I've read many similar topics here but not one helps me with problem.
i'm trying to get sections_id value from query in controller.
$query_id = "SELECT sections_id FROM sections WHERE name='".$table_name."'";
debug($id = $this->Info->query($query_id)); die();
there is debug result
array(
(int) 0 => array(
'sections' => array(
'sections_id' => '14'
)
)
)
and i tried in controller get value of id typing $id['sections']['sections_id'], or
$id['sections_id'] and many other types, nothing works. Do you have any idea ?
use $id[0]['sections']['sections_id'] to access it
For one result use
$data[0]['sections']['sections_id']
If query returns more than one result the use below code:
foreach($id as $data){
echo $data['sections']['sections_id']
}
Look for array index's carefully! Iinjoy
I have a model A that
$hasMany = array('B' => array('order' => 'B.created DESC'),
'C' => array('order' => 'C.created ASC')
);
So now i want to fetch the 5 most desolated entries of A(B,C) from my database that did not get modified the longest time(including B and C 'modified'-field).
How do i do that?
/EDIT: To explain it in other words:
Assume i have a model User that has information like name, adress, email etc.
A User also has many model Post on his page. And he has many model Other on his page too.
So we have a relationship of 1:n in User:Post and User:Other
Now i want to fetch those 5 User of my User, whose highest 'modified' value (User.modified, Post.modified or Other.modified) is one of they five lowest of all Users.
In other words: The 5 User who did not update their profile for the longest time.
I foresee some data inconsistencies in the future with this technique. I would do it this way.
Add field to user table called lastactivity
In AppController.php add beforeSave function
public function beforeSave() {
$save_data = $this->data;
if(isset($save_data['User']['modified']) || isset($save_data['Post']['modified'] || isset($save_data['Other']['modified']))){
$data = array();
$data['User']['id'] = $this->Auth->user('id');
$data['User']['lastactivity'] = date("Y-m-d H:i:s");
$this->User->save($data);
}
}
Can't make sort links work. (concrete or virtual fields).
Vitual fields for my sum() field on this action:
$this->Qca->virtualFields['comps'] = 'Sum(CASE WHEN Qca.qca_tipcode = 1 THEN 1 END)';
$this->Qca->virtualFields['production'] = 'Sum(qca_end - qca_start)';
$this->Qca->virtualFields['idle'] = 'Sum(Qca.qca_durend)';
My find(), works fine:
$hoursvalues = $this->Qca->find('all', array('conditions' => $conditions,
'fields' => array('Qca.dir_id', 'Qca.name', 'Sum(CASE WHEN Qca.qca_tipcode = 1 THEN 1 END) AS Qca__comps', 'Sum(qca_end - qca_start) as Qca__production', 'Sum(Qca.qca_durend) as Qca__idle'),
'group' => array('Qca.dir_id')
)
);
and then:
$this->paginate('Qca' );
$this->set('hoursvalues', $hoursvalues);
What extra settings does $this->paginate('Qca' ); needs? Please note I have all data I need via find().
What is it I'm missing that sorting does not work for either concrete or virtual fields?
Thansk a lot!
Carlos
Part of your problem is the following:
$this->paginate('Qca');
$this->set('hoursvalues', $hoursvalues);
$this->paginate() returns an array with sorted values. What you need to do is specify your extra settings in $this->paginate array and then
$this->set('hoursvalues', $this->paginate('Qca'));
For your other fields, making them virtualFields will make them easier to work with.
I have two tables,Forms and Attributes. I'm tring to retrieve the last inserted id from the Forms table and insert it to the form_id column of the Attributes table, along with the other field columns.
Earlier I retrieved the form Id from the Forms table and used it to update the value of the Form name column. It worked fine.The code for that is given below:
function saveFormName($data)
{
$this->data['Form']['formname']=$data['Form']['formname'];
$this->data['Form']['id']=$this->find('all', array(
'fields' => array('Form.id'),
'order' => 'Form.id DESC'
));
$this->id=$this->data['Form']['id'][0];
$this->saveField('name',$this->data['Form']['formname']);
}
But When I tried to do it in a similar way for updating the attributes table,the row is not saved in the database,since the value of $this->data['Form']['id'][0] is an 'Array'.
Even in the saveFormName function, the value of $this->data['Form']['id'][0] is an 'Array',but the form name gets updated correctly. Someone explain me the concept.
function saveFieldEntries($data)
{
$this->data['Form']['id']=$this->find('all', array(
'fields' => array('Form.id'),
'order' => 'Form.id DESC'
));
$this->data['Attribute']['form_id'] = $this->data['Form']['id'][0];
$this->data['Attribute']['label']= 'Label';
$this->data['Attribute']['size']='20';
$this->data['Attribute']['type']=$data['Attribute']['type'];
$this->data['Attribute']['sequence_no'] = $data['Attribute']['sequence_no'];
$this->Attribute->save($this->data);
}
EDIT:
Ok, here is the corresponding code in the controller.
function insertFormName()
{
$this->data['Form']['formname']=$this->params['form']['formname'];
$this->Form->saveFormName($this->data);
}
function insertFieldEntry()
{
$this->data['Attribute']['type']=$this->params['form']['type'];
$this->data['Attribute']['sequence_no'] = $this->params['form']['sequence_no'];
$this->Form->saveFieldEntries($this->data);
}
The parameters form name,type and sequence no are passed to the controller from the corresponding view file.
$this->data['Form']['id'][0] holds an array because find('all') returns an array.
So if you need first ID from this array, you need to pick it properly in function saveFieldEntries:
...
$this->data['Attribute']['form_id'] = $this->data['Form']['id'][0]['Form']['id'];
...
