[SOLVED] Only 1 results shows

[Lamez]

alright similar to my last post, but different problem. My script shows who is online, their account name, and their character name. But when there are more than one account logged in, it only shows one person online, when there are more than one.

 

I have no idea what so ever why this is doing this, perhaps you guys do.

code:

<?php
define(ACC_SERVER, "localhost");
define(ACC_USER, "user");
define(ACC_PASS, "pass");
define(ACC_NAME, "name");

$connection_a = mysql_connect(ACC_SERVER,
                            ACC_USER,
                            ACC_PASS);

	 $sql = mysql_query("SELECT * FROM ".ACC_NAME.".characters WHERE `online` = '1'")or die(mysql_error());
	 $char = mysql_fetch_array($sql);
	 $id = $char['acct'];
	 $name = $char['name'];
         $sql = mysql_query("SELECT * FROM ".ACC_NAME.".accounts WHERE `acct` = '$id'")or die(mysql_error());
     $on_num = mysql_num_rows($sql);

?>
<font face="Arial, Helvetica, sans-serif">
<h1 align="center">People Online</h1>

<table width="50%" border="0" bgcolor="#666666" align="center">
  <tr>
    <td>
<?php

if($on_num == (0)){
echo "There are no players online.";
}
              while($char = mysql_fetch_array($sql)){
	  echo $char['login'];
	  echo "(".$name.")";
	  echo " ";
	 }

?></td>
  </tr>
  <tr>
    <td><strong>Total Online</strong>: <font color="#FFFFFF"><?php echo $on_num; ?></font></td>
  </tr>
</table>
</font>

[Lamez]

also the total number, shows one as well.

[Lamez]

I cleaned up the code a bit. Nothing Different.

 

<?php
define(ACC_SERVER, "localhost");
define(ACC_USER, "usr");
define(ACC_PASS, "pass");
define(ACC_NAME, "name");

mysql_connect(ACC_SERVER, ACC_USER, ACC_PASS);

$sql = mysql_query("SELECT * FROM ".ACC_NAME.".characters WHERE `online` = '1'")or die(mysql_error());
        $char = mysql_fetch_array($sql);

$id = $char['acct'];
$char_name = $char['name'];

        $sql_r = mysql_query("SELECT * FROM ".ACC_NAME.".accounts WHERE `acct` = '$id'")or die(mysql_error());
    $on_num = mysql_num_rows($sql_r);      
?>
<font face="Arial, Helvetica, sans-serif">
<h1 align="center">People Online</h1>

<table width="50%" border="0" bgcolor="#666666" align="center">
  <tr>
    <td>
<?php
  if($on_num == (0)){
   echo "There are no players online.";
  }
      while($acc = mysql_fetch_array($sql_r)){
    $acc_name = $acc['login'];
         echo $acc_name."(".$char_name.") ";
       }
?></td>
  </tr>
  <tr>
    <td><strong>Total Online</strong>: <font color="#FFFFFF"><?php echo $on_num; ?></font></td>
  </tr>
</table>
</font>

[trq]

If your only getting one result and the total number also states one then your query is only finding one matching result.

[sader]

try to clean upper <?php ?> part a bit

 

<?php
define(ACC_SERVER, "localhost");
define(ACC_USER, "usr");
define(ACC_PASS, "pass");
define(ACC_NAME, "name");

mysql_connect(ACC_SERVER, ACC_USER, ACC_PASS);

$sql = mysql_query("SELECT * FROM ".ACC_NAME.".characters WHERE `online` = '1'")or die(mysql_error());
$on_num = mysql_num_rows($sql);      
?>

[dezkit]

<?php

  if($on_num == (0)){

  echo "There are no players online.";

  }

      while($acc = mysql_fetch_array($sql_r)){

    $acc_name = $acc['login'];

        echo $acc_name."(".$char_name.") ";

      }

?>

 

the whole code is messed up

 

[Lamez]

well dezkit, what is messed up about it?

 

-Thanks Guys!

[sader]

Did you solve it?

BTW code is realy messy a bit

 

[Lamez]

ya I am working on cleaning up my code, and no I have not solved it. I think I might need to make another loop.

[Lamez]

ok so what I am trying to do is see who is online, but I have a problem. The name of the account is in one table, and to tell if they are online is in a different table

 

accounts-

acct(id) login(acc name)

 

characters

id char_name online

 

so I wanna take the the id, and find the account name. Then I wanna find out which char is online.

So I think I might have to make 2 loops

[sader]

I think it would be much easyer just add one more column into accounts table

 

accounts-

id(auto_increment) 

login(char)

logged(tinyint)

 

when user logs in u upadte table and set logged to 1

when user logs out u set it to 0

[Laogeodritt]

Just a note, I'm trying to figure everything out based on what you say works in your code and the PHP manual, since I haven't started working with MySQL in PHP.

 

By the looks of it, to iterate through ALL the rows, you'd have to loop for the first mysql_fetch_array() call. But consider that $id and $char_name would be overwritten each time — you'll have to nest the second loop into the first one, I think.

 

You're also checking for no players online but still going through with outputting the users afterwards.

 

I'd do something like this.

 

<?php
define(ACC_SERVER, "localhost");
define(ACC_USER, "usr");
define(ACC_PASS, "pass");
define(ACC_NAME, "name");

mysql_connect(ACC_SERVER, ACC_USER, ACC_PASS);

$sql = mysql_query("SELECT * FROM ".ACC_NAME.".characters WHERE `online` = '1'")or die(mysql_error());	
$on_num = mysql_num_rows($sql_r);
if($on_num == (0)){
$on_list = "There are no players online.";
}
else {
while($char = mysql_fetch_array($sql)) {
	$id = $char['acct'];
	$char_name = $char['name'];

	$sql_r = mysql_query("SELECT * FROM ".ACC_NAME.".accounts WHERE `acct` = '$id'")or die(mysql_error());
	while($acc = mysql_fetch_array($sql_r)){
		$acc_name = $acc['login'];
		$on_list .= $acc_name." (".$char_name.") ";
	}
}
}?><font face="Arial, Helvetica, sans-serif">
<h1 align="center">People Online</h1>

<table width="50%" border="0" bgcolor="#666666" align="center">
 <tr>
   <td><?php echo $on_list; ?></td>
 </tr>
 <tr>
   <td><strong>Total Online</strong>: <font color="#FFFFFF"><?php echo $on_num; ?></font></td>
 </tr>
</table>
</font>

 

Also, your HTML code is invalid for HTML 4.01. You can't use block elements (h1, table, etc.) in an inline element (in this case, font); if you want the whole page to use those fonts, use CSS on <body> tag; otherwise, use a <div>, which is a block element, and CSS.

 

@sader: I assume he wants to keep the data separate because char_name != login, and perhaps that a given user account can have multiple characters (though probably only one "active" at a time if that's the case).

[Lamez]

Gosh thank you so much. I am too lazy to mess with CSS, lol.

 

This is for my use any ways.

 

I did not setup the database, but Account name is a login, and they also have characters, they can have more than one character.

 

Thank you so much!