RE: [PHP-DB] A newbie in deep need for help

Yes, first you use: $result = mysql_result($query2); That will return something like resourceID #3. That's right. What you need to do now is to use a function to access the resource. For example, if you have a table named User with the following columns: ID username passwd and you want to perform a query that just returns the password, your code would look like something like this: $query2 = "select passwd from user where username = '$username'";//query $result = mysql_query($query2);//perform the query and return result set $thisPass = mysql_result($result);//return one column (passwd) from result set echo"<p>".$thisPass."</p>"; //this line would just echo out the variable $thisPass to the browser to show you what it contains. Try that code and see if it works for you. If not, respond and paste all of the code you're working with and I'll see if I can find what's going wrong. Rich -----Original Message----- From: Ehab El Maraghy [mailto:e_elmaraghy@yahoo.com] Sent: Friday, December 27, 2002 12:24 PM To: php-db@lists.php.net Subject: Re: [PHP-DB] A newbie in deep need for help Rich thanx for answering I did what you said yesterday before using mysql_resut it returns resource id#3 didn't know what is it and how can i transform it it the data inside the cell i corrected the situation and i waited to give me res.id#3 but it doesn't write anything any suggestions ? from my part i will revise the code once again waiting for your reply thanx -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php