[RESOLVED] login question

pnarendran

Hi friends:
Earlier I was able to fix a login code with your help. Now, I wanted to try if after verifying the login credentials using email and password of session user, I can display the users first name from a database, so in the code that we fixed earlier where I had a page called login.php I tried to make some changes to get the desired result.

The altered and current login.php file code is:

<!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">
   <body>

<?php include("header.php");include("config.php"); ?>

<!-- END HEADER --->

<?php if (isset($_SESSION['user'])){

$result = mysql_query("SELECT 'fname' FROM userprofile
 WHERE username=='(isset($_SESSION['user']))'")' or die(mysql_error());

echo "Welcome ".'$result');

?>

<!-- BEGIN FOOTER -->

<?php include("footer.php"); ?>

</body>
</html>

The error I get is:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in D:\Hosting\3420630\html\test\login.php on line 14

The original working code is:

<?php
session_start();
require_once 'config.php';


if (isset($_SESSION['user'])){


echo "Welcome ".$_SESSION['user'];

?>




<form name="logout" method="post" action="logout.php">


<input type="submit" name="logout" id="logout" value="Logout">


</form>





<?php

}else{
?>






<table align="left">
<tr><td>
 <form name="login_form" method="post" action="login2.php">
<div id="login-box-name" style="margin-top:10px;">Email:</div><div id="login-box-field" style="margin-top:20px;"><input name="user" id="user" class="form-login" title="Username" value="" size="30" maxlength="2048" /></div>
<div id="login-box-name">Password:</div><div id="login-box-field"><input name="pass" type="password" id="pass" class="form-login" title="Password" value="" size="30" maxlength="2048" /></div>

<span class="login-box-options"><input type="checkbox" name="1" value="1"> Remember Me <a href="#" style="margin-left:10px;">Forgot password?</a></span>
<br />
<br />
<a href="#"><input type="image" name="login" src="/india/images/login-btn.png" width="103" height="42" style="margin-left:90px;" /></a>

</div>

</div></form></table>

<?php
 }
 ?>

Can someone please help me see my mistake? Thanks in advance.🙂

Derokorian

$result = mysql_query("SELECT 'fname' FROM userprofile
WHERE username=='(isset($_SESSION['user']))'")' or die(mysql_error());

Extraneous quote. Remove.

$result = mysql_query("SELECT 'fname' FROM userprofile 
WHERE username=='(isset($_SESSION['user']))'") or die(mysql_error());

bradgrafelman

Note that you've got more problems with those 2 lines than just a missing quote...

Single quotes in a SQL query string are used to denote strings, not identifiers (such as the name of a column). As such, doing a SELECT 'fname' is going to do nothing but return the literal string 'fname' in the result set.
The comparison operator to test for equality in a SQL query string is a single equal sign, not two.
[man]isset/man is a PHP function, yet you've thrown it the middle of a string and I'm guessing expect PHP to see it and somehow do something with it? PHP doesn't look inside of strings for what looks like function calls it can execute...
... even if it did, [man]isset/man returns a boolean value, so you'd end up with something like "WHERE username=='1'" anyway.

pnarendran

I first removed the quote a shown - didn't work - it gave the same error. So I tried altering the code like this based on whatever little I understood from what was said by bradgrafelman.

New code:

<!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">
   <body>
<?php include("header.php");include("config.php"); ?>
<!-- END HEADER --->
<?php if (isset($_SESSION['user'])){
$user= (isset($_SESSION['user']))
$result = mysql_query("SELECT fname FROM userprofile
WHERE username= $user or die(mysql_error());
echo "Welcome ".$user);
?>
<!-- BEGIN FOOTER -->
<?php include("footer.php"); ?>
</body>
</html>

The error shifted to next line as seen below:

[PHP
Parse error: syntax error, unexpected T_VARIABLE in D:\Hosting\7520570\html\test\login.php on line 15
[/code]

Weedpacket

If you look at the code you posted you'll see how the syntax highlighting goes haywire around the middle. Specifically [font=monospace]...WHERE username=$user[/font] looks right, but [font=monospace]or die(mysql_erorr());...[/font] is wrong. There needs to be a [font=monospace]"[/font] in there.

pnarendran

Now I changed a little bit again with this code:

<!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">
   <body>
<? php include("header.php");include("config.php"); ?>
<!-- END HEADER --->
<? php if (isset($_SESSION['user'])){
$user= (isset($_SESSION['user']))
$result = mysql_query("SELECT fname FROM userprofile
WHERE username= $user or die(mysql_error());
echo "Welcome ".$user);
} 
?>
<!-- BEGIN FOOTER -->
<?php include("footer.php"); ?>
</body>
</html>

Now this error:

