PHP Logged In As:

Butters

Butters

Butters

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Joined
22 Apr 2007
Posts
1,805
Me again,

I have a login which checks username and password stored in a MySQL db.

When the user logs in and gets redirected to the right page, I'd like to show a

"You are logged in as: xxxxxxxxxx"

thingy

Any ideas?
 
Erm, session variables?

Code: 
echo 'You are logged in as: ' . $_SESSION['username'];

You haven't really told us what your problem is.
 
No thats it. Cool. Thanks

One other thing. I have a field in a MySQL table that is classed as Unique. When a form submits to the MySQL database (with PHP) how can I get it to check for uniqueness, and, if its not unique, display something on the site to tell the user that?

If you know, that would be great.
 
I'm not great with SQL but I think you can do something like:

Code: 
$username = 'Joe'
$query = SELECT * FROM site_users WHERE username='$username'
$users = mysql_num_rows($query);

if ($users > 0){
//Username isn't unique
}

There may be an shorter way though. I believe that the MySQL 'unique' flag only checks data as it is inserted into the database, resulting in a query failure.
 
Code: 
$uniqueValue = mysql_real_escape_string($_POST['unique_value']);
$query = "SELECT `someuniquefield` FROM `sometable` WHERE `someuniquefield` = '$uniqueValue'";
$result = mysql_query($query);
$rowCount = mysql_num_rows($result);

if ($rowCount > 0)
{
    // Error logic.
}
else
{
    // Carry on with the script.
}
 
Cool thanks. So under //Error Logic you'd have something like

echo 'sorry, blah blah already exits', etc, etc?

Also, where you have:

($_POST['unique_value']);

does that have to be the field that is unique? (in my case, hotel_name)
 
Last edited:
Butters said:
echo 'sorry, blah blah already exits', etc, etc?
Click to expand...

You would probably want to display a message as well as the form again with all of the data from the previous page exept the username, this saves re-entering it all again.
 
to fill the form in with the submitted info is easy. you just echo the $_POST data back on screen like this. notice it's just a normal form but we're using the "value" to set it.......

Code: 
echo '<input type="text" name="username" value="'.htmlentities($_POST['username']).'">';

rather than echo the raw $_POST value, you should use the htmlentities function when displaying any user input on screen. this stops them from running malicious code.

and with the error messages, just create a string when an error is found and output that on the screen along with the form.

Code: 
<?php
//check user input
if error
     $error = 'sorry there was error. please check your details and try again.';
.....
.....
//output form
.....
.....
if($error) echo $error; 
?>

as an example of how that works, see my page here. enter your name but leave everything else blank.
 
Cool, thanks marc

I'll give this a go. Funny I was just reading up about HTMLentities on Tizag.

thanks again, I'll reply if I have any problems.
 
ok, I've got this and can't seem to find out whats wrong. Perhaps its a has of lots of different code areas.

Code: 
<?php
session_start();
if(!$_SESSION['username']) {
       //not logged in. redirect to login page
       header("Location: login.php");
       exit;
}

$address = "localhost";
$username = "myuser";
$password = "mypass";
$database = "mydb";

mysql_connect($address,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
?>

<!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>

</head>

<body>


<?php
echo "<strong>You are logged in as:</strong> " . $_SESSION['username'];
?>


<p>Please answer all questions. For blank fields, please enter <strong>N/A</strong>.</p>
<p>Fields marked with an asterisk (<span class="warning">*</span>) must be completed.</p>

<?php
if (!isset($_POST['submit'])) {
?>

<form action="" method="post">

//HTML FORM HERE
</form>



<?php

$uniqueValue = mysql_real_escape_string($_POST['hotel_name']);
$query = "SELECT `hotel_name` FROM `survey` WHERE `hotel_name` = '$uniqueValue'";
$result = mysql_query($query);
$rowCount = mysql_num_rows($result);

if ($rowCount > 0)
{
    // Error logic.
	echo '<input type="text" name="hotel_name" value="'.htmlentities($_POST['hotel_name']).'">';
}
else
{
    // Carry on with the script.
}




