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[SOLVED] My first SQL trial... but where am i going wrong?


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My first SQL trial... but where am i going wrong?

[Code]

<?php // connect to database
$host = "localhost";
$dbname = "###"; $dbuser = "###";
$dbpass = "###";
$connection = mysql_connect($host, $dbuser, $dbpass) or die(mysql_error()); $db = mysql_select_db($dbname) or die(mysql_error());
// Bah Humbug! i will have this work!!!
$max = mysql_query("SELECT MAX(uid) FROM user") or die ("nope. Didn't work...!");
echo $max;
?>
[/Code]


Ok so this is my first SQL trial... but where am i going wrong? i get the message "Resource id #3" where i was expecting to see "14"
the returned result of of a query is the Resource id number, what you need to do is to is create another variable to store the answers to your query such as:-
[code]
<?php // connect to database
$host = "localhost";
$dbname = "###"; $dbuser = "###";
$dbpass = "###";
$connection = mysql_connect($host, $dbuser, $dbpass) or die(mysql_error()); $db = mysql_select_db($dbname) or die(mysql_error());
// Bah Humbug! i will have this work!!!
$query = "SELECT MAX(uid) FROM user"; // i find it easier to break it down to its parts
$result = mysql_query($query) or die ('Error in query: $query. ' . mysql_error()); //shows a more informative error then "nope didnt work"
$row = mysql_fetch_assoc($max); // all informations store in an associative array
$max = $row['MAX(uid)']; //pull what you need from the array
echo $max; //print it out
?>
[/code]
Paul
when it is broken down into the four lines of code it sort of runs in order

query_variable = query
returned_variable =  mysql_query(query_variable)
array_variable = mysql_fetch_assoc(returned_variable)

all other variables set from the array/....
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