I have two tables as post and gallery, and i have made a relationship gallery to post table.
My requirement is,
When user upload content it store in the post table(content field) ,
If user upload the images are video i want to store the images/video name in, gallery table and the gallery id refers to the post table. I dont know how to do it. please any one help me?
post table:
CREATE TABLE IF NOT EXISTS `post` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`user_id` bigint(20) unsigned NOT NULL,
`gallery_id` bigint(20) unsigned NOT NULL,
`content` longtext,
`photo` varchar(128) DEFAULT NULL,
`video` varchar(128) DEFAULT NULL,
`created` timestamp NULL DEFAULT NULL,
`updated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `fk_forum_post_user` (`user_id`),
KEY `fk_forum_post_gallery` (`gallery_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=10 ;
ALTER TABLE `post`
ADD CONSTRAINT `fk_post_gallery` FOREIGN KEY (`gallery_id`) REFERENCES `gallery` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
gallery table
CREATE TABLE IF NOT EXISTS `gallery` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`user_id` bigint(20) unsigned NOT NULL,
`type` int(11) NOT NULL DEFAULT '1' COMMENT '1- Photo, 2-Video, 3-Documents, 4-Unknown',
`profile_picture` varchar(50) DEFAULT NULL,
`forum_image` varchar(200) DEFAULT NULL,
`forum_video` varchar(200) DEFAULT NULL,
`forum_video_link` varchar(200) DEFAULT NULL,
`created` timestamp NULL DEFAULT NULL,
`updated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `fk_gallery_user` (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=60 ;
is any other idea is to do or how can i move forward?
There are a lot of ways you can perform this tasks. you can perform logic on code level as well as on database level. however here is a quick answer . hope this could help you
First Remove the column (gallery_id) as it is possible that user may have more than one images or video for single post or user may not want to upload any image / video in this case your gallery id would be null.
In this case your Pk would be only postId
your gallery table is fine
Make a third table name PostGalleryRealation
make 2 column in this table PostId as fk from post table and galleryid as fk from gallery table.
this is basically for one to many relation as one post may have more than one gallery
insert the post id and gallery id in PostGalleryRealation table
Finally you can write this query to fetch the result for you.
I did not test the query, so it's just a basic idea
Select p.id, p.content, p.created, g.type, g.profilepicture, g.forum_image
from post p, gallery g, postgalleryrelation pgr
where p.id = pgr.postid
and g.id = pgr.galleryid
and p.id = 1
It's just an idea. You can do much better.
YII have relation option while creating CRUD using gii. For that, you have to create tables with foreign key relationship. So YII will automatically create the relation in coding level. You have to choose the option Build Relation while creating model for both tables.
Check the Yii relation tutorials for more information
Related
I've been thinking on this problem for fews days and I still can't find a way to do what I want.
Below is how my database is currently designed (it's where I'm stuck) :
This is what I want :
a User can create multiple PriceSheets. A User can give a PriceSheet any name he wants. There are two PriceSheets types : "Lab Fulfillment", or "Self Fulfillment".
if the User chooses "Lab Fulfillment", he can import all or part of the Products of one of the predefined Labs. (I rephrase : there are few Labs that come with a predefined list of Products). The User will only be able to customize the price. He can't add custom products to this PriceSheet.
if the User chooses "Self Fulfillment", he can add his own products, and can personalize each field (name, cost, price, dimension_h, dimension_l).
I don't know how to link the tables between them. If I put the predefined Products in the Products table and set a Many-to-Many relationship between PriceSheets and Product, the default price of a predefined Product will be overwritten when a User customizes it, which is not what I want.
Also, I want the default values of my predefined Products to be only once in my database. If 100 users uses the predefined Products, I don't want the default cost to be in my database 100 times.
Don't hesitate to ask for precisions, I had trouble making this question clear and I think it's still not totaly clear.
Thanks in advance for your help
OK, database normalization 101. Lots of ways to do this, would take me a day to really optimize all this, this should help:
User
Lab
Product
id name cost dimension .....
1 a
2 b
3 c
4 d
So those three tables are fine. All your products will go in the Product table. No foreign keys in any of those tables.
