I'm using 2 tables with 1 to many relation: campaign and group
This code returns only the relevant campaigns in the group.
$models = Campaigns::model()->with(array(
'Campgroupassoc' => array('condition' => "groupid=$id"),
))->findAll();
while this code:
$dataProvider = new CActiveDataProvider('Campaigns', array(
'criteria' => array(
'with' => array(
'Campgroupassoc' => array(
'condition' => "groupid=$id"
)
),
)
));
returns campaigns which are not in the same group..
What am i doing wrong?
Thx
'Campgroupassoc' => array(
'condition' => "groupid=:id",
'params'=>array(':id'=>$id)
)
Also it is better to specify to what model the groupid belongs to. And of course you can see what SQL dataprovider is generating and analize it.
Edit-----
After consulting with the Yii live chat, i found that even when using eager loading it will result in 2 queries instead of 1...
in order to solve this you need to add:
'together' => TRUE
Related
I have the following problem with CakePHP:
Two tables are joined (filters and accounts). Then I am building conditions and only the second condition Account.active =>1 gets executed. If I print the result, there are still showing filters that are having another mode_id than 3.
$joins= array(
array('table' => 'filters',
'alias' => 'Filter',
'type' => 'right',
'conditions' => array(
'Filter.account_id = Account.id',
)
),
);
Then I execute the request including joins and conditions
$activeAccounts = $this->Account->find('all',array(
'conditions'=>array('AND'=>array('Filter.mode_id'=>3,'Account.active'=>1)),
'joins'=>$joins));
The models were checked and no problems identified. Filter belongs to Account. Account has many Filter.
Below the query that is generated. The results are still showing filters with Filter.mode_id other than 3
Here is the query that is generated. The results are still containing rows with Filter.mode_id other than 3 despite the fact that one condition is 'Filter.mode_id'=>3
SELECT `Account`.`id`, `Account`.`user_id`, `Account`.`name`,
`Account`.`api_key`, `Account`.`account_number`, `Account`.`remaining_balance`,
`Account`.`investment_size`, `Account`.`active`
FROM `baseline_db`.`accounts` AS `Account`
right JOIN `baseline_db`.`filters` AS `Filter`
ON (`Filter`.`account_id` = `Account`.`id`)
WHERE ((`Filter`.`mode_id` = 3) AND
(`Account`.`active` = '1'))
Like say Oldskool, use the Model associations
and for your condition, The "AND" is not necessary,
you cant put :
$activeAccounts = $this->Account->find('all',array(
'conditions' => array(
'Filter.mode_id'=>3,
'Account.active'=>1
)
));
the request you want to make with the type of relation you have, seem to me weird.
If i understand, perhaps with something like that :
$this->loadModel('Filter');
$filters =$this->Filter->find("list", array(
'conditions' => array('Filter.mode_id' => 3),
'fields' => array('Filter.account_id')
));
$activeAccounts = $this->Account->find('all',array(
'conditions' => array(
'Account.account_id'=>$filters,
'Account.active'=>1
)
));
I have hasMany Through table which is Chats table with Chat model and I'm using loadModel in User controller to load Chat model then ran below query to bindModel with Chat.user_id and User.id :
$this->loadModel('Chat');
$this->Chat->bindModel(array(
'belongsTo' => array(
'User' => array(
'foreignKey' => false,
'conditions' => array('Chat.user_id = User.id')
)
)
));
$lastChat = $this->Chat->find('all', array(
'conditions' => array(
'Chat.receiver_id' => $user_id['User']['id']
),
'order' => array('Chat.id DESC'),
'fields' => array(
'Chat.id',
'Chat.chat',
'Chat.user_id',
'Chat.receiver_id',
'Chat.read',
'Chat.created'
),
'group' => array('Chat.user_id')
));
I want to join those tables together but this does not seem to work in Cake way I tried with normal SQL query and it works fine.
What could be wrong here?
