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Editorial Team
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Asked: 2026-05-26T19:37:16+00:00

I need some help with echoing an SQL field but it keeps showing Resource

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I need some help with echoing an SQL field but it keeps showing Resource #4 or does not show anything at all.

My code is

mysql_connect("$host", "$username", "$password") or die("cannot connect"); 
mysql_select_db("$db_name") or die("cannot select DB");
$myusername=$_POST['myusername']; 
session_register("myusername");
$result = mysql_query("SELECT statsight from playerinfo WHERE username = 'myusername'") or die(mysql_error());
$row = mysql_fetch_array($result);
if(!session_is_registered('myusername')){
    header("location:mainlogin.php");
}

echo $row['statsight'];
?>

<html>
    <body>
        Login Successful
        <form name="form2" method="post" action="stat.php">
           Stat1: <?php echo "<td>".$row['statsight']."<td>"?>
           <input type="submit" value="+">
       </form>
    </body>
</html>

And it does not show anything. It only shows Stat1: with the + button.

The value of statsight in the database is 3 if it matters, and the myusername information comes from a form.

Got it working using

session_start();
$myusername=$_SESSION['myusername'];
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1 Answer

  1. Editorial Team
    username = 'myusername'

    you are not suing variable here.

    I am no PHP expert, but this might be the proper way

    mysql_query("SELECT statsight from playerinfo WHERE username = ' ". $myusername."'")            or die(mysql_error());

    Although this is not a proper way, you should be using mechanism of Prepared statements while executing queries.

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