Member Avatar for din3sh

Hi .. I'm not so experienced with php .. I'm getting this error on the bolded line .. Can neone help me figure it out ..

Regards,
Dinesh Ramchandani

<?php
$con = mysql_connect("localhost","username","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
$db_selected = mysql_select_db('lasagnia_user', $con);
if (!$db_selected) {
    die ('Can\'t use user : ' . mysql_error());
}
?>
<?php
session_start();
if (isset($_REQUEST['Button'])) 
{
[B]$sqlquery = “SELECT email FROM userdetails WHERE email=(’".$_POST[email]."’)”;[/B]
$result = mysql_query($sqlquery,$con) or die(“Query died: email”);
$num = mysql_num_rows($result); 
if($num > 0) 
{
$sql = “SELECT email FROM userdetails WHERE email=’$_POST[email]’ AND pass=md5(‘$_POST[pass]’)”;
$result2 = mysql_query($sql,$con) or die(“Query died: pass”);
$num2 = mysql_num_rows($result2); 
if($num2 > 0) //password matches 
{
$_SESSION[‘auth’]=”yes”; 
$_SESSION[‘logname’] = $_POST[‘email’]; 
//header(“Location: SecretPage.php”); 
}
else // password does not match 
{
$message_1=”The Login Name, ‘$_POST[email]’
exists, but you have not entered the
correct password! Please try again.”;
$email = strip_tags(trim($_POST[email]));
//include(“form_login_reg.inc”);
} 
} // end if $num > 0
elseif($num == 0) // login name not found 
{
$message_1 = “The User Name you entered does not
exist! Please try again.”;
//include(“form_login_reg.inc”);
}



}
else
{?>





<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">
<head><title>Customer Login Page</title></head>
<div id='wrapper'> 
<div id='login'>
<form>
<fieldset style=’border: 2px solid #000000’>
<legend><b>Login Form</b></legend>
<div id=’field1’>
<label ><b>Email:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</label>
<input id=’email’ name=’email’ type=’text’
value=’ ’ size=’20’ maxlength=’255’ />
</div><br>
<div id=’field2’>
<label ><b>Password:</label>
<input id=’pass’ name=’pass’ type="password"
 size=’20’ maxlength=’255’ />
</div><br>
<input type='submit' name='Button' 
style=’margin-left: 45%; margin-bottom: .5em’
value='Login' />
</fieldset>
</form>
<p style=’text-align: center; margin: 1em’>
If you already have an account, log in.</p>
<p style=’text-align: center; margin: 1em’>
If you do not have an account, <a href="/regform/index.php">Register now.</a></p>
</html>
<?php }?>
  • 4 Contributors
  • 6 Replies
  • 248 Views
  • 6 Days Discussion Span
  • Latest Post Latest Post by rcollins

Recommended Answers

All 6 Replies

Member Avatar for jkon

You use three symbols for closing quoting marks” " and ' there are really only two in programming " and ' . Sometimes in your code you open with” and close with» that will cause PHP not to know what to do

Member Avatar for din3sh

u mean

$sqlquery = “SELECT email FROM userdetails WHERE email=’$_POST[email].’”;

i tried almost everything .. But no help ..

Member Avatar for din3sh

widout the end dot ..

Member Avatar for Shanti C

try this:

$sqlquery = "SELECT email FROM userdetails WHERE email='".$_POST['email']."'";
Member Avatar for din3sh

still the same error .. is there a problem with above lines ?

Member Avatar for rcollins

You need to tell your form element to use post as it's method. <form method="post"> Either that or use the $_REQUESTor $_GET variables

Edited by rcollins because: n/a
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