PriceSheet
user_id custom_price product_id type
1 1.99 1 lab-fulfillment
0 NULL 2 self-fulfillment
1 5.99 3 lab-fulfillment
So a user can have as many price sheets as they want, and they can only adjust the price of a product. This can actually be normalized further if you so wish:
PriceSheet (composite key on id, user_id, FK user_id)
id user_id
0 0
1 1
2 1
LabPriceSheet (you could add an id, might be better, or you could use a composite key, stricter)
PriceSheet_id custom_price lab_product_id
0 1.99 0
2 5.99 1
CustomPriceSheet
PriceSheet_id custom_product_id
1 0
With foreign keys as appropriate. This now makes MySQL restrict the custom_price, rather than in PHP (although you would still have to deal with ensuring correct INSERT!).
Now, to deal with who adds the products:
CustomProduct
id user_id product_id timestamp
0 3 2 ...
LabProduct
id lab_id product_id timestamp
0 0 1 ...
1 0 3 ...
So let's double check:
This is what I want :
a User can create multiple PriceSheets. check A User can give a PriceSheet
any name he wants. check There are two PriceSheets types : "Lab
Fulfillment", or "Self Fulfillment". check
if the User chooses "Lab Fulfillment", he can import all or part of the Products of one of the predefined Labs. (I rephrase : there are few Labs that come with a predefined list of Products). The User will only be able to customize the price. He can't add custom products to this PriceSheet.
Yup, because he would create a LabPriceSheet that can only add lab_product_id. Custom price is there too, that overrides the default price in product table.
if the User chooses "Self Fulfillment", he can add his own products, and can personalize each field (name, cost, price, dimension_h, dimension_l).
Yup, he would add a product (you would need to check if a similar one exists, else return the id of the existing product in the product table), and then that would also be an entry in CustomProduct.
I don't know how to link the tables between them. If I put the predefined Products in the Products table and set a Many-to-Many relationship between PriceSheets and Product, the default price of a predefined Product will be overwritten when a User customizes it, which is not what I want.
Yeah that won't happen :) Never (very very rarely) implement many-many rels.
Also, I want the default values of my predefined Products to be only
once in my database. If 100 users uses the predefined Products, I
don't want the default cost to be in my database 100 times.
Of course.
Let me know if you want the MySQL code, I assume you're good! Remember to use InnoDB and properly configure your MySQL configuration!
EDIT
I felt like helping you out with a copy and paste thing. I like copy and paste things. Also, there's a redundant user_id column in the blurb above which I fixed in an earlier edit.
SET GLOBAL innodb_file_per_table = 1;
SET GLOBAL general_log = 'OFF';
SET FOREIGN_KEY_CHECKS=1;
SET GLOBAL character_set_server = utf8mb4;
SET NAMES utf8mb4;
CREATE DATABASE SO; USE SO;
ALTER DATABASE SO CHARACTER SET = utf8mb4 COLLATE = utf8mb4_unicode_ci;
CREATE TABLE `User` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`email` VARCHAR(555) NOT NULL,
`password` VARBINARY(200) NOT NULL,
`username` VARCHAR(100) NOT NULL,
`role` INT(2) NOT NULL,
`active` TINYINT(1) NOT NULL,
`created` DATETIME NOT NULL DEFAULT CURRENT_TIMESTAMP,
`modified` DATETIME ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
CREATE TABLE `Lab` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`name` VARCHAR(1000) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
CREATE TABLE `Product` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`name` VARCHAR(1000) NOT NULL,
`password` VARBINARY(200) NOT NULL,
`cost` DECIMAL(10, 2) NOT NULL,
`price` DECIMAL(10, 2) NOT NULL,
`height` DECIMAL(15, 5) NOT NULL,
`length` DECIMAL(15, 5) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
CREATE TABLE `CustomProduct` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`user` BIGINT(20) UNSIGNED NOT NULL,
`product` BIGINT(20) UNSIGNED NOT NULL,
`created` DATETIME NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
FOREIGN KEY (`user`) REFERENCES `User`(`id`),
FOREIGN KEY (`product`) REFERENCES `Product`(`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
