Jordan Goh Jordan Goh - 3 months ago 6
MySQL Question

Echoing an SQL Field in PHP

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'];

Answer
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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