Have you tried setting the recursive property before the find? Eg:
$this->Chat->recursive = 3;
You may need to set this after $this->Chat->bindModel, but I am not sure if this will make a difference or not. You will also need to re-bind the User model before each find, if, for example, your find queries are run within a loop ...
I have a model template which hasmany themes.I want to show the list of templates with count of themes.I am using this
$this->Template->bindModel(
array(
'hasMany' => array(
'TemplateTheme' => array(
'className' => 'TemplateTheme',
'fields' => 'count(TemplateTheme.id) AS themes'
)
)
), false ...
it gives me 2 templates.But it gives me all the 3 themes count in the first template whereas 2 themes belongs to template 1 and the third theme belongs to template 2
in the query it is using id IN(template_id1,template_id2)
Any idea how to do this?
You are doing a common mistake, you are counting everyrow each time since you are not using group by, you should do is group by Template.id when you do your search. Butttttttt.... has many wont do a join :( so you have to force it a littleor use something like linkable component
example
$join = array(
array('table' => 'templateThemes',
'alias' => 'TemplateTheme',
'type' => 'LEFT',
'conditions' => array(
'Template.id = TemplateTheme.Template_id',
)
)
);
$fields = array('Template.id','count(TemplateTheme.id) AS themes');
$this->Template->find('all', array('fields'=>$fields, 'joins'=>$join', $group =>array('Template.id')));
You may also do it in reverse since belongsTo does the join something like this
in your model (it is always recommended to put it static in your model unless is not a normal association)
var belongsTo = array(
'Template'=> array(
'classname' => 'Template',
'foreign_key' => 'template_id'
);
and in controller
$fields = array('Template.id','count(TemplateTheme.id) AS themes');
$this->Template->find('all', array('fields'=>$fields, $group =>array('Template.id')));
Hope this helps you, if not just comment
I am stuck on pagination in CakePHP 1.3. I am trying to paginate feePayment records based on certain criteria.
feePayments belongs to Students which in turn belongs YearGroups.
I want to paginate 'unpaid' feePayments for each year group. The problem I am having is that the SQL query seems to only take into account the conditions I specified for the FeePayment model and ignores the YearGroup criteria so only overdue unpaid records are returned regardless of the year group specified.
Here is my code:
function unpaidClass($id) {
$this->paginate = array(
'FeePayment' => array ('recursive' => 1, 'conditions' => array('FeePayment.status' => 'Unpaid', 'FeePayment.due_date <= ' => date("Y-m-d"))),
'YearGroup' => array ('recursive' => 1, 'conditions' => array('YearGroup.school_year' => $id))
);
$this->set('feePayments', $this->paginate());
}
Hope this makes sense, appreciate any help.
Thanks,
Sid.
You should consider using the Containable behavior. This behavior allows you to group the necessary data you want without relaying on the "recursiveness" of your query. You can place conditions on your contained data similar to the way you would in your queries and is a more permanent way to structure data across your application instead of specifying conditions over and over again in each query.
Paginate should automatically pick up these associations when you Paginate your main model. Here's an example of what I mean here: http://cakephp.1045679.n5.nabble.com/Paginate-with-Containable-td1300971.html#a1300971
These should make your task easier.
After a lot of searching the net and reading CakePHP's documentation, here is the solution I came up with:
function unpaidClass($id) {
$this->FeePayment->unbindModel(array(
'belongsTo' => array('Student')
), $reset = 0);
$this->FeePayment->bindModel(array(
'belongsTo' => array(
'Student' => array(
'foreignKey' => false,
'conditions' => array('Student.id = FeePayment.student_id')
),
'YearGroup' => array(
'foreignKey' => false,
'conditions' => array('YearGroup.id = Student.year_group_id')
)
)
), $reset = 0);
$this->paginate = array(
'contain' => array('Student','YearGroup'),
'conditions' => array('YearGroup.school_year' => $id,
'FeePayment.status' => 'Unpaid',
'FeePayment.due_date <= ' => date("Y-m-d")));
$this->set('feePayments', $this->paginate());
}
I'm using the Containable behavior to get a list of Comments (belongsTo Post, which belongs to Question; Question hasMany Post, and Post hasMany Comments; all of these belong to Users).