CREATE TABLE `LabProduct` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`lab` BIGINT(20) UNSIGNED NOT NULL,
`product` BIGINT(20) UNSIGNED NOT NULL,
`created` DATETIME NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
FOREIGN KEY (`lab`) REFERENCES `Lab`(`id`),
FOREIGN KEY (`product`) REFERENCES `Product`(`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
CREATE TABLE `PriceSheet` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`name` VARCHAR(1000) NOT NULL,
`user` BIGINT(20) UNSIGNED NOT NULL,
PRIMARY KEY (`id`,`user`),
FOREIGN KEY (`user`) REFERENCES `User`(`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
CREATE TABLE `LabPriceSheet` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`price_sheet` BIGINT(20) UNSIGNED NOT NULL,
`lab_product` BIGINT(20) UNSIGNED NOT NULL,
`custom_price` DECIMAL(10, 2) NOT NULL,
PRIMARY KEY (`id`),
FOREIGN KEY (`price_sheet`) REFERENCES `PriceSheet`(`id`),
FOREIGN KEY (`lab_product`) REFERENCES `LabProduct`(`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
CREATE TABLE `CustomPriceSheet` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`price_sheet` BIGINT(20) UNSIGNED NOT NULL,
`custom_product` BIGINT(20) UNSIGNED NOT NULL,
PRIMARY KEY (`id`),
FOREIGN KEY (`price_sheet`) REFERENCES `PriceSheet`(`id`),
FOREIGN KEY (`custom_product`) REFERENCES `CustomProduct`(`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
Using MySQL workbench, I copied the create statement from a few related tables to put into a clean schema. They all reference each other in some way so there is no inherent order I can create them in. How can I just force MySQL to create the tables while ignoring any warning that may occur, just until the rest of the tables are created?
Would I have to group it inside a transaction of some sort?
A very simple example would be:
CREATE TABLE `vehicle` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`gallery_id` int(11) DEFAULT NULL,
`make_id` int(11) NOT NULL,
`model_id` int(11) DEFAULT NULL,
`make` varchar(100) DEFAULT '',
`model` varchar(100) DEFAULT '',
`colour_id` int(11) DEFAULT NULL,
`currency_id` int(11) NOT NULL DEFAULT '1',
`fuel_id` int(11) DEFAULT NULL,
`status_id` int(11) DEFAULT NULL,
`stock_code` varchar(100) DEFAULT NULL,
`registration` varchar(20) DEFAULT NULL,
`title` varchar(100) DEFAULT NULL,
`description` text,
`month` tinyint(4) DEFAULT NULL,
`public` tinyint(4) NOT NULL DEFAULT '0',
`sold` tinyint(4) NOT NULL DEFAULT '0',
PRIMARY KEY (`id`),
KEY `vehicle_fk_idx` (`status_id`),
CONSTRAINT `vehicle_fk` FOREIGN KEY (`status_id`) REFERENCES `vehicle_status` (`id`) ON DELETE SET NULL ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
CREATE TABLE `vehicle_status` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`slug` varchar(100) DEFAULT NULL,
`title` varchar(100) DEFAULT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `slug_UNIQUE` (`slug`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=latin1;
vehicle references vehicle_status which would mean that vehicle_status would have to be created first. How would I create vehicle first and then vehicle_status without adding the reference afterwards?
You do not define foreign keys when creating tables, you would have them in a separate query like this:
ALTER TABLE `vehicle`
ADD CONSTRAINT `vehicle_ibfk_1` FOREIGN KEY (`status_id`)
REFERENCES `vehicle_status` (`id`) ON DELETE SET NULL ON UPDATE CASCADE;
So you would first create tables and then create foreign keys.
The way I personally do is I have all CREATE TABLE queries in one file that I know I can simply just import without any errors. I have all CONSTRAINT queries in a separate file that I import after all the tables have been created and I have all INSERT INTO queries in a separate file that adds data after all constraints and tables have been set.
It looks like vehicle_status is a lookup table, while vehicle is a primary transaction table.
In general, lookup tables do not reference other tables, although there can be designs where one lookup table references another lookup table. In your case, it's simple: just create vehicle_status first. If your schema has a hundred tables in it, it's going to involve a little work to order the create commands in the right sequence.
