Only show Mysql data from a certain user

winmastergames · February 5, 2008

Well just to say this is different to my other post...

Im trying to find a way to get data from a certain username only like email or something. The user will login and i need it to display their email address how do i go about trying to do this I tried in mysql just as a guess didnt work but i tried typing in

mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());

it just came up with nothing but i guess it will be something like that

Moon-Man.net · February 5, 2008

Providing the data in the MySql database field `username` matches $username, this should work.

Echo $username, and match it to the `username` field in the database.

If they don't match, then there is your problem. if they do, then it is somewhere in your display code.

PHP Monkeh · February 5, 2008

Remove the ' ' around $username. You could use '{$username}' or just do:

WHERE username = '".$username."'

winmastergames · February 5, 2008

It still doesnt show anything if i dont have WHERE username =....blah.blah stuff it shows the Data but when i do Where username =..blah.blah stuff..... it shows nothing?? what would be wrong i did Moon-Man.net post and everything was fine but does somehow the username and let say email do they have to be connected or something to be read??

Moon-Man.net · February 5, 2008

can you please post the complete code?

Thanks

haku · February 5, 2008

$email = mysql_query("SELECT email FROM users WHERE username = '" . $username . "''")or die(mysql_error());
$email = mysql_fetch_array($email);
echo "email: " . $email[0];

This will only output the email for one user, so if more than one person has the same username, its not going to work.

winmastergames · February 6, 2008

Heres the complete code

<?php 
//Defines Vars

// Connects to your Database 
mysql_connect("localhost", "root", "com12345") or die(mysql_error()); 
mysql_select_db("test") or die(mysql_error()); 

//checks cookies to make sure they are logged in 
if(isset($_COOKIE['ID_my_site'])) 
{ 
$username = $_COOKIE['ID_my_site']; 
$pass = $_COOKIE['Key_my_site']; 
$check = mysql_query("SELECT * FROM users WHERE username = '{$username}'")or die(mysql_error()); 
while($info = mysql_fetch_array( $check )) 
{ 

//if the cookie has the wrong password, they are taken to the login page 
if ($pass != $info['password']) 
{ header("Location: login.php"); 
} 

//otherwise they are shown the admin area	
else 
{ 
$yrwebsite = $_POST["weburl"];
echo "Website Administration<p>"; 
if (isset($_COOKIE["ID_my_site"]))
  echo "Welcome " . $_COOKIE["ID_my_site"] . " to SWD<br />";
else
  echo "Welcome guest!<br />";
echo "Your website URL is";
$sql = "SELECT * FROM users WHERE username = '$username'";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query)) {
echo "<td>".$row['weburls']."</td><br>";
}
echo "Your FTP Username is: ";
$sql = "SELECT * FROM users";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query)) {
echo "<td>".$row['ftpuser']."</td>";
}
echo "<br>";
echo "Your FTP Password is: ";
$sql = "SELECT * FROM users";
$query = mysql_query($sql);
while($row = mysql_fetch_array($query)) {
echo "<td>".$row['ftppass']."</td>";
}
echo "<br><a href=logout.php>Logout</a>"; 
} 
} 
} 
else 

//if the cookie does not exist, they are taken to the login screen 
{	
header("Location: login.php"); 
} 
?>

Moon-Man.net · February 6, 2008

Please try this

<?php 
//Defines Vars

// Connects to your Database 
mysql_connect("localhost", "root", "com12345") or die(mysql_error()); 
mysql_select_db("test") or die(mysql_error()); 

//checks cookies to make sure they are logged in 
if(isset($_COOKIE['ID_my_site'])){ 
$username = $_COOKIE['ID_my_site']; 
$pass = $_COOKIE['Key_my_site']; 
$check = mysql_query("SELECT * FROM users WHERE username = '{$username}'")or die(mysql_error()); 
while($info = mysql_fetch_array( $check )){ 
	//if the cookie has the wrong password, they are taken to the login page 
	if ($pass != $info['password']){
		header("Location: login.php"); 
	}else{ 
		$yrwebsite = $_POST["weburl"];
		echo "Website Administration<p>"; 
		if (isset($_COOKIE["ID_my_site"])){
		  echo "Welcome " . $_COOKIE["ID_my_site"] . " to SWD<br />";
		}else{
			echo "Welcome guest!<br />";
		}
		$sql = "SELECT * FROM users WHERE username = '$username'";
		$query = mysql_query($sql);

		while($row = mysql_fetch_array($query)) {
			echo "<td>Your website URL is: ".$row['weburls']."</td>";
			echo "<td>Your FTP Username is: ".$row['ftpuser']."</td>";
			echo "<td>Your FTP Password is: ".$row['ftppass']."</td>";
		}
		echo "<br><a href=logout.php>Logout</a>"; 
		echo "<br>";
		echo "<br>DEBUG RESULT:"
		print_r(mysql_fetch_array($query));
	} 
} 
}else{
//if the cookie does not exist, they are taken to the login screen	
header("Location: login.php"); 
} 
?>

winmastergames · February 6, 2008

Tried it and this error comes up

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\xampp\htdocs\userauthttt\members.php on line 36

