Display mysql values with php help.

[user51]

I am trying to display a single value on my page and i cant seem to display it for the life of me. can someone tell me if im doing something wrong?

this is what i have so far:

 

<?php
require_once('inc/common.inc.php');

// Connect to the database
db_connect();

// Check and update the authentication.
check_auth();

db("select * from ${db_name}_users where login_id = '$login_id'");
echo "$login_id";

?>

 

obviously there is a field in the database that corresponds to turns and it is int(4),

 

..... Now, i am able to display $login_id, but when i try to display another field, named turns, it just displays a blank page.

Any pointers would be appreciated

thank you.

 

[Pikachu2000]

Echoing $login_id has nothing to do with the database query result. Its value is obviously set before the query attempts to run. Since you are able to echo a value from $login_id, either that code isn't the code you're actually using, or you've not included all the relevant parts.

[user51]

that is all the code i have on the page, my goal is to just display that single field value, and use the same code to display other values on the top of the page in separate cells of a table.

 

if you dont mind, how can i call to display that?

again it does echo the user id but no the other cells.

[user51]

<?php
require_once('inc/config.inc.php');



$sql = "SELECT `turns` FROM `exodus_users` WHERE 1 LIMIT 0, 30 ";
echo  "turns";

?>

 

the above $sql i pulled from phpmyadmin,

config.inc.php has the login info.

i still get just the word turns echoed, help

[jcbones]

<?php
require_once('inc/config.inc.php');

$sql = "SELECT `turns` FROM `exodus_users` WHERE 1 LIMIT 0, 30 ";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0) {
while($r = mysql_fetch_assoc($result)) {
echo $r['turns'] . '<br />';
}
}
else {
echo 'No rows returned from the query.';
}

?>