Can you please check this script???

spyros

In to the first file I have the form that the user fills the username and the password.
And then I have the second php page that checks in to the database if the username and the password is correct.
I run the form I enter the user name and the password and then in the psscript.php I have this error message:

Parse error: syntax error, unexpected $end in c:\diphp\webroot\pscript.php on line 52
What am I doing wrong ?

Form Page:

<html>
<body>
<h1>Home page</h1>
<?php	

echo "<form method=post action=\"pscript.php\">";
echo "<table>";
echo "<tr><td>Userid:</td>";
echo "<td><input type=text name=first_name></td></tr>";
echo "<tr><td>Password:</td>";
echo "<td><input type=text name=last_name></td></tr>";
echo "<tr><td colspan=2 align=center>";
echo "<input type=submit value=\"Log in\"></td></tr>";
echo "</table></form>";    

?>

<br>
</body>
</html>

pscript.php

<?php
session_start(); 

if (!$first_name || !$last_name)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	{


 require($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php");
 $connection = mysql_connect($db_host, $db_user, $db_password) or die("error connecting");
 mysql_select_db($db_name, $connection);


 $query = "select * from guestsbook "
          ."where name='$first_name' "
           ." and comment='$last_name'";
  $result = mysql_query($query, $connection);
  if (mysql_num_rows($result) > 0 )
  {

$valid_user = $first_name;

  }


?>



<?php 

  if (session_is_registered("valid_user"))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first_name))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>

samudasu

The closing brace "}" on this if statement is backwards. Turn it the other way.
{
echo "You have not entered all the required details . <br>";
exit;
{

thorpe

It would help if you let us know which is line 52. You might also want to tidy up your indentation, i can hardly read that.

spyros

Thank you
I haven’t seen that brace

Now the script it not showing any error but I can’t log in with the password and the user name. The script stops to the first if and it echo out “You have not entered all the required details”

if (!$first_name || !$last_name)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	}

Is the form script ok because it seem that is not passing the values to the verifying script????
Thank You!

spyros

pscript.php

<?php
session_start();
if (!$first_name || !$last_name)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	}


 require($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php");
 $connection = mysql_connect($db_host, $db_user, $db_password) or die("error connecting");
 mysql_select_db($db_name, $connection);


 $query = "select * from guestsbook "
          ."where name='$first_name' "
           ." and comment='$last_name'";
  $result = mysql_query($query, $connection);
  if (mysql_num_rows($result) > 0 )
  {

$valid_user = $first_name;

  }
?>
<http>
<head>
<title> Script </title>
</head>
<body>

<?php 

  if(isset($_SESSION['valid_user']))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first_name))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>
</body>
</html>

Weedpacket

This might be due to the issue covered in Question One of the Newbies' Forum FAQ.

thepeccavi

thorpe wrote:
It would help if you let us know which is line 52. You might also want to tidy up your indentation, i can hardly read that.

line 52 is the last line (?>)

i just counted :eek:

spyros

In the first IF I have two open braces that’s why I had this error.

I did some changes on the script

I replace the
if (session_is_registered("valid_user"))
with
if(isset($_SESSION['valid_user']))

in the psscript.php

if (session_is_registered("valid_user"))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else

Now the script it not showing any error but I can’t log in with the password and the user name. The script stops to the first if and it echo out “You have not entered all the required details”


if (!$first_name || !$last_name)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	}

I am not sure if the form sends the values to the pscript.php


<?php
session_start();
if (!$first_name || !$last_name)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	}


 require($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php");
 $connection = mysql_connect($db_host, $db_user, $db_password) or die("error connecting");
 mysql_select_db($db_name, $connection);


 $query = "select * from guestsbook "
          ."where name='$first_name' "
           ." and comment='$last_name'";
  $result = mysql_query($query, $connection);
  if (mysql_num_rows($result) > 0 )
  {

$valid_user = $first_name;

  }
?>
<http>
<head>
<title> Script </title>
</head>
<body>

<?php 

  if(isset($_SESSION['valid_user']))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first_name))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>
</body>
</html>

Thank you !!!

thorpe

Where do you define the variables $first_name and $last_name?

Answer... You dont. If your trying to retrieve these from your form, you need to use the $POST array. eg;

$_POST['first_name']

spyros

I have done the changes that you told me.
Now I can echo the values on the psctipt.php but I get this error:

first name: test
Last name: test

connection made

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\diphp\webroot\pscript5.php on line 27
Could not log you in

Something is wrong in the confirmation on the values and the DATABASE.

