I'm a bit of a newbie and I'm having trouble getting my cakephp query correct.
I have a users table and users hasmany rotas. The rotas table looks like
monday_am
monday_pm
tuesday_am
etc
and if the user is due to be in on monday_pm for example, then that field will have 1. Otherwise 0.
I want to get all the users who belong to a particular room, have deregistered=0 and - and this is where I'm having the problem - who have a rota where (for example) tuesday_am = 1
So I have the code:
$options['conditions'] = array('User.room_id'=>$room_id, 'User.deregistered'=>0, 'User.request'=>0);
$options['joins'] = array(
array('table' => 'rotas',
'alias' => 'rota',
'type' => 'INNER',
'conditions' => array('rota.'.$todaysDay.'_'.$amPm => 1)
)
);
$usersToday = $this->User->find('all', $options);
If I don't try the join then I get a nice small list of users like I expect. If I try the join I get 100s of results which are mostly duplicates!
How can I get just one record of a user who has the required conditions in the Users table as well as the Rotas table, and how come doing the join as I have causes so many results?!?
thanks
Assuming you have a column in your rota table called user_id, you can make your Rota model like this:
class Rota extends Model
{
public $belongsTo = array('User');
}
Then in your controller, you can query the Rotas and get the user from it.
$users = $this->Rota->find('all', array(
'fields' => array('DISTINCT (User.id)' /* add more fields for selection */ ),
'conditions' => array(
'User.room_id'=>$room_id,
'User.deregistered'=>0,
'User.request'=>0,
'Rota.'.$todaysDay.'_'.$amPm => 1
)
));
Related
I am new to CakePHP (2.x) and am creating a post and comments feature. Everything works except I cannot figure out how to get the user's username out of the registrations (third) table (linked with "registration_id"). My associations currently look like:
class Article extends AppModel {
public $hasMany = array('ArticleComment');
public $belongsTo = array(
'ArticleRegistration' => array(
'className' => 'Registration',
'foreignKey' => 'Article.registration_id' //(doesn't work)
),
'ArticleCommentRegistration' => array(
'className' => 'Registration',
'foreignKey' => 'ArticleComment.registration_id' //(doesn't work)
)
);
class ArticleComment extends AppModel {
public $belongsTo = array('Registration','Article');
I am not sure if the associations from ArticleComment are being applied since it is being called through the Article model. I am retrieving the data by:
$this->set('articles', $this->Article->find('all', array('order' => 'Article.created desc', 'limit' => '3')));
I have tried a join and passing two separate variables for the articles and comments array but then I have to remove my associations which leads me to believe it's not proper coding.
Tables are:
articles
__________
id
registration_id
body
article_comments
__________
id
article_id
registration_id
body
registration
__________
id
username
I am fetching the information with:
$this->set('articles', $this->Article->find('all', array('order' => 'Article.created desc')));
TIA!
The generated SQL query from cakephp when you make a find() call is as follows
SELECT `Article`.`id`,
`Article`.`registration_id`,
`Article`.`body`,
`Registration`.`id`,
`Registration`.`username`
FROM `test`.`articles` AS `Article`
LEFT JOIN `test`.`registration` AS `Registration`
ON (`Article`.`registration_id` = `Registration`.`id`)
WHERE 1 = 1 LIMIT 20
So, in order to access the username you just need to write something like this in your view
$articles['Registration']['username'];
I recommend you to use the cakephp bake commands, I could make this proyect like in 10 minutes with them. And also I recommend using Netbeans as IDE, the cakephp plugin is awesome.
Here is a link to an example project that you can clone with git.
I have 2 tables joined by a HABTM relationship. portfolios and assets. The join table is portfolios_assets.
I wish to have an extra field in the join table, rebalance_date such that I can query the assets in a portfolio for a given date. How would I construct a find such that I can determine the most recent date and only return the assets for that date.
So, in my Portfolio model I might have:
$params = array(
'conditions' => array(
'Portfolio.id' => 5,
'?.rebalance_date' => '2013-11-01' //no model name for this field as it's in the join table
),
'order' => array(...)
);
$result = $this->find('all', $params);
In the above example, I just keyed in a date. I'm not sure how I would retrieve the latest date without writing a raw query in Cake. (I could do SELECT rebalance_date FROM portfolios_assets ORDER BY rebalance_date DESC LIMIT 1; but this is not following Cake's convention)
You need to use the hasMany through model: http://book.cakephp.org/2.0/en/models/associations-linking-models-together.html#hasmany-through-the-join-model
You would need to create another model eg PortfolioAssests and the table would need to be portfolio_assets not portfolios_assets.
