My name is a... number?

[JelvinJS7]

I'm trying to add a very simple feature: if a user is logged in, say hello (i don't know why I'm wasting my time on this right now. Not important really—but now it's frustrating!). I'm using the MySQL class from the BetterPHP library.
for some reason, it outputs "Hello 1". I understand why 1--it's my user id. I don't understand why it's not outputting "Jonathan" or "James" (I don't know what name I used). Here's my code:
[syntax=php]
<?php
If (isset($_SESSION['uid'])){
     include("includes/mysql.class.inc.php");
     $mysql = new mysql("mysql**.000webhost.com", "******", "*****^", "*******");
    $q = $mysql->query("SELECT `name` FROM `users` WHERE `id` = {$_SESSION['uid']}");
    $name = $mysql->fetch_cell($q);
?>
<p style="float:center">Hello <?php echo $name; ?></p>
[/syntax]
My table `users` has 5 columns: name, username, password, email, and user_id (not in order) Witt the usual settings (unique email/username, A_I/primary id, etc.) can someone explain what's wrong?
And yes, it is connecting correctly.

[bowersbros]

well if you have user_id as your field, then change id to user_id

Screenshot of the table will help

[JelvinJS7]

well if you have user_id as your field, then change id to user_id

Screenshot of the table will help

No, it's id in the table, but uid in the session. `user_id` results in an error.
The table is exactly as I said: id, name, username, password, email. I'll post a pic later; can't right now.
The thing worked for me before..... In a different part of Ye website...

[jacek]

The 1 is probably true. I think you are meant to do

[syntax=php]$mysql = new mysql("mysql**.000webhost.com", "******", "*****^", "*******");
$mysql->query("SELECT `name` FROM `users` WHERE `id` = {$_SESSION['uid']}");
$mysql->fetch_cell($name);[/syntax]
been a while since I looked at the code though

[JelvinJS7]

Doesn't work.
Here's the code:
[syntax=php]
public function fetch_cell(&$result){
$result = mysql_result($this->result, 0);

return ($result !== false);
}
[/syntax]

I also tried this minor edit:
[syntax=php]
public function fetch_cell(&$result, $cell = null){
$result = mysql_result($this->result, 0, $cell);

return ($result !== false);
}
[/syntax]

[JelvinJS7]

here's a screen shot of the table structure

`users` table

Screen Shot 2011-12-19 at 8.12.34 PM.png (92.49 KiB) Viewed 2643 times

[jacek]

It should be used as I said above

Try this and see what the exact value you are getting is

[syntax=php] $mysql = new mysql("mysql**.000webhost.com", "******", "*****^", "*******");
$mysql->query("SELECT `name` FROM `users` WHERE `id` = {$_SESSION['uid']}");
$mysql->fetch_cell($name);

var_dump($name);
[/syntax]

[JelvinJS7]

It should be used as I said above

Nope

Try this and see what the exact value you are getting is

[syntax=php] $mysql = new mysql("mysql**.000webhost.com", "******", "*****^", "*******");
$mysql->query("SELECT `name` FROM `users` WHERE `id` = {$_SESSION['uid']}");
$mysql->fetch_cell($name);

var_dump($name);
[/syntax]

[syntax=text]
string(8) "Jonathan"
[/syntax]
I don't really understand the var_dump function, but that just adds to confusion; the result.
However, $name was never defined, so how'd that work?

[JelvinJS7]

Alright! It's working now, but I'm extremely confused!
Code:
[syntax=php]
 <?php
If (isset($_SESSION['uid'])){
     include_once("includes/mysql.class.inc.php");
     $mysql = new mysql("mysql**.000webhost.com", "a********_^^", "****%}^", "a**^+^%^_#%#^");
    $mysql->query("SELECT `name` FROM `users` WHERE `id` = {$_SESSION['uid']}");
    $mysql->fetch_cell($name); //not assigned to a value
?>
<p style="float:center">Hello, <?php echo $name; ?>!</p>
<?php
}
?>
[/syntax]
Why does this work now? I'm kinda confused. I know it's at the part I commented, but still....

[jacek]

The variable is passed by reference to the fetch_cell method, basically this means that anything you do to the variable inside of the function is also done outside of it. so that's not it's defined.