There are designs where the reference chain forms a circle. In such a case, you'll have to do what GGio suggests: add the constraints later. There are other problems with designs involving circular references, and those problems may or may not be present in your schema.
When you go to populate the tables, you'll have to worry about order as well. In general, you'll have to populate the lookup tables first, before you begin to populate the transaction tables. Otherwise you'll get reference violations at load time.
I have a project where the admin needs to create multiple newsletters with some crawled posts from the web.
I insert the posts in posts table after crawling has completed and assign them a feed_id to identify the source. this is the structure of posts table (truncated):
CREATE TABLE `posts` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`feed_id` int(11) NOT NULL,
`created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`updated_at` timestamp NULL DEFAULT NULL,
`identifier` varchar(255) DEFAULT NULL,
`published` timestamp NULL DEFAULT NULL,
`content` longtext,
...
...
`is_unread` int(1) NOT NULL DEFAULT '1',
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
Every admin (user) has access to one or more "feeds". So in Newsletter creation page I want to show them a list of posts from the feeds they are allowed to see and also, I show a button to put the posts in specifict categories of that newsletter, if the user previously selected that post, I should show him that and let him remove it from the category. So I have some other tables too: newsletters, categories, newsletter_post, category_post. Here is their structures:
newsletters:
CREATE TABLE `newsletters` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`updated_at` timestamp NULL DEFAULT NULL,
`sent_at` timestamp NULL DEFAULT NULL,
`title` varchar(255) DEFAULT NULL,
`date` date DEFAULT NULL,
`topic_id` int(11) NOT NULL,
`user_id` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
categories:
CREATE TABLE `categories` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`topic_id` int(11) NOT NULL,
`title` varchar(255) DEFAULT NULL,
`slug` varchar(255) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
newsletter_post:
CREATE TABLE `newsletter_post` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`updated_at` timestamp NULL DEFAULT NULL,
`newsletter_id` int(11) NOT NULL,
`post_id` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
category_post:
CREATE TABLE `category_post` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT,
`created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`updated_at` timestamp NULL DEFAULT NULL,
`category_id` int(11) NOT NULL,
`post_id` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
So I'm using this query to find posts for the allowed feeds and check the status if a post is in a specific category of this specific newsletter:
SELECT DISTINCT `posts`.`id`, `published`, `posts`.`title`, `posts`.`content`, `source_name`, `category_id`, `newsletter_id`, `link_href`, categories.title as category_title
FROM `posts`
LEFT JOIN `category_post` ON `posts`.`id` = `category_post`.`post_id`
LEFT JOIN `categories` ON `categories`.`id` = `category_post`.`category_id`
LEFT JOIN `newsletter_post` ON `posts`.`id` = `newsletter_post`.`post_id`
LEFT JOIN `newsletters` ON `newsletters`.`id` = `newsletter_post`.`newsletter_id`
WHERE `feed_id` IN (6, 7) ORDER BY `posts`.`published` DESC LIMIT 40 OFFSET 0
but the problem is this is horrible and not optimized. My posts table contains up to 50,000 rows each month, and each row with 3~10kbs of data in avg., so sometimes when I try to run the query (which is frequently run by the admin to make the newsletter, pagination etc) mysql shows this error: too much rows to join, etc. and most of the times its really slow.
and the reason I'm doing all this in one query is because I want the result to be in one json response so I can show them the user quickly without doing additional requests.
I wanna know if there is a better way to do this query or use indexes or something else.
Thanks you in advance for your help.
index your posts table on
( feed_id, published )
so the data is already optimized for your WHERE clause, and pre-sorted to help your ORDER BY.
For reading querys that have a lot of demand, InnoDB is very inefficient. I recommend you to use a NoSQL Database but if you don't want or the cost of change is too much... you can try this:
1) LIKE Sallar Kaboli told you, you have to index your tables in columns that use in JOIN querys. For example:
CREATE INDEX index1 ON newsletter_post (post_id);
2) USE only important columns for JOINS.
I mean, you have to only use the columns that use in SELECT part of query.
I hope this'd be helpful.
To complete other answers, I suggest to change this types on posts table:
1) Change feed_id to int(4). Really you have more than int(4) feeds?
