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Iam stuck on my login screen.  I have no clue what the deal is.  I don't know if I am not connecting to the database properly or what.  Here is what I have code wise and the screen shot is what I get when I run it in my browser.

 

<!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>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<?php
$con = mysql_connect("localhost", "tutorial", "LsGQmFYsMBjUwvL4");
mysql_select_db("tutorial", $con);
$q = "SELECT * FROM `logintable` WHERE user='$usernamefield' AND pass='$passwordfield'";
$logincheck = mysql_query($q, $con) or die(mysql_error($con));

if(!isset($_POST['username']) || !isset($_POST['password'])){
   print "No username/password defined";}
print_r($logincheck)."<br/>";
print_r($rows);

$usernamefield = $_POST['username'];
$passwordfield = $_POST['password'];

$rows = mysql_fetch_array($logincheck);

if ($rows != "") {
echo "Welcome please select your lake";
}
else {

?>


<form id="form1" name="form1" method="post" action="">
  <p>
    <label>
      User Name:  
      <input type="text" name="username" id="username" />
    </label>
  </p>
  <p>Password: 
    <label>
      <input type="text" name="password" id="password" />
    </label>
  </p>
  <p>
    <label>
      <input type="submit" name="submit" id="submit" value="Submit" />
    </label>
  </p>
</form>
</body>
</html>

 

screen1-1.jpg

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Hi,

If you have php installed (which you do) but are getting a white screen, this means that you have a php error but the server has it's warnings turned off.

 

Glancing through your code I can see the following errors:

 

line 15.

You are missing various brackets.

You have this:

if(!isset($_POST['username']) || !isset($_POST['password'])){

it should probably be this:

if( (!isset($_POST['username']) || (!isset($_POST['password'])) ){

 

Then, you open an "if" but don't close it:

if ($rows != "") {
   echo "Welcome please select your lake";
}
else {

?>


<form id="form1" name="form1" method="post" action="">
  <p>
    <label>
      User Name: 
      <input type="text" name="username" id="username" />
    </label>
  </p>
  <p>Password:
    <label>
      <input type="text" name="password" id="password" />
    </label>
  </p>
  <p>
    <label>
      <input type="submit" name="submit" id="submit" value="Submit" />
    </label>
  </p>
</form>

After that closing form tag you need to open php again and close the "if":

 

</form>
<?php
}
?>

 

That should help you on your way.

 

Chris

 

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You have this:

if(!isset($_POST['username']) || !isset($_POST['password'])){

it should probably be this:

if( (!isset($_POST['username']) || (!isset($_POST['password'])) ){

 

Actually you don't need to add in extra brackets like that. His was fine, yours has a syntax error.

 

Also to add though, you're trying to use a variable before you've declared it. The $usernamefield and $passwordfield variables need to be moved before the query. You should also looking securing those inputs from XSS attacks. 1 more thing as well, to determine if the query successfully returned a matching row, use mysql_num_rows.

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Actually you don't need to add in extra brackets like that. His was fine, yours has a syntax error.

Opps, I missed a closing bracket in the middle there, not very helpful  of me :-[ sorry.

Of course you are right about not needing the extra brackets, I should have mentioned them as they aren't part of the problem.

 

Other than what you suggest about the correct way to check if the query has returned a result, I still think that the main problem here is the lack of the closing brackets on the "if" clause.

 

Sorry if I have caused any confusion, trying to do too many things at once, I need to step away from the keyboard for a while  :P

 

Chris

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