I am writing a CakePHP 1.2 app. I have a list of people that I want the user to be able to filter on different fields. For each filterable field, I have a drop down list. Choose the filter combination, click filter, and the page shows only the records that match.
In people_controller, I have this bit of code:
$first_names = $this->Person->find('list', array(
'fields'=>'first_name',
'order'=>'Person.first_name ASC',
'conditions'=> array('Person.status'=>'1')
));
$this->set('first_names', $first_names);
(Status = 1 because I am using a soft delete.)
That creates an ordered list of all first_names. But duplicates are in there.
Digging around in the Cookbook, I found an example using the DISTINCT keyword and modified my code to use it.
$first_names = $this->Person->find('list', array(
'fields'=>'DISTINCT first_name',
'order'=>'Person.first_name ASC',
'conditions'=> array('Person.status'=>'1')
));
This gives me an SQL error like this:
Query: SELECT `Person`.`id`, DISTINCT `Person`.` first_name` FROM `people` AS `Person` WHERE `Person`.`status` = 1 ORDER BY `Person`.`first_name` ASC
The problem is obvious. The framework is adding Person.id to the query. I suspect this comes from using 'list'.
I will use the selected filter to create an SQL statement when the filter button is clicked. I don't need the is field, but can't get rid of it.
Thank you,
Frank Luke
Try to use 'group by', it works perfectry:
$first_names = $this->Person->find('list', array(
'fields'=>'first_name',
'order'=>'Person.first_name ASC',
'conditions'=> array('Person.status'=>'1'),
'group' => 'first_name'));
You're right, it seems that you cannot use DISTINCT with list. Since you don't need id but only the names, you can use find all like above and then $first_names = Set::extract($first_names, '/Person/first_name');. That will give you a array with distinct first names.
Here's how I did it in CakePHP 3.x:
$query = $this->MyTables->find('all');
$result = $query->distinct()->toArray();
You can try this.
Here this takes Person id as key, so there is no chance for duplicate entries.
$first_names = $this->Person->find('list', array(
'fields' => array('id','first_name'),
'conditions' => array('Person.status' => '1'),
));
$this->set('first_names', $first_names);
Using SQL grouping will also produce a distinct list. Not sure of the negative consequences if any, but it seems to work fine for me.
$first_names = $this->Person->find('list', array(
'fields' => 'first_name',
'order' => 'first_name',
'group' => 'first_name',
'conditions' => array('Person.status' => '1'),
));
$this->set('first_names', $first_names);
I know it is a question for CakePHP 1.2, but I was searching for that too with CakePHP version 3. And in this version there is a method to form the Query into a distinct one:
$first_names = $this->Persons->find(
'list',
[
'fields'=> ['name']
]
)
->distinct();
;
This will generate a sql query like this:
SELECT DISTINCT Persons.name AS `Persons__name` FROM persons Persons
But the distinct method is a bit mightier than just inserting DISTINCT in the query.
If you want to just distinct the result on one field, so just throwing away a row with a duplicated name, you can also use the distinct method with an array of the fields as parameter:
$first_names = $this->Persons->find(
'list',
[
'fields'=> ['id', 'name'] //it also works if the field isn't in the selection
]
)
->distinct(['name']); //when you use more tables in the query, use: 'Persons.name'
;
This will general a sql query like this (for sure it is a group query):
SELECT DISTINCT
Persons.id AS `Persons__id`, Persons.name AS `Persons__name`
FROM persons Persons
GROUP BY name #respectively Persons.name
Hope I will help some for CakePHP 3. :)
Yes the problem is that you are using a listing which designed for a id / value output. You probably will have to do a find('all') and then build the list yourself.
Yes I also tried to fetch unique results with 'list' but its not working. Then I fixed the problem by using 'all'.
In the example now on the book for version 2 it states the following:
public function some_function() {
// ...
$justusernames = $this->Article->User->find('list', array(
'fields' => array('User.username')
));
$usernameMap = $this->Article->User->find('list', array(
'fields' => array('User.username', 'User.first_name')
));
$usernameGroups = $this->Article->User->find('list', array(
'fields' => array('User.username', 'User.first_name', 'User.group')
));
// ...
}
With the above code example, the resultant vars would look something like this:
$justusernames = Array
(
//[id] => 'username',
[213] => 'AD7six',
[25] => '_psychic_',
[1] => 'PHPNut',
[2] => 'gwoo',
[400] => 'jperras',
)
$usernameMap = Array
(
//[username] => 'firstname',
['AD7six'] => 'Andy',
['_psychic_'] => 'John',
['PHPNut'] => 'Larry',
['gwoo'] => 'Gwoo',
['jperras'] => 'Joël',
)
$usernameGroups = Array
(
['User'] => Array
(
['PHPNut'] => 'Larry',
['gwoo'] => 'Gwoo',
)
['Admin'] => Array
(
['_psychic_'] => 'John',
['AD7six'] => 'Andy',
['jperras'] => 'Joël',
)
)
You have to plan your query in a slightly different way and plan your database to accommodate a list find.
In some cases, you wish to group by using some key, but you want unique element within the results.
For example, you have a calendar application with two types of events. One event on day 1rst and the other one on the 2nd day of month 1. And you want to show or rather count all the events, grouped by day and by type.
If you use only DISTINCT, it is quite difficult. The simplest solution is to group twice:
$this->Events->virtualFields['count'] = 'COUNT(*)';
$acts = $this->Activity->find('all',array(
'group' => array('DAY(Event.start)','EventType.id'),
));
Just group by the fields that you want to get distinct.. or use Set::extract() and then array_unique()
In 2.7-RC version, this is working
$this->Model1->find('list', array(
'fields' => array('Model1.field1', Model1.field1')
));