MySQL always return expects parameter 1 to be mysqli_result, boolean given

Question

I have 3 tables. In table_1 and table_2 I have field called plevel. Third table is to join them and is. cars_user and have 2 columns - table1_plevel and table2_plevel The query that I want to select is

$q = mysqli_query($con, "SELECT * FROM `cars` AS c
      LEFT JOIN `cars_user` AS c2u ON c.cars_plevel = c2u.car_plevel
      LEFT JOIN `users` AS u ON c2u.user_plevel = u.users_plevel");

When I run it I get that the column is Unknown but I know that this column is there.

 #1054 - Unknown column 'c.cars_plevel' in 'on clause'

I have put references also in cars_user to cars.plevel and users.plevel

ALTER TABLE `cars_user` ADD FOREIGN KEY ( `cars_plevel` ) 
     REFERENCES `app`.`cars` (`plevel`) ON DELETE RESTRICT ON UPDATE RESTRICT;
ALTER TABLE `cars_user` ADD FOREIGN KEY ( `user_plevel` ) 
     REFERENCES `app`.`users` (`plevel`) ON DELETE RESTRICT ON UPDATE RESTRICT;

What is the problem here?

This is what exactly the error says: you have no column like `c.cars_plevel`. So, the query fails, and the result will false, not a resource. So you can not fetch that, or use any other operation. — vaso123, Dec 04 '14 at 13:21
@Machavity, it is not duplicate. I saw a lot of thread with same name but none is helped. — Jason Paddle, Dec 04 '14 at 13:24
@AdrianNasui `#1054 - Unknown column 'r.car_plevel' in 'on clause'` — Jason Paddle, Dec 04 '14 at 13:24
Please, show all the columns of all the tables in your query. — IROEGBU, Dec 04 '14 at 13:32
If you have a column called 'plevel' in the table 'cars' then it isnt referenced 'cars_plevel' its referenced by 'cars.plevel' or with your aliasing 'c.plevel' — Mike Miller, Dec 04 '14 at 13:38

score 0 · Accepted Answer · answered Dec 04 '14 at 13:35

0

I like error said you don't have c.cars_plevel you have c.plevel and you don't have u.users_plevel you have u.plevel or in other words:

$q = mysqli_query($con, "SELECT * FROM `cars` AS c
  LEFT JOIN `cars_user` AS c2u ON c.plevel = c2u.car_plevel
  LEFT JOIN `users` AS u ON c2u.user_plevel = u.plevel");

answered Dec 04 '14 at 13:35

Goro

519
1
11
30

1

Yes, now is working. Thank you! – Jason Paddle Dec 04 '14 at 14:03

MySQL always return expects parameter 1 to be mysqli_result, boolean given

1 Answers1