2007 If its in another file, are you including that file? Whatever your doing $tableName is empty or is out of scope. Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222684 Share on other sites More sharing options...
supermerc Posted April 6, 2007 Author Share Posted April 6, 2007 heh, well no i forgot to include... anyways I fixed that now im getting Parse error: syntax error, unexpected T_VARIABLE in /home/randomy/public_html/star_rating.php on line 4 Line 4 is $rating_posted=$_GET['vote'];//pased variable by the the stars value Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222687 Share on other sites More sharing options...
AndyB Posted April 6, 2007 Share Posted April 6, 2007 Bet there's a problem with line 1 or line 2 or line 3 - none of which we know - like a missing ; Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222688 Share on other sites More sharing options...
supermerc Posted April 6, 2007 Author Share Posted April 6, 2007 you were right there was, but now theres a new error, No results found Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/randomy/public_html/star_rating.php on line 16 <?php include'config.php'; $tableName=ratings; $rating_posted=$_GET['vote'];//pased variable by the the stars value $id=(INT)$_GET['member_id']; $sql = "SELECT total_votes, total_value, used_ips FROM $tableName WHERE id = $id"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { $row = mysql_fetch_assoc($result); } else { echo "No results found"; } } else { echo "Query failed<br />$sql<br />". mysql_error(); } $numbers=mysql_fetch_assoc($query); $checkIP=unserialize($numbers['used_ips']); $count=$numbers['total_votes'];//how many votes total $current_rating=$numbers['total_value'];//total number of rating added together and stored $sum=$rating_posted+$current_rating;// add together the current vote value and the total vote value $tense=($count==1) ? "vote" : "votes";//plural form votes/vote $voted=@mysql_fetch_assoc(@mysql_query("SELECT title FROM $tableName WHERE used_ips LIKE '%".$_SERVER['REMOTE_ADDR']."%' AND id='$id' ")); //Pattern match ip:suggested by Bramus! http://www.bram.us/ - this variable searches through the previous ip address that have voted and returns true or false if($voted){ echo "<div class=\"rating\">". "<ul class=\"star-rating\">". "<li class=\"current-rating\" style=\"width:". @number_format($current_rating/$count,2)*20 ."px;\">Current rating.</li>". "<li class=\"one-star\">1</li>". "<li class=\"two-stars\" >2</li>". "<li class=\"three-stars\">3</li>". "<li class=\"four-stars\">4</li>". "<li class=\"five-stars\">5</li>". "</ul>". "<p>Rating: <strong>".@number_format($current_rating/$count,2)."</strong> {".$count." ".$tense." cast} <br />You have previously voted.</p></div>";//show the current value of the vote with the current numbers }else{ if(isset($_GET['vote'])){ if($sum==0){ $added=0;//checking to see if the first vote has been tallied }else{ $added=$count+1;//increment the current number of votes } if(is_array($checkIP)){ array_push($checkIP,$_SERVER['REMOTE_ADDR']);//if it is an array i.e. already has entries the push in another value }else{ $checkIP=array($_SERVER['REMOTE_ADDR']);//for the first entry } $insert=serialize($checkIP); mysql_query("UPDATE $tableName SET total_votes='".$added."', total_value='".$sum."', used_ips='".$insert."' WHERE id='".$_GET['id']."'"); echo "<div class=\"rating\"><p>Rating: <strong>".@number_format($sum/$added,2)."</strong> {".$added." ".$tense." cast} <span>Thank you for your vote!</span></p></div>";//show the updated value of the vote }else{ ?> <link href="star_rating.css" rel="stylesheet" type="text/css" /> <div class="rating"> <p>How clear was this tutorial?</p> <ul class="star-rating"> <li class="current-rating" style="width:<?php echo @number_format($current_rating/$count,2)*20 ?>px;">Current rating</li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 1 star out of 5" class="one-star">1</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 2 stars out of 5" class="two-stars" >2</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 3 stars out of 5" class="three-stars" >3</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 4 stars out of 5" class="four-stars" >4</a></li> <li><a href="<?php echo $_SERVER['PHP_SELF'] . "?" .$_GET['section'] . "&id=" . $_GET['id'] . "&vote=1";?>" title="Rate this 5 stars out of 5" class="five-stars" >5</a></li> </ul> <?php echo "<p>Rating: <strong>".@number_format($sum/$count,2)."</strong> {".$count." ".$tense." cast}</p></div>";//show the current updated value of the vote } // end isset get vote } //end voted true, false ?> Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222691 Share on other sites More sharing options...
AndyB Posted April 6, 2007 Share Posted April 6, 2007 $tableName = "ratings"; If you used sensible error-trapping like echoing the query when there's an error you would save lots of time. Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222692 Share on other sites More sharing options...
trq Posted April 6, 2007 Share Posted April 6, 2007 Just copying and pasting code without looking at how it works won't help you. The code I posted was s simple example, you need to make it work within the context of what YOU already have. $sql = "SELECT total_votes, total_value, used_ips FROM $tableName WHERE id = $id"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { $numbers = mysql_fetch_assoc($result); $checkIP = unserialize($numbers['used_ips']); $count = $numbers['total_votes'];//how many votes total $current_rating = $numbers['total_value'];//total number of rating added together and stored $sum = $rating_posted+$current_rating;// add together the current vote value and the total vote value $tense = ($count==1) ? "vote" : "votes";//plural form votes/vote } else { echo "No results found"; } } else { echo "Query failed<br />$sql<br />". mysql_error(); } Link to comment https://forums.phpfreaks.com/topic/45832-rating-system-sql-syntax-error/#findComment-222697 Share on other sites More sharing options...
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