2) Change is_unread to bit instead of int(1). I should say that this may not improve your given query in the question but according to the field name, the correct type is bit.
Another more improvement to this answer is that never use default int(11) for numeric or id fields, assign types more specific. Using smaller size of types will improve your indexes also. I don't think you need more than int(4) for fields id.
For example indexing and querying int(3) column is more faster than int(11).
Please create the following indexes indexes on ::
1) `post_id` in `category_post`
2) `post_id` in `newsletter_post`
Multiple Tables...
CREATE TABLE IF NOT EXISTS `pictures` (
`id` bigint(20) NOT NULL auto_increment,
`title` varchar(255) NOT NULL,
`url` varchar(255) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `propPictures` (
`id` bigint(20) NOT NULL auto_increment,
`picture_id` varchar(255) NOT NULL,
`property_id` varchar(255) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
OR
CREATE TABLE IF NOT EXISTS `pictures` (
`id` bigint(20) NOT NULL auto_increment,
`title` varchar(255) NOT NULL,
`url` varchar(255) NOT NULL,
`property_id` varchar(255) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
????
thank you
It depends. Will every picture have exactly one property_id? If so then a single table is fine.
A Property can and probably will have multiple pictures, so what you want is a property table and then a picture table that includes property id and has a foreign key to the porperty table.
Incidentally ID is a terrible choice for the id field. It creates problems with reporting and can cause accidental join problems. Two tables should almost never use the same name for fields which mean something different. Use tablenameID instead.
It looks like you're suggesting table1 for pictures, and table2 for any properties it may have. This would usually be better than a single table design as it makes it easy to handle any given picture having, 0, 1, or many more properties.
With a one-to-one relationship between pictures.id and property_id, a single table would suffice. If this relationship changes to one to many, a second table will become necessary to relate the one pictures.id to the many property ids.
I really need some help with forming a MySQL query that I just cannot work out. On my website I have a system in place that will hopefully remember some selections that user made when they last visited the site.
On the site the user can select which category they wish to read the content of next time they come to site. That setting will be remembered but the menu should be displayed slightly different. It should show all the other categories minus the one that has been saved.
So if I have these categories,
Blog
Inspiration
Case Studies
and the user saved Blog, the next time they come to the site the categories list should just be
Inspiration
Case Studies.
How can this data be pulled from the database?
Currently I have a table that identifies the user via a unique cookie id:
CREATE TABLE IF NOT EXISTS `cookieTable` (
`cookieEntryId` int(11) NOT NULL AUTO_INCREMENT,
`cookieId` varchar(32) NOT NULL,
`expirationDate` int(10) NOT NULL,
PRIMARY KEY (`cookieEntryId`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ;
I have a category table
CREATE TABLE IF NOT EXISTS `categoryTable` (
`categoryId` int(11) NOT NULL AUTO_INCREMENT,
`categoryTitle` varchar(25) NOT NULL,
`categoryAbstract` varchar(150) NOT NULL,
`categorySlug` varchar(25) NOT NULL,
`categoryIsSpecial` int(1) DEFAULT NULL,
`categoryOnline` int(1) DEFAULT NULL,
`dashboardUserId` int(11) NOT NULL,
`categoryDateCreated` int(10) NOT NULL,
PRIMARY KEY (`categoryId`),
KEY `dashboardUserId` (`dashboardUserId`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ;
And I have the table that saves what categories the user has saved,
CREATE TABLE IF NOT EXISTS `userMenuTable` (
`menuEntryId` int(11) NOT NULL AUTO_INCREMENT,
`categoryId` int(11) NOT NULL,
`cookieId` varchar(32) NOT NULL,
PRIMARY KEY (`menuEntryId`),
KEY `categoryId` (`categoryId`,`cookieId`),
KEY `cookieId` (`cookieId`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=6;
The following query should get the categories the user hasn't saved assuming the cookieId stays constant for a user. If it doesn't you should put a userId into the userMenuTable instead. Just replace USERSCOOKIEID with their actual cookie ID.
SELECT * FROM categoryTable WHERE categoryId not in
(SELECT categoryId FROM userMenuTable WHERE cookieId = 'USERSCOOKIEID') as x