Resource id #5???

[loureiro]

im trying to query the database to get the customerID according to the email that was submitted but i keep getting this returned! whats going on?

Resource id #5

this is the code:

$dbh=mysql_connect ("localhost", "***", "***");

			mysql_select_db ("ziggy_Lifestyles");

			$sql = 'SELECT `customerID` FROM `Customer` WHERE 1 AND `email` = \'[e-mail]\' LIMIT 0 , 30';
			$result = mysql_query($sql, $dbh);

			echo "<strong>" . $result . "</strong>";

			mysql_close($dbh);

Database: ziggy_Lifestyles

Table: Customer

Fields: customerID
email

Return: customerID

please help!

[stolzyboy]

$dbh=mysql_connect ("localhost", "******", "******"); 

mysql_select_db ("ziggy_Lifestyles"); 

$sql = 'SELECT `customerID` FROM `Customer` WHERE 1 AND `email` = \'[e-mail]\' LIMIT 0 , 30'; 
$result = mysql_query($sql, $dbh); 
$row = mysql_fetch_object($result);

echo "<strong>" . $row->customerID. "</strong>"; //or whatever fieldnames you want to echo

mysql_close($dbh);

[kevinsequeira]

do this

$result = mysql_query($sql, $dbh);
$row = mysql_fetch_array($result);
echo $row[0];

reg
kevin

[loureiro]

i actually wanted to change the email address to be dynamic so this is what i have:

mysql_select_db ("ziggy_Lifestyles");

			$sql = 'SELECT `customerID` FROM `Customer` WHERE 1 AND `email` = ' . $_POST["Email"] . ' LIMIT 0 , 30';  
			$result = mysql_query($sql, $dbh);  
			$row = mysql_fetch_object($result); 

			echo "<strong>" . $row->customerID . "</strong>"; //or whatever fieldnames you want to echo 

and this is what it returns:

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/ziggy/public_html/lifestyles/final.php

[loureiro]

nevermind! i figured it out! thanks a lot guys!

[kevinsequeira]

.
.
.
.
$result = mysql_query($sql, $dbh);
if(mysql_error()){
echo "error in sql ".$sql."<BR>";
echo mysql_error();
}else{
if(mysql_num_rows($result)<1){
echo "no matching record";
}
}
.
.
.
.
.
.

reg
kevin