Development More MySQL Problems!

right, my program is whining about this code...

PHP:

        include 'db-info.php';
        $select=="1";
    
        $query = "SELECT * from UKAB_Candidate WHERE no='$select'";

        while($a_row = mysql_fetch_object($query)) {
            $FORENAME == $a_row->forename;
            $SURNAME == $a_row->surname;
            $DOB == $a_row->dob;
            $CAND_NO == $a_row->cand_no;
            $CENT_NO == $a_row->cent_no;
        }


the error reports this line...

PHP:

while($a_row = mysql_fetch_object($query))


to be causing the problems! but i cannot work out why! any help is appreciated, thnks!

jafefr

 

Not to go after the obvious, but there's a mysql_connect() and a mysql_select_db() in there somewhere before you blast away with mysql_fetch_object(), right?

What's the error message you get? It's usually kinda helpful.

 

yes there is

sorry, forgot to give you there error it is...

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/usr/abunting/public_html/CWK/candidate.php on line 27

 

that means mysql_connect() failed usually, try echoing mysql_error() right after you connect, and if you have one of those @ signs in front of mysql_connect() remove it, and more info should appear magically.

This is an excellent example of a common mistake in PHP programming, assuming that a function call succeeded. In fact, it's number 12.

 

PHP:

include 'db-info.php';
        $select=="1";
    
        $query = "SELECT * from UKAB_Candidate WHERE no='$select'";

        while($a_row = mysql_fetch_object($query)) {


try

PHP:

        $query = "SELECT * from UKAB_Candidate WHERE no='$select'";
        $result = mysql_query(mysql_query);
        while($a_row = mysql_fetch_object($result)) {


1) you need to get the query before getting the objects with mysql_query($query).
2) did you mean to have
$DOB == $a_row->dob; instead of
$DOB = $a_row->dob; single '=' insead of 2