[RESOLVED] Quick question

gszauer

Quick question
PHP

<?php 
$location = "localhost"; 
$username = "nrcvcfh"; 
$password = "ci893i4yen"; 
$database = "nrcvcfh_vexour"; 

mysql_connect($hostname,$username, $password) OR DIE (" ");
mysql_select_db($database); 
$conn = @mysql_connect($location,$username,$password); 

mysql_select_db($database,$conn) or die ("Could not open database"); 
$insert = "INSERT INTO vexour (track_title,band_name,song_name,song_url,image_url,band_url,stars) 
    VALUES ('".$a1."','".$a2."','".$a3."','".$a4."','".$a5."','".$a6."','3')"; 

mysql_query($insert) or die ("An error has occured, and the data can not be added to the list, please contact the administrator. <Br />Error: " . mysql_error());
?>

HTML

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>index</title>
<style type="text/css">
<!--
.style2 {font-size: smaller}
.style4 {font-size: 9px}
.style8 {font-size: 14px}
-->
</style>
</head>
<body bgcolor="#ffffff">
<form name="form1" method="post" action="insert.php">
  <table width="259" border="0" cellspacing="0" cellpadding="0">
    <tr>
      <td width="284"><p class="style2"> Track Title
              <input name="a1" type="text" class="style4" id="a1" size="60">
              Band Name
              <input name="a2" type="text" class="style4" id="a2" size="60">
        Song Name
        <input name="a3" type="text" class="style4" id="a3" size="60">
        Song URL
        <input name="a4" type="text" class="style4" id="a4" size="60">
        Image URL
        <input name="a5" type="text" class="style4" id="a5" size="60">
        <span class="style8">Website URL</span>
        <input name="a6" type="text" class="style4" id="a6" size="60">
        </p>
          <p>
            <input type="submit" class="style4" value="Submit">
        </p></td>
    </tr>
  </table>
  </form>
</body>
</html>

When i click Submit there is a blank row added to the databse, with notihng in it... Is there something wrong with my script?
Any help is appriciated, and thank you for your time.
~Gabor Szauer

herve

There was a time where "magically" the value inputed in the HTML form were directly created
in PHP variables, but not today, so add :

$a1 = $POST["a1"];
$a2 = $POST["a2"];
$a3 = $POST["a3"];
$a4 = $POST["a4"];
$a5 = $POST["a5"];
$a6 = $POST["a6"];

at the beginning of your script and it should be ok.

Roger_Ramjet

And do some input validation while you are at it. Look up mysql_real_escape_string() in the manual and google for sql injection attacks if you don't know why this is required.