de innerjoin werkt niet echt en ik snap niet waarom:
PHP
SELECT car_users.car_id car_users.count,
cars_overview.car_name cars_overview.car_price
cars_overview.cars_max_speed cars_overview.car_weight car_0_100
FROM car_users
INNER JOIN cars_overview USING (car_id)error message:
CitaatErrormessage: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '.count, cars_overview.car_name ' at line 1
EDIT:
kan dit werken:
PHP
SELECT car_users.*,
cars_overview.*
FROM car_users
INNER JOIN cars_overview USING (car_id)EDIT 2:
Bij de 2 sql lijn krijg ik de error:
Citaat
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/patrick/domains/noxxie.nl/public_html/beta/ingame/garage.php on line 44
PHP
while ($jFetch = mysql_fetch_assoc($jQuery) )
{
echo $jFetch['count'] . ' ' . $jFetch['car_name'];
}
Greetz,
Patick