Parse error: syntax error, unexpected T_INCLUDE in D:\Hosting\7423570\html\test\login.php on line 5

Please help me someone. 🙂

Weedpacket

See my previous post; it still applies.

pnarendran

HI; I added the quote and still it gives me error:
Code with Correction as suggested

<td><img src="images/logo_s1.png" /></td>
<?php
session_start();
require_once 'config.php';

if (isset($_SESSION['user'])){
$user= (isset($_SESSION['user']))
$result = mysql_query("SELECT fname FROM userprofile
WHERE username= $user or die(mysql_error())");
echo "Welcome ".$user);
}
?>

<br>
<td><img src="images/menubar1.png" /></td><td><img src="images/menubar2.png" /></td>
<form name="logout" method="post" action="logout.php">

<input type="submit" name="logout" id="logout" value="Logout">

</form>

<?php

}else{
?>

<head>

<table align="left">
<tr><td>
 <form name="login_form" method="post" action="login2.php">
<div id="login-box-name" style="margin-top:10px;">Email:</div><div 

id="login-box-field" style="margin-top:20px;"><input name="user" id="user" class="form-login" title="Username" value="" size="30" maxlength="2048" /></div>
<div id="login-box-name">Password:</div><div id="login-box-field"><input name="pass" type="password" id="pass" class="form-login" title="Password" value="" 

size="30" maxlength="2048" /></div>

<span class="login-box-options"><input type="checkbox" name="1" value="1"> Remember Me <a href="#" style="margin-

left:10px;">Forgot password?</a></span>
<br />
<br />
<a href="#"><input type="image" name="login" src="/india/images/login-btn.png" width="103" height="42" 

style="margin-left:90px;" />  </a>

</div>

</div>
</body>

<?php
 }
 ?>

Error:

Parse error: syntax error, unexpected T_VARIABLE in D:\Hosting\7520570\html\india\loggedin.php on line 8

Weedpacket

You're missing a semicolon on the end of the previous line.

And you probably want $user to be the actual session value, not just a boolean that says whether it's set or not.

pnarendran

Hello friends;
Just when I thought I was close to making it work here is a new problem;

Here is my new code in login.php

<td><img src="images/logo_s1.png" /></td>
<?php
session_start();
require_once 'config.php';
if (isset($_SESSION['user'])){
$user=$_SESSION['user'];
$fname = mysql_query("SELECT fname and username FROM userprofile WHERE username = '$user'") or die (mysql_error());
echo "Welcome ".$fname;
?>
<br>
<td><img src="images/menubar1.png" /></td><td><img src="images/menubar2.png" /></td>
<form name="logout" method="post" action="logout.php">
<input type="submit" name="logout" id="logout" value="Logout">
</form>
<?php
}else{
?>
<head>
<table align="left">
<tr><td>
 <form name="login_form" method="post" action="login2.php">
<div id="login-box-name" style="margin-top:10px;">Email:</div><div 
id="login-box-field" style="margin-top:20px;"><input name="user" id="user" class="form-login" title="Username" value="" size="30" maxlength="2048" /></div>
<div id="login-box-name">Password:</div><div id="login-box-field"><input name="pass" type="password" id="pass" class="form-login" title="Password" value="" 
size="30" maxlength="2048" /></div>
<span class="login-box-options"><input type="checkbox" name="1" value="1"> Remember Me <a href="#" style="margin-
left:10px;">Forgot password?</a></span>
<br />
<br />
<a href="#"><input type="image" name="login" src="/india/images/login-btn.png" width="103" height="42" 
style="margin-left:90px;" />  </a>
</div>
</div>
</body>
<?php
 }
 ?>

With this code I don't get error but instead of echoing Welcome firstname of the person who logged in with his email id (username) which is the desired result it says: Welcome Resource id #3, which I don't understand where it is getting from - but I believe it is accessing the db because if I take out the ' ' in the SELECT statement before '$user' I get this error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@tomcox.net' at line 1

where @tomcox.net is part of the emailid(username) of person logged in.
I'm really going crazy with this. Please please help. Thanks in advance.🙂

Weedpacket

SELECT fname and username FROM userprofile WHERE username = '$user'"

You have the word "and" instead of a comma. As you've seen, computers usually don't cooperate when you don't pay attention to the rules of the language you're using to communicate with them.

bradgrafelman

Also note that you're trying to output the value returned by [man]mysql_query/man. That function does not return the data you're after - it returns a resource that you can use to fetch that data (e.g. using one of the mysql_fetch_*() functions).

pnarendran

thanks guys. I added these 2 lines and made all the fixes you all told me and it works great.🙂

[PHP
$data = mysql_fetch_assoc($res);
$fname = htmlspecialchars($data['fname']);
[/code]