在执行基本SELECT时，SQL语法中存在错误

Question

I'm trying to search a MySQL database for the user specific website activity, to do this I created a form and process to search through a statistics table and return every record with a userID matching the query. However I keep getting this message and not sure why:

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 '@hotmail.com' at line 1

This is the form I'm using

<form method='get' action='searchuser.php'  id='searchuser'>
<input type='text' input name='txtSearch' id='txtSearch'>
    <input type='submit' name='submit' value='Search'>
</form>";

and this is the process file

<?php
require_once( "Functions.php" );
$header = makeHeader();

$con= connect();


$user = $_GET['txtSearch'];

$query = "SELECT * FROM statistics WHERE userID = $user";
$result=mysql_query($query) or die (mysql_error());

echo"<table border='1'><th>User</th><th>IP</th>
    <th>Date</th><th>Page visited</th><th>Page from</th>";

while($row = mysql_fetch_assoc($result))
{
$username = $row['userID'];
$ip =  $row['ipAddress'];
$date = $row['dateOfVisit'];
$pagev = $row['pageVisited'];
$pagef = $row['pageFrom'];

echo "<tr><td>".$row->UserID."</td><td>".$row->ipAddress."</td><td>".$row->dateOfVisit."</td><td>".
    $row->pageVisited."</td><td>".$row->pageFrom . "<br/>
"."</td></tr>";
    }
IF (mysql_num_rows($queryresult) == "") 
            {
                Echo "<p>Sorry there were no results for your         search<p> <br /><br /> <p><A HREF='javascript:javascript:history.go(-1)'>Click here to go back to previous page</A></p>";
            } 


$footer = makeFooter();

?>

Answer 1

The SQL problem was with

userID = $user

The correct way of putting a string into sql query is

• to delimit it with quotes
• and use mysql_real_escape_string to escape possible delimiters that may happen inside.

So. the right code would be

  $user =  mysql_real_escape_string($user);
  $query = "SELECT * FROM statistics WHERE userID = '$user'";