$data = $this->Question->find ( 'first',
array ('contain' =>
array ('User',
'Post' => array ('User', /* 'order' => 'User.created DESC'*/ )
)
)
);
It works, when I comment out the section in comments above. I suppose this is to be expected, but what I want is all of the Posts that are found, should be sorted in order of the 'created' field of the 'User' they belong to. How do I accomplish this deeper level sorting in CakePHP? I always get, "Warning (512): SQL Error: 1054: Unknown column 'User.created' in 'order clause'"
Thanks for your help!
Also, you might be trying to group on a related table from a find call that doesn't use joins.
Set your debug level to something greater than 1 so you can see the query log and make sure that Cake isn't doing two queries to fetch your data. If that is the case then the first query is not actually referencing the second table.
If you want to manually force a join in these situations you can use the Ad-Hoc joins method outlined by Nate at the following link.
http://bakery.cakephp.org/articles/view/quick-tip-doing-ad-hoc-joins-in-model-find
I have found two ways to get around this.
The first is to define the second level associacion directly in the model.
Now you will have access to this data everywhere.
It should look something like this.....
var $belongsTo = array(
'Foo' => array(
'className' => 'Foo', //unique name of 1st level join ( Model Name )
'foreignKey' => 'foo_id', //key to use for join
'conditions' => '',
'fields' => '',
'order' => ''
),
'Bar' => array(
'className' => 'Bar', //name of 2nd level join ( Model Name )
'foreignKey' => false,
'conditions' => array(
'Bar.id = Foo.bar_id' //id of 2nd lvl table = associated column in 1st level join
),
'fields' => '',
'order' => ''
)
);
The problem with this method is that it could make general queries more complex than they need be.
You can thus also add the second level queries directly into te find or paginate statement as follows: (Note: I found that for some reason you can't use the $belongsTo associations in the second level joins and will need to redefine them if they are already defined. eg if 'Foo' is already defined in $belongsTo, you need to create a duplicate 'Foo1' to make the association work, like the example below.)
$options['joins'] = array(
array('table' => 'foos',
'alias' => 'Foo1',
'type' => 'inner',
'conditions' => array(
'CurrentModel.foo_id = Foo1.id'
)
),
array('table' => 'bars',
'alias' => 'Bar',
'type' => 'inner',
'foreignKey' => false,
'conditions' => array(
'Bar.id = Foo1.bar_id'
)
)
);
$options['conditions'] = array('Bar.column' => "value");
$this->paginate = $options;
$[modelname] = $this->paginate();
$this->set(compact('[modelname]'));
I hope this is clear enough to understand and that it helps someone.
Check your recursive value. If it's too limiting, it will ignore the containable links, IIRC. I remember bumping into this a few times. I'd try containing multiple models, but my recursive option was set to 0 and nothing would get pulled. For your example, I'd think that a value of 1 (the default) would suffice, but maybe you've explicitly set it to 0 somewhere?
You can add before your call to find() the following:
$this->Question->order = 'Question.created DESC';
Yeah, I couldn't work out how to sort based on the related/associated model, so ended up using the Set::sort() method. Checkout this article for a good explanation.
// This finds all FAQ articles sorted by:
// Category.sortorder, then Category.id, then Faq.displaying_order
$faqs = $this->Faq->find('all', array('order' => 'displaying_order'));
$faqs = Set::sort($faqs, '{n}.Category.id', 'ASC');
$faqs = Set::sort($faqs, '{n}.Category.sortorder', 'ASC');
...And yes, it should probably be a Category->find() but unfortunately the original developer didn't code it that way, and I didn't wanna rework the views.