Pscript.php From line 27 to 32

 line 27---- if (mysql_num_rows($result) > 0 )
  {

$valid_user = $first;

  }

pscript5.php

<?php
session_start();
$first = $_POST["first1"];
$last = $_POST["last1"];
echo "first name: $first<br>";
echo "Last name: $last<br><br>";



if (!$first || !$last)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	}


 require($_SERVER["DOCUMENT_ROOT"]."/config/db_config.php");
 $connection = mysql_connect($db_host, $db_user, $db_password) or die("error connecting");
 mysql_select_db($db_name, $connection);
 echo "connection made<br><br>";


 $query = "select * from guestsbook "
          ."where name='$first'"
           ." and comment='$last'";
  $result = mysql_query($query, $connection);
  if (mysql_num_rows($result) > 0 )
  {

$valid_user = $first;

  }
?>
<http>
<head>
<title> Script </title>
</head>
<body>

<?php 

  if(isset($_SESSION['valid_user']))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>
</body>
</html>

Form script:

<html>
<head>
	<title> PASS </title>
</head>
<body>
<h1>Home page</h1>

<form action="pscript5.php" method="POST">
      <font face="arial" size="1">
            Name: <input type="text" name="first1"><br>
            Email: <input type="text" name="last1"><br>
            <input type="submit" value="Submit">
      </font>
</form>

<br>
</body>
</html>

Thank you

thorpe

Your query failed.

darkerstar

If you ever get a "not a valid MySQL result resource" error message, this means your SQL statement failed, if SQL statement run fine, but returned no result, you won't get this error and you will be able to use mysql_num_rows().

To fix it, this is what I always do.

Print your SQL statement, say echo $SQL;
Print out MySQL error, echo mysql_error()
If you have a very long and complicated SQL statement, it is quite difficult to spot an error this way. In this case, Enter phpMyAdmin, run your SQL in phpMyAdmin. It help you read the SQL statement by colouring the SQL code.

William

spyros

Thank you very mach darkerstar

I found the error I had a mistyping on the name of the table.
Now I don’t get any error

The psctipt.php print out:

first name: user
Last name: pass

connection made

Found a Valid User: user
Could not log you in

The script verifies the user name and the password from the FORM.
The problem is that I don’t gent the right message.

Something is not ok on this part of the script.


<?php 

  if (isset($_SESSION['valid_user']))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>

pscript5.php


<?php
session_start();
$first = $_POST["first1"];
$last = $_POST["last1"];
echo "first name: $first<br>";
echo "Last name: $last<br><br>";



if (!$first || !$last)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	}


 require($_SERVER["DOCUMENT_ROOT"]."/db_config2.php");
 $connection = mysql_connect($db_host, $db_user, $db_password) or die("error connecting");
 mysql_select_db($db_name, $connection);
 echo "connection made<br><br>";


 $query = "select * from auth "
          ."where name='$first'"
           ." and pass='$last'";
  $result = mysql_query($query, $connection);
  if (mysql_num_rows($result) > 0 )
  {

$valid_user = $first;
echo "Found a Valid User: $valid_user";
  }
  else
  {
     echo "No valid User";
  }


  echo "<br>";
  echo $SQL;
  echo mysql_error()
?>
<http>
<head>
<title> Script </title>
</head>
<body>

<?php 

  if (isset($_SESSION['valid_user']))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>
</body>
</html>

THANK YOU

thorpe

Something is not ok on this part of the script.

What does that meen? Please read the link in my signiture and come back with a well formed question.

spyros

Thank you very mach darkerstar

I found the error I had a mistyping on the name of the table.
Now I don’t get any error

The psctipt.php print out:

first name: user
Last name: pass

connection made

Found a Valid User: user
Could not log you in

The script verifies the user name and the password from the FORM.
The problem is that I don’t gent the right message.

There is probably a mistake on this part of the script


<?php 

  if (isset($_SESSION['valid_user']))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>

pscript5.php


<?php
session_start();
$first = $_POST["first1"];
$last = $_POST["last1"];
echo "first name: $first<br>";
echo "Last name: $last<br><br>";



if (!$first || !$last)
	{
	echo "You have not entered all the required details . <br>";
	exit;
	}


 require($_SERVER["DOCUMENT_ROOT"]."/db_config2.php");
 $connection = mysql_connect($db_host, $db_user, $db_password) or die("error connecting");
 mysql_select_db($db_name, $connection);
 echo "connection made<br><br>";


 $query = "select * from auth "
          ."where name='$first'"
           ." and pass='$last'";
  $result = mysql_query($query, $connection);
  if (mysql_num_rows($result) > 0 )
  {

$valid_user = $first;
echo "Found a Valid User: $valid_user";
  }
  else
  {
     echo "No valid User";
  }


  echo "<br>";
  echo $SQL;
  echo mysql_error()
?>
<http>
<head>
<title> Script </title>
</head>
<body>

<?php 

  if (isset($_SESSION['valid_user']))
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first))
    {

  echo "Could not log you in";
}
else 
{

  echo "You are not logged in.<br>";
  }
}
?>
</body>
</html>

THANK YOU

wilsonodk

You're getting the "Could not log you in" message for two reasons.

if (isset($_SESSION['valid_user'])) // Is not set
  {
    echo "You are logged in as: $valid_user <br>";
    echo "<a href=\"logout.php\">Log out</a><br>";
  }
  else
  {
    if (isset($first)) // Is set
    {
      // Therefore this prints
      echo "Could not log you in";
    }
    else 
    {
      echo "You are not logged in.<br>";
          }
    }

spyros

Thank you