Error MySQL server version for the right syntax to use near '-20, 20' at line 1

Question

How to fix this 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 '-20, 20' at line 1

The same script in my website is working properly but I uploaded to another server which is Windows server and received this error.

<?php

//connect to database
mysql_connect('xyz.ipowermysql.com','itomi','password.');
mysql_select_db('itomi');

$max_results = 20;
$from = (($page * $max_results) - $max_results);


if(empty($_POST)) {
    $query = "SELECT * FROM `itomi` WHERE `ntitle` LIKE '".$letter."%' ORDER BY `ntitle` ASC LIMIT $from, $max_results";
} 
$result = mysql_query("SET NAMES utf8"); //the main trick
$result = mysql_query($query) or die(mysql_error());
$rows = mysql_num_rows($result);

echo "<table class='hovertable' border='1' cellpadding='0' cellspacing='0'>";
echo "<tr><th>Keyword</th><th>Title</th><th>Detail</th></tr>";

if ($rows > 0) {
    while($row = mysql_fetch_array($result)) {
        echo "<tr><td>";
        echo '<a href="detail.php?id=' . $row['id'] . '" class="style1">' .$row['ntitle'].' </a>';
        echo "</td><td>";
        echo $row['ndetails'];
        echo "</td><td>";
        echo $row['counter'];
        echo "</td></tr>";
    }
} else {
    echo "<tr><td colspan=\"5\">No results found!</td></tr>";
}

echo "</table>";

// Figure out the total number of results in DB: 
$total_results = mysql_result(mysql_query("SELECT COUNT(*) as ntitle FROM itomi ORDER BY ntitle ASC"),0);

// Figure out the total number of pages. Always round up using ceil() 
$total_pages = ceil($total_results / $max_results);

// Build Page Number Hyperlinks 
echo "<p class=\"style1\">Pages: ";

// Build Previous Link 
if($page > 1){ 
    $prev = ($page - 1); 
    echo "<a href=\"".$_SERVER['php_SELF']."?page=$prev&letter=$letter\" class=\"style1\">Previous</a> "; 
}

for($i = 1; $i <= $total_pages; $i++){ 
    if(($page) == $i){ 
        echo "$i "; 
    } else { 
        echo " "; 
    }
}


// Build Next Link 
if($page < $total_pages){ 
    $next = ($page + 1); 
    echo "<a href=\"".$_SERVER['php_SELF']."?page=$next&letter=$letter\" class=\"style1\">Next</a>"; 
} 
echo "</p>";

mysql_close();

Seems like `$page` is empty or zero where it should be one or greater. — Salman A, Mar 30 '13 at 10:06

score 2 · Answer 1 · edited May 23 '17 at 12:30

2

Reading on the title which is part of the whole mysql syntax error message, the value you are passing on the LIMIT clause is less than the Minimum value of 0.

LIMIT 0~, 1~

the problem is caused here:

$from = (($page * $max_results) - $max_results)

try this:

$partVal = (($page * $max_results) - $max_results);
$from = ($partVal <= 0 ? 0 : $partVal)

As a sidenote, the query is vulnerable with SQL Injection if the value(s) of the variables came from the outside. Please take a look at the article below to learn how to prevent from it. By using PreparedStatements you can get rid of using single quotes around values.

How to prevent SQL injection in PHP?

edited May 23 '17 at 12:30

Community

1
1

answered Mar 30 '13 at 10:06

John Woo

258,903
69
498
492

so how can i fix this – user2226586 Mar 30 '13 at 10:06
i update my code but now page is not opening and shwoing Server error – user2226586 Mar 30 '13 at 10:13
what is exactly the error? – John Woo Mar 30 '13 at 10:13
$max_results = 20; $partVal = (($page * $max_results) - $max_results); $from = ($partVal <= 0 ? 0 : $partVal) i update this code – user2226586 Mar 30 '13 at 10:14
Server error The website encountered an error while retrieving http://xyz.com/page.php It may be down for maintenance or configured incorrectly. Here are some suggestions: Reload this webpage later. HTTP Error 500 (Internal Server Error): An unexpected condition was encountered while the server was attempting to fulfill the request. – user2226586 Mar 30 '13 at 10:15

Error MySQL server version for the right syntax to use near '-20, 20' at line 1

1 Answers1