Moon-Man.net · February 6, 2008

My Mistake, forgot a ';' on line 35

<?php 
//Defines Vars

// Connects to your Database 
mysql_connect("localhost", "root", "com12345") or die(mysql_error()); 
mysql_select_db("test") or die(mysql_error()); 

//checks cookies to make sure they are logged in 
if(isset($_COOKIE['ID_my_site'])){ 
$username = $_COOKIE['ID_my_site']; 
$pass = $_COOKIE['Key_my_site']; 
$check = mysql_query("SELECT * FROM users WHERE username = '{$username}'")or die(mysql_error()); 
while($info = mysql_fetch_array( $check )){ 
	//if the cookie has the wrong password, they are taken to the login page 
	if ($pass != $info['password']){
		header("Location: login.php"); 
	}else{ 
		$yrwebsite = $_POST["weburl"];
		echo "Website Administration<p>"; 
		if (isset($_COOKIE["ID_my_site"])){
		  echo "Welcome " . $_COOKIE["ID_my_site"] . " to SWD<br />";
		}else{
			echo "Welcome guest!<br />";
		}
		$sql = "SELECT * FROM users WHERE username = '$username'";
		$query = mysql_query($sql);

		while($row = mysql_fetch_array($query)) {
			echo "<td>Your website URL is: ".$row['weburls']."</td>";
			echo "<td>Your FTP Username is: ".$row['ftpuser']."</td>";
			echo "<td>Your FTP Password is: ".$row['ftppass']."</td>";
		}
		echo "<br><a href=logout.php>Logout</a>"; 
		echo "<br>";
		echo "<br>DEBUG RESULT:";
		print_r(mysql_fetch_array($query));
	} 
} 
}else{
//if the cookie does not exist, they are taken to the login screen	
header("Location: login.php"); 
} 
?>

winmastergames · February 6, 2008

I tried it no errors but none of the information showed up heres what exactly came up

Website Administration
Welcome ttt to SWD

Your website URL is: Your FTP Username is: Your FTP Password is: 
Logout

DEBUG RESULT:

haku · February 6, 2008

Did you try my code?

You don't need to select * from the database, that gives you way more than you need. You just need to select email.

Moon-Man.net · February 6, 2008

Haku, he is not trying to display the email at all, he wants the ftp username, ftp password and weburl for the specific user out of the database....

<?php 
//Defines Vars

// Connects to your Database 
mysql_connect("localhost", "root", "com12345") or die(mysql_error()); 
mysql_select_db("test") or die(mysql_error()); 

//checks cookies to make sure they are logged in 
if(isset($_COOKIE['ID_my_site'])){ 
$username = $_COOKIE['ID_my_site']; 
$pass = $_COOKIE['Key_my_site']; 
$check = mysql_query("SELECT * FROM users WHERE username = '{$username}'")or die(mysql_error()); 
while($info = mysql_fetch_array( $check )){ 
	//if the cookie has the wrong password, they are taken to the login page 
	if ($pass != $info['password']){
		header("Location: login.php"); 
	}else{ 
		$yrwebsite = $_POST["weburl"];
		echo "Website Administration<p>"; 
		if (isset($_COOKIE["ID_my_site"])){
		  echo "Welcome " . $_COOKIE["ID_my_site"] . " to SWD<br />";
		}else{
			echo "Welcome guest!<br />";
		}
		$sql = "SELECT * FROM users";
		$query = mysql_query($sql);
	        $result=mysql_fetch_array($query);
		while($result as $row ) {
			echo "<td>Your website URL is: ".$row['weburls']."</td>";
			echo "<td>Your FTP Username is: ".$row['ftpuser']."</td>";
			echo "<td>Your FTP Password is: ".$row['ftppass']."</td>";
		}
		echo "<br><a href=logout.php>Logout</a>"; 
		echo "<br>";
		echo "<br>DEBUG RESULT:";
                        echo "Username: " .$username ."<br>";
		print_r(mysql_fetch_array($query));
	} 
} 
}else{
//if the cookie does not exist, they are taken to the login screen	
header("Location: login.php"); 
} 
?>

Let me know what that outputs, I'm guessing $username isnt being set correctly

haku · February 6, 2008

Really? Cause his first message said this:

Im trying to find a way to get data from a certain username only like email or something. The user will login and i need it to display their email address how do i go about trying to do this

Sure looks to me like he was looking to get the email address...

Moon-Man.net · February 6, 2008

Haku, I apologize, my mistake. I was judging what he wanted via the code, I must have misread his first post.

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Only show Mysql data from a certain user

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