A about php and mysql of problem

[Anon]

I write some code to count!There are some wrong,but I don't know how to settle!
code!

<?
function counter() {
$conn=mysql_connect("localhost","phpbook","");
$sql="select *from counter";
$result=mysql_query($sql,$conn);
$objResult=mysql_fetch_object($result);
$count=$objResult->counter;

$sql="update counter set counter=".($count+1)." where id= 1";
mysql_query($sql,$conn);
return $count+1;
}

echo counter();

?>

Warning: Supplied argument is not a valid MySQL result resource in
C:\Inetpub\wwwroot\mysqlcount.php3 on line 6
1

[Anon]

Well, the warning clearly states that $result doesn't refer to a valid result resource. There's something wrong with the query. Hmm. Try mysql_db_query, or use mysql_select_db to select the database to use.

-j

[Anon]

O.K.

This is the way I would write the code...

once you do the query
$result=mysql_query(...);

Do this

while($res=mysql_fetch_array($result))
{
$count=$res["counter"]
}
remember, that you can have several results of the query, and mysql_fetch_object() or mysql_fetch_array() retrieve only one element at a time from the array variable $result, that is why you need looping...

This is how I would write your second sql statement

$num=$count+1;
$sql="update counter set counter=$num where id=1;

Di