What is wrong with this MYSQL statement?

smproph · December 4, 2010

The problem lies in this line

$sql2="SELECT picture, email, participant_name FROM $tbl_name WHERE team_no='$team'";

I copied the rest of my code below so you can see the rest of my code. Any help would be great.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<script type="text/javascript">


  var _gaq = _gaq || [];
  _gaq.push(['_setAccount', 'UA-19722573-1']);
  _gaq.push(['_trackPageview']);

  (function() {
    var ga = document.createElement('script'); ga.type = 'text/javascript'; ga.async = true;
    ga.src = ('https:' == document.location.protocol ? 'https://ssl' : 'http://www') + '.google-analytics.com/ga.js';
    var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(ga, s);
  })();

</script>
<meta content="yes" name="apple-mobile-web-app-capable" />
<meta content="index,follow" name="robots" />
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<link href="../pics/homescreen.gif" rel="apple-touch-icon" />
<meta content="minimum-scale=1.0, width=device-width, maximum-scale=0.6667, user-scalable=no" name="viewport" />
<!--[if IE]> <link rel="stylesheet" type="text/css" href="../css/styleIE.css" /> 
<![endif]-->
<link href="../css/style.css" rel="stylesheet" media="screen" type="text/css" />
<script src="../javascript/functions.js" type="text/javascript"></script>
<title>SNU Intramurals</title>
<LINK REL="SHORTCUT ICON" HREF="images/favicon.ico">
</head>

<body>
<div id="outer">

<div id="topbar">

<div id="title">
SNU Intramurals</div>

<div id="rightbutton">
	<a href="../index.php">Home</a> </div>

<div id="leftnav">
	<a href="../gender.html">Back</a> </div>


</div>


<div id="content">
<ul class="pageitem">
	<li class="textbox3"><span class="header3">Teams</span></li>


<?php

$host="localhost"; // Host name 
$username="*****"; // Mysql username 
$password="*******"; // Mysql password 
$db_name="********"; // Database name 
$tbl_name="participants"; // Table Name

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// Retrieve data from database 
$sql="SELECT DISTINCT team_no FROM $tbl_name";
$result=mysql_query($sql);

// Start looping rows in mysql database.
while($rows=mysql_fetch_array($result)){

$team=$rows['team_no'];


?>

        
	<li class="menu">
       <a href="#"; 
       onclick = "
       if(document.getElementById('<?=$team?>').style.display=='inline') 
       {document.getElementById('<?=$team?>').style.display='none';} 
       else if (document.getElementById('<?=$team?>').style.display='none') 
       {document.getElementById('<?=$team?>').style.display='inline';}"
       >
        <span class="name">
        <table>
        <tr><td><?=$team?></td></tr>
        </table>
   		 </span>
         </a>
          </li>
        
        


<?
        
[b]$sql2="SELECT picture, email, participant_name FROM $tbl_name WHERE team_no='$team'";[/b]
$result2=mysql_query($sql2);
$team2=$team;
?>
<li class="textbox"><span class="name">

<table id="<?=$team?>" style="display:none;">
<?
$num=1;
while(($rows2=mysql_fetch_array($result2)) && ($num >= 1)){
$picture=$rows2['picture'];
$email=$rows2['email'];
$name=$rows2['participant_name'];


if ($num == 1) {
      echo "<tr>";
    }
    else if ($num == 4) {
      echo "</tr>";
      $num = 1;
    }

?>

<? 
echo "<td align='center'>";
echo "$name"; 
echo"<br \>";

if ($picture==T)

		{ 

		 echo '<SCRIPT LANGUAGE="javascript">
var url;
var email = "' . $email . '";
function emailsplit () 
{ 
var userid = email.split("@"); 
var url = userid[0]; 
var imgid = "http://my.snu.edu/images/idpictures/" + url + "-S.jpg"; 
return imgid;
}  </SCRIPT>';
		 echo "<SCRIPT LANGUAGE='javascript'>document.write ('<img src=\"' + emailsplit() + '\" width=\"64\" height=\"64\" alt=\"Picture ID\" style=\"-webkit-border-radius: 8px; -moz-border-radius: 1em;\" />'); </SCRIPT>";


}
else {
 echo "<SCRIPT LANGUAGE='javascript'>document.write ('<img src=\"http://dangerouslycareful.com/images/noimage.jpg\" width=\"64\" height=\"64\" alt=\"Picture ID\" />'); </SCRIPT>";
};
echo"<br \>";
echo"</td>";
$num++;				   
			   
?>


<?
// close while loop 

}
?>
</table>
</span></li>
<?

}

// close connection 
mysql_close();
?>
</ul>
</body>

</html>

smproph · December 4, 2010

I figured it out. My value was Men's 1. MYSQL does not like " ' " apparently. Just had to mysql_real_escape_string($team); and it works great now.

Pikachu2000 · December 4, 2010

You should be sanitizing ALL form data before using it in a database query anyhow. Not doing so is like begging people to hack your site.

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What is wrong with this MYSQL statement?

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