Then you should be able to use:
$assets = $this->Assets->find('all', array(
'conditions' => array(
'Portfolio.id' => 5,
'PortfolioAsset.rebalance_date' => '2013-11-01'
)
));
System
CakePHP 2.x
Database
Houses --> id, name
Visits --> id, date, house_id
Models
Visit belongsTo House
House hasMany Visit
View
I have a view for all houses called houses/index.ctp.
It lists each house with their id and their name.
I also use the PaginatorHelper to sort my array.
Issue
I'd like to see for each house last visit they had.
This have to be sortable by the PaginatorHelper
Tries
I thought that I could do it with virtualFields but I failed (it always returns the max value of all houses).
public $virtualFields = array('lastvisit' => 'SELECT MAX(visits.date) FROM visits, houses WHERE houses.id = visits.house_id GROUP BY house_id');
I tried to "cache" each max visit in a new column of Houses table but it's very dirty.
Thank you for helping.
Declaring a virtual field for this may be tricky, so I suggest you to add it on the fly whenever you need it.
So your HousesController's index action will look like this:
public function index() {
$this->House->virtualFields['last_visit'] = 'MAX(Visit.date)';
$this->paginate = array(
'House' => array(
'group' => array('house_id'),
'joins' => array(
array(
'table' => 'visits',
'alias' => 'Visit',
'type' => 'INNER',
'conditions' => array(
'House.id = Visit.house_id',
)
)
)
)
);
$this->set('houses', $this->paginate());
}
Notice you have to remove the $publicFields declaration from your model, and that I changed the name of the field to from lastvisit to last_visit to make it more Cake.
Hope it helps.
This is pretty simple if you use the Containable behavior.
$this->paginate['House'] = array(
'contain' => array('Visit' => array('order' => 'date DESC')
);
$houses = $this->paginate('House');
I haven't tested this yet but it should be pretty close to what you need. You may want to add a limit to contain so you only see the last x visits.
When I was working on my current project, I ran into a rather complex issue. I'll point out my problem much more detailed right now:
There are three Models: User, UsersExtendedField and UsersExtended.
UsersExtendedField contains custom fields that can be managed manually. All custom fields can be shown in the Users view as well, and be filled out of course. The values are then stored in UsersExtended, which has got two foreignKeys: user_id and field_id.
The relations look like this: User hasMany UsersExtendedField, UsersExtendedField hasMany UsersExtended, UsersExtended belongsTo User, UsersExtendedField.
The problem: When accessing the Users view, a form with user information input is shown. Any UsersExtendedFields are available as well, and since these hasMany UsersExtended, they've got plenty of UsersExtended values. But I want to reduce those to only the value(s) that belong to the User, whose view is shown at the moment. Here are my (desired) relations:
Croogo::hookBehavior('User', 'Crooboard.ExtendedUser', array(
'relationship' => array(
'hasMany' => array(
'UsersExtendedField' => array(
'className' => 'Crooboard.UsersExtendedField',
'foreignKey' => '',
'conditions' => array('status' => 1)
),
),
),
));
class UsersExtendedField extends AppModel {
var $name = 'UsersExtendedField';
var $displayField = 'fieldname';
var $hasMany = array(
'UsersExtended' => array(
'className' => 'Crooboard.UsersExtended',
'foreignKey' => 'field_id',
'conditions' => array(
'UsersExtended.user_id = User.id'
)
),
);
}
This is not the full code, these are the important parts. The problem starts right where I wrote 'UsersExtended.user_id = User.id'. Obviously, this won't work. But I do not have any idea how to access the User.id here. I also could not imagine a HABTM structure to solve this task. Do you have any idea how to get the semantics of this 'UsersExtended.user_id = User.id' to work?
Thank your very much for taking the time to read through this and helping me!
It sounds like you need to set up your HABTM relationship properly.
You already have the join table, UsersExtended, which contains your foreign keys.
Remove all previous relationships and set up HABTM in each of your User and UserExtendedField models.
The relationship code in your User model would look like this:
var $hasAndBelongsToMany = array(
'UsersExtended' => array(
'className' => 'UsersExtended',
'joinTable' => 'UsersExtended', //assuming this is the
//name of that model's table
'foreignKey' => 'user_id',
'associationForeignKey' => 'field_id'
)
);
For more information check out the page in the cakephp book
In addition, this blog post helped me grasp the relationship concepts when I was learning cakephp.
