php giving an error when sanitizing input

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When I do an insert the magic quotes work, but when I try to update a row it gives 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 'Road 12, St Julian\'s, UK'', lat = '35.911179', lng = '14.487689', ca' at line 1

Any ideas please?

PHP

<?php

function check_input($value){
    // Stripslashes
    if (get_magic_quotes_gpc())
      {
      $value = stripslashes($value);
      }
    // Quote if not a number
    if (!is_numeric($value))
      {
      $value = "'" . mysql_real_escape_string($value) . "'";
      }
    return $value;
}

//Get database credentials
require 'config.php';

$link = mysql_connect($dbhost, $dbusername,$dbpass);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
    mysql_select_db($dbname) or die('Cannot select database');

    $id = $_GET['id'];
    $riskCategory = $_GET['riskCategory'];
    $EventAccidentSubject = $_GET['EventAccidentSubject'];
    $description = $_GET['description'];
    $peopleInvolved = $_GET['peopleInvolved'];
    $hazards = $_GET['hazards'];
    //$address = $_GET['address'];
    $address = check_input($_GET['address']);
    $lat = $_GET['lat'];
    $lng = $_GET['lng'];
    $caseStatus = $_GET['caseStatus'];

    $query = "UPDATE dataentry" .
        " SET riskCategory = '".$riskCategory."'," .
        " EventAccidentSubject = '".$EventAccidentSubject."'," .
        " description = '".$description."'," .
        " peopleInvolved = '".$peopleInvolved."'," .
        " hazards = '".$hazards."'," .
        " address = '".$address."',".
        " lat = '".$lat."'," .
        " lng = '".$lng."'," .
        " caseStatus = '".$caseStatus."'" .
" WHERE id = '".$id."'";

//Run the query
$result = mysql_query($query) or die(mysql_error());

//link variable is equal to the referring page
$link = $_SERVER['HTTP_REFERER'];
//sends a header directly to the browser telling it to redirect the user to the referring page
header("Location: $link");
?>

Posted 16-Apr-13 20:57pm

datt265

Updated 18-Apr-13 5:28am

v2

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Comments

Prasad Khandekar 18-Apr-13 11:33am

Please use bind parameters (parameterised Queries), the current method you are using is a perfect way to disaster. See this link for more info (http://php.net/manual/en/mysqli-stmt.bind-param.php) or this (http://php.net/manual/en/security.database.sql-injection.php)

Regards,

2 solutions

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datt265 · Answer 1 · 2013-04-18T08:02:00

I have implemented this by I am getting an error, I can't figure out whats wrong

PHP

<?php

//Get database credentials
require 'config.php';



    $mysqli = new mysqli($dbhost,  $dbusername, $dbpass, $dbname);



$query = "UPDATE dataentry SET riskCategory = ?, "  .
        "EventAccidentSubject = ? , " .
        "description = ? , " .
        "peopleInvolved = ? , " .
        "hazards = ?, " .
        "address = ?, ".
        "lat = ?, " .
        "lng = ?, " .
        "caseStatus = ? " .
        "WHERE id = ?";
$stmt = $mysqli->prepare(query);
$stmt->bind_param("s", $riskCategory);
$stmt->bind_param("s", $EventAccidentSubject);
$stmt->bind_param("s", $description);
$stmt->bind_param("s", $peopleInvolved);
$stmt->bind_param("s", $hazards);
$stmt->bind_param("s", $address);
$stmt->bind_param("i", $lat);
$stmt->bind_param("i", $lng);
$stmt->bind_param("s", $caseStatus);
$stmt->bind_param("s", $id);
$stmt->execute();
$stmt->close();
$mysqli->close();

//link variable is equal to the referring page
$link = $_SERVER['HTTP_REFERER'];
//sends a header directly to the browser telling it to redirect the user to the referring page
header("Location: $link");
?>

Prasad Khandekar · Answer 2 · 2013-04-18T05:47:00

Hello,

Try changing your update code to something similar to one shown below.

PHP

$mysqli = new mysqli("server", "username", "password", "database_name");
$query = "UPDATE dataentry SET riskCategory = ?, "  .
        "EventAccidentSubject = ? , " .
        "description = ? , " .
        "peopleInvolved = ? , " .
        "hazards = ?, " .
        "address = ?, ".
        "lat = ?, " .
        "lng = ?, " .
        "caseStatus = ? " .
        "WHERE id = ?";
$stmt = $mysqli->prepare($query);
$stmt->bind_param("s", $riskCategory);
$stmt->bind_param("s", $EventAccidentSubject);
$stmt->bind_param("s", $description);
$stmt->bind_param("s", $peopleInvolved);
$stmt->bind_param("s", $hazards);
$stmt->bind_param("s", $address);
$stmt->bind_param("i", $lat);
$stmt->bind_param("i", $lng);
$stmt->bind_param("s", $caseStatus);
$stmt->bind_param("s", $id);
$stmt->execute();
$stmt->close();
$mysqli->close();

Regards,

php giving an error when sanitizing input

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