} else {

$hotel_name = mysql_real_escape_string($_POST['hotel_name']);
$contact = mysql_real_escape_string($_POST['contact_name']);
$telephone = mysql_real_escape_string($_POST['telephone']);
$do = mysql_real_escape_string($_POST['do']);
$have = mysql_real_escape_string($_POST['have']);



mysql_query("INSERT INTO `survey` (hotel_name, contact, telephone, do, have) VALUES 

('$hotel_name','$contact','$telephone','$do','$have')");
echo "Thank you! Your survey is now complete and the information has been submitted";
}


mysql_close();

?>

</body>
</html>

When I hit submit, the "thank you your data has been entered" message appears, but nothing enters the Database.

NOTE:(I am forcing the same username as I already have stored in the DB, if I create a unique entry, all is well).

EDIT:

and what about this......


Quote:
My last php question ever!

lies.
Click to expand...

Teehee :p
 
oh dear. :p you need to output the whole form - not just the one field with the $_POST values. also your error checking is all wrong.

try this.....

Code: 
<?php
session_start();
if(!$_SESSION['username']) {
       //not logged in. redirect to login page
       header("Location: login.php");
       exit;
}

$address = "localhost";
$username = "myuser";
$password = "mypass";
$database = "mydb";

mysql_connect($address,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

if (isset($_POST['submit'])) {
	$hotel_name = mysql_real_escape_string($_POST['hotel_name']);
	$contact = mysql_real_escape_string($_POST['contact_name']);
	$telephone = mysql_real_escape_string($_POST['telephone']);
	$do = mysql_real_escape_string($_POST['do']);
	$have = mysql_real_escape_string($_POST['have']);
	
	$query = "SELECT `hotel_name` FROM `survey` WHERE `hotel_name` = '$hotel_name'";
	$result = mysql_query($query);
	if (mysql_num_rows($result) == 1) {
		$message = 'That hotel name is already in the database. Please use another.';
	} else {
		mysql_query("INSERT INTO `survey` (hotel_name, contact, telephone, do, have) VALUES ('$hotel_name','$contact','$telephone','$do','$have')");
		$message = "Thank you! Your survey is now complete and the information has been submitted";
	}
	mysql_close();
?>
<!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>

</head>

<body>


<?php
echo "<strong>You are logged in as:</strong> " . $_SESSION['username'];
?>


<p>Please answer all questions. For blank fields, please enter <strong>N/A</strong>.</p>
<p>Fields marked with an asterisk (<span class="warning">*</span>) must be completed.</p>

<form action="" method="post">
<?php
echo '<input type="text" name="hotel_name" value="'.htmlentities($_POST['hotel_name']).'">';
//repeat that for each input
if($message) echo '<p>'.$message.'</p>';
?>
</form>
</body>
</html>
 
Cool,

Ok, two things:

1), I'm getting this error when the page loads now:

Code: 
Parse error: syntax error, unexpected $end in survey.php on line 72

2) in my form I have drop down boxes eg
Code: 
<select name="do">
<option value="Yes" selected="selected">Yes</option>
<option value="No">No</option>
</select></p>

how do I code that wih the html entities?

Thanks again
 
Last edited:
1) probably just a typo somewhere with the php code. good practice for you to find it. :p
2) this isn't the best method but for a one off it should be ok......

Code: 
<select name="do">
<?php
if($_POST['do'] == 'No') {
        echo '<option value="Yes">Yes</option>';
        echo '<option value="No" selected="selected">No</option>';
} else {
        echo '<option value="Yes" selected="selected">Yes</option>';
        echo '<option value="No">No</option>';
}
?>
</select>

if you had multiple drop down boxes in the same form, you'd create a function and use arrays instead of bodging it like that.
 
Havent used PHP/MySQL in a while but surely attempting to insert an existing value into a unique field results in the insert query failing and thus affecting no rows? If thats the case the check with the select query is completely redundant.
 
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