mysql insert problem - php

Question

ive got a really weird problem. i have no clue why its not working. i create an user and get the id and insert a row based on that id in another table. the row gets inserted with that id but the other values however for that row are not inserted!
$user_id = mysqli_insert_id($this->connection);
$query = "INSERT INTO selections
(user_id, language_id, country_id, region_id, city_id, gender_id, age_id, category_id)
VALUES ($user_id, 1, 1, 0, 0, 0, 20, 0)";
so the user_id gets inserted, but not the other values (they are all 0 in the table). i have really checked the columns and deleted all foreign keys to debug this problem. but i have no clue at all.
the columns are all INT. the weird part is sometime when i replace $user_id with a literal number it works, sometimes it doesnt. but the row is always created. and i have checked that $user_id is an integer.
i know this is a hard problem and that it can be caused of a lot of things, but i have tried to solve this tiny issue for 3 hours now. so would be great if someone just gave me something i could do to debug this problem.
UPDATE: even when i have set default values and just insert the first column (user_id) it doesnt work. every other field is 0. So weird!
| selections | CREATE TABLE `selections` (
`user_id` int(11) NOT NULL,
`language_id` int(11) NOT NULL DEFAULT '1',
`country_id` int(11) NOT NULL DEFAULT '1',
`region_id` int(11) NOT NULL DEFAULT '0',
`city_id` int(11) NOT NULL DEFAULT '0',
`gender_id` int(11) NOT NULL DEFAULT '0',
`age_id` int(11) NOT NULL DEFAULT '0',
`category_id` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 |
$query = "INSERT INTO selections
(user_id)
VALUES ('$user_id')";
the user_id shows 178 and other fields are 0:(
UPDATE:
It worked in the sql command line. but not in php. but mysqli generated no error and the row was indeed inserted but why are the other fields 0?
ANSWER: My fault. i had a jquery script that changed it back to 0 0 0 0 0 0 0. There's a lot of AJAX on my page so it was tricky to find it...sorry my bad!

Answer 1

When you run into situations like this, print the query to screen before it is executed:
$query = "INSERT INTO ...";
echo $query
Try:
$query = "INSERT INTO selections
(user_id, language_id, country_id, region_id, city_id, gender_id, age_id, category_id)
VALUES
({$user_id}, 1, 1, 0, 0, 0, 20, 0)";
You need to wrap PHP variables in {} when referencing them in SQL string statements.
Use Your DEFAULT Constraints
If you have defaults then you don't need to set the values in your INSERT statement:
INSERT INTO selections
(user_id)
VALUES
({$user_id})
Referencial Integrity
You're getting the last inserted id and using it in a subsequent insert into another table, but you don't have a foreign key defined on the user_id column to ensure that the value going into that column actually exists in the other table. If you provide the name of the table & column you are getting for your last insert id, I'll provide the ALTER TABLE statement.

Answer 2

$query = "INSERT INTO selections
(user_id, language_id, country_id, region_id, city_id, gender_id, age_id, category_id)
VALUES ('$user_id', 1, 1, 0, 0, 0, 20, 0)";
Single quotes around $user_id might do it.