I'm having trouble composing a CakePHP find() which returns the records I'm looking for.
My associations go like this:
User ->(has many)-> Friends ,
User ->(has many)-> Posts
I'm trying to display a list of all a user's friends recent posts, in other words, list every post that was created by a friend of the current user logged in.
The only way I can think of doing this is by putting all the user's friends' user_ids in a big array, and then looping through each one, so that the find() call would look something like:
$posts = $this->Post->find('all',array(
'conditions' => array(
'Post.user_id' => array(
'OR' => array(
$user_id_array[0],$user_id_array[1],$user_id_array[2] # .. etc
)
)
)
));
I get the impression this isn't the best way of doing things as if that user is popular that's a lot of OR conditions. Can anyone suggest a better alternative?
To clarify, here is a simplified version of my database:
"Users" table
id
username
etc
"Friends" table
id
user_id
friend_id
etc
"Posts" table
id
user_id
etc
After reviewing what you have rewritten, I think I understand what you are doing. Your current structure will not work. There is no reference in POSTS to friends. So based on the schema you have posted, friends CANNOT add any POSTS. I think what you are trying to do is reference a friend as one of the other users. Meaning, A users FRIEND is actually just another USER in the USERS table. This is a self referential HABTM relationship. So here is what I would propose:
1- First, make sure you have the HABTM table created in the DB:
-- MySQL CREATE TABLE users_users ( user_id char(36) NOT NULL,
friend_id char(36) NOT NULL );
2- Establish the relationships in the User model.
var $hasAndBelongsToMany = array(
'friend' => array('className' => 'User',
'joinTable' => 'users_users',
'foreignKey' => 'user_id',
'associationForeignKey' => 'friend_id',
'unique' => true,
),
);
var $hasMany = array(
'Post' => array(
'className' => 'Post',
'foreignKey' => 'user_id'
),
);
3- use the scaffolding to insert a few records, linking friends and adding posts.
4- Add the view record function to the Users controller:
function get_user($id)
{
$posts = $this->User->find('first', array(
'conditions' => array('User.id' => $id),
'recursive' => '2'
));
pr($posts);
}
5- Now you can query the User table using recursive to pull the records using the following command:
http://test/users/get_user/USER_ID
6- Your output will show all of the records (recursively) including the friends and their posts in the returned data tree when you pr($posts)
I know this is a long post, but I think it will provide the best solution for what you are trying to do. The power of CakePHP is incredible. It's the learning curve that kills us.
Happy Coding!
If Post.user_id points to Friend.id (which wouldn't follow the convention btw) then it would be
$posts = $this->Post->find('all',array(
'conditions' => array(
'Post.user_id' => $user_id_array
)
);
which would result in .. WHERE Post.user_id IN (1, 2, 3) ..
Depending on your setup, it might be quicker to run two queries rather than trying to chain them together via the Cake stuff. I'd recommend adding something like getFriendsPosts() in the Users model.
<?php
class UserModel extends AppModel {
// ... stuff
function getFriendsPosts( $user_id )
{
$friends = $this->find( ... parameters to get user IDs of all friends );
// flatten the array or tweak your params so they fit the conditions parameter. Check out the Set class in CakePHP
$posts = $this->find( 'all', array( 'conditions' => array( 'User.id' => $friends ) ) );
return $posts;
}
}
?>
Then to call it, in the controller just do
$friends = $this->User->getFriendsPosts( $this->Auth->User('id') );
HTH,
Travis
Isn't CakePHP already generating the efficient code of:
SELECT * from Posts WHERE user_id IN (id1, id2 ...)
if not, you can do
$conditions='NULL';
foreach($user_id_array as $id) $conditions.=", $id";
$posts = $this->Posts->find('all', array(
'conditions' => "Post.user_id IN ($conditions)",
));
If your models are properly associated, Cake will automatically retrieve related model records. So, when you search for a specific user, Cake will automatically retrieve related friends, and related posts of these friends. All you need to do is set the recursion level high enough.
$user = $this->User->find('first', array('conditions' => array('User.id' => $id), 'recursive' => 2));
debug($user);
// gives something like:
array(
User => array()
Friend => array(
0 => array(
...
Post => array()
),
1 => array(
...
Post => array()
)
)
)
All you need to do is extract the posts from the user's friends, which is as easy as:
$postsOfFriends = Set::extract('/Friend/Post/.', $user);