Resource id #3

[janet24]

I'm missing something in this code to make it work. Right now the output is Resource id #3

My database has a username which is janet and a scorer which is 4. I want the results of the scorer where the user name is janet to be displayed. This should be easy for somebody but not me.

<?

$conn = mysql_connect("localhost","xxx","xxx") or die(mysql_error());
mysql_select_db('score',$conn) or die(mysql_error());


$q = "select scorer from scoreuser where username='janet' ";
$res = mysql_query($q,$conn);


if ($res)
{
echo "".$res;
}
else
{echo "Thank you";
}

mysql_close($conn);
?>

 

[janet24]

I got this to work but I can't image this is the best way to write this code or isn't it?


<?

$conn = mysql_connect("localhost","xxx","xxx") or die(mysql_error());
mysql_select_db('score',$conn) or die(mysql_error());


$q = "select scorer from scoreuser where username='janet' ";
$res = mysql_query($q,$conn);
$row=mysql_fetch_array($res);
$result=$row['scorer'];

if ($result==4){
echo "It works and outputs number " .$result;
}
mysql_close($conn);
?>

 

[sleipnir214]

Best" way? As far as connecting, selecting the database and issuing a query goes, aside from adding more error checking, what other way is there?

Have you looked at the example code in the PHP online manual:

http://www.php.net/manual/en/function.mysql-fetch-array.php

http://www.php.net/manual/en/function.mysql-fetch-assoc.php

?


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[janet24]

I guess I don't understand why you would need an array if your only looking for one result. I'm just so new at it I thought maybe I was in left field somewhere.

That's where I got the results is from the PHP online manual.

Thanks...

 

[sleipnir214]

It's simply because the functions return arrays. Sometimes, yes, arrays of single elements, in the cases that the resultset only has one column.


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