Help kickstart my code!

whtitus

Just got a new site and already I've hit a brick wall.

I used cPanelX to create a new MySQL database called "mydb." (It is thereafter called "artpros_mydb" in cPanel and in phpMyAdmin. ) An employees table was added. It contains an id and four fields.

I'm using the following script within an html page to generate the results.

<?php
$db = mysql_connect("localhost", "artpros_mydb");
mysql_select_db("mydb",$db);
$result = mysql_query("SELECT * FROM employees",$db);
printf("First Name: %s<br>\n", mysql_result($result,0,"first"));
printf("Last Name: %s<br>\n", mysql_result($result,0,"last"));
printf("Address: %s<br>\n", mysql_result($result,0,"address"));
printf("Position: %s<br>\n", mysql_result($result,0,"position"));
?>

But, it fails.

The error message I receive for each printf is:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in home/artpros/public_html/test.php on line [nn] [fieldname]

I'm sure this problem is trivial, but I'm stumped. Hey, I'm a Newbie!

Bill

drew010

if your database in cpanel was named artpros_mydb then your problem is in the line:
mysql_select_db("mydb",$db);
change it to
mysql_select_db("artpros_mydb",$db);

anytime there is a bad link to mysql you will see
supplied argument is not a valid MySQL result resource in ...
just means you are trying to issue a command to the sql server when you are not connected or are trying to retrieve data from something that isnt a mysql link. usually when you see that you will know you issued the query wrong or for some reason did not successfully connect.

and just for extra debugging, change your opening code to
$db = mysql_connect("localhost", "artpros_mydb") or die("Could not connect to sql because: " . mysql_error());
mysql_select_db("mydb",$db) or die("Could not select database because " . mysql_error());

$result = mysql_query("SELECT * FROM employees",$db);
if(!$result) {
echo "didnt get query because " . mysql_error();
}