In Laravel 4, how do you use a table from another schema (other than the default schema defined in database.php) to query against?
The L4 docs do not show any sign of being able to do this...
http://laravel.com/docs/database
define another schema config in the connections array
'pgsql' => array(
'driver' => 'pgsql',
'host' => 'localhost',
'database' => 'database',
'username' => 'root',
'password' => '',
'charset' => 'utf8',
'prefix' => '',
'schema' => 'public',
),
'newConnection' => array(
'driver' => 'pgsql',
'host' => 'localhost',
'database' => 'database',
'username' => 'root',
'password' => '',
'charset' => 'utf8',
'prefix' => '',
'schema' => 'public',
),
then just switch to it, per the docs:
$users = DB::connection('foo')->select(...);
the question was: How to use an alternate schema in queries?
the answer was to define another schema, then choose to connect to it.
example:
$schema_conn = DB::connection('{ARRAY_KEY_FOR_OTHER_SCHEMA_CONFIG}');
$query = $schema_conn->where('property', 'value')->get();
Related
Can you help me. How to relate it and make a join
. i get error has no relation named "project".
i using ActiveRecord with my code :
$posts = MaKantor::find()
->leftJoin('project.track', '`track`.`id_office` = `m_kantor`.`kantor_id`')
->with('project.track')->where(['collecting_id' => $model->collecting_id])
->all();
and config
'db' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=project',
'username' => 'root',
'password' => '',
'charset' => 'utf8',
],
'db2' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=db_master',
'username' => 'root',
'password' => '',
'charset' => 'utf8',
],
When you use with('relationName') in query, relation function needs to be defined in MaKantor model.
For example :
public function getProject()
{
return $this->hasOne(Project::className(), ['id' => 'project_id']);
}
I'm using CakePHP 2.8.5. It's not letting me log in "Username or password is incorrect". This seems totally straightforward in the docs but it isn't working for me. I wonder if my model/data structure might be confusing CakePHP. I have a Users model, but the logins are associated with an Admins model. The login form and action are in the Pages model (it has forms for multiple models).
in AppController:
public $components = array(
'DebugKit.Toolbar',
'Flash',
'Session',
'Auth' => array(
'userModel' => 'Admin',
'authenticate' => array(
'Form' => array(
'fields' => array(
'username' => 'email',
'password' => 'password'
),
'passwordHasher' => 'Blowfish'
)
),
'loginAction' => array(
'controller' => 'pages',
'action' => 'login',
),
'logoutRedirect' => array(
'controller' => 'pages',
'action' => 'login',
),
'authError' => 'Please log in',
'authorize' => array('Controller')
)
);
My login view, in /View/Pages. "email" is the username field:
<?php
echo $this->Form->create('Admin');
echo $this->Form->input('email');
echo $this->Form->input('password');
echo $this->Form->end('Submit');
?>
PagesController:
public function login() {
if ($this->request->is('post')) {
if ($this->Auth->login()) {
return $this->redirect($this->Auth->redirect());
} else {
$this->Flash->error(__('Username or password is incorrect'));
}
}}
Top of Admin model:
App::uses('BlowfishPasswordHasher', 'Controller/Component/Auth');
Automatic Blowfish encryption in Admin model:
public function beforeSave($options = array()) {
if (isset($this->data['Admin']['password'])) {
$passwordHasher = new BlowfishPasswordHasher();
$this->data['Admin']['password'] = $passwordHasher->hash(
$this->data['Admin']['password']
);
}
return true;
}
I notice if I enter the same password for different Admins, I get a different encryption result, but I've read that's normal.
If you want to see anything else, I'll add it.
The userModel key is in the wrong place
Compare the config in the question:
public $components = array(
'DebugKit.Toolbar',
'Flash',
'Session',
'Auth' => array(
'userModel' => 'Admin',
'authenticate' => array(
'Form' => array(
'fields' => array(
'username' => 'email',
'password' => 'password'
),
'passwordHasher' => 'Blowfish'
)
),
To the config in the docs:
$this->Auth->authenticate = array(
'Basic' => array('userModel' => 'Member'),
'Form' => array('userModel' => 'Member')
);
In the question userModel is a top-level key, in the docs it is part of the individual authenticate keys. Looking at the api examples (or the doc blocks in the source code) the error is more clear:
... you can define settings that should be set to all authentications objects using the 'all' key:
$this->Auth->authenticate = array(
'all' => array(
'userModel' => 'Users.User',
'scope' => array('User.active' => 1)
),
'Form',
'Basic'
);
It is possible to define a global userModel for all authenticate objects to use, but the syntax is simply different than the question.
Use the all key
Therefore to define a user model to use for all authenticate options, use the all key:
public $components = array(
'DebugKit.Toolbar',
'Flash',
'Session',
'Auth' => array(
//'userModel' => 'Admin', // <- no
'authenticate' => array(
'all' => array(
'userModel' => 'Admin' // <- yes
),
'Form' => array(
'fields' => array(
'username' => 'email',
'password' => 'password'
),
'passwordHasher' => 'Blowfish'
)
),
I'm developing a API in YII2 with multiple databases. I want to choose the database in real time. The idea is to read one variable available in controler (API key) to identify the correct connection database in model.
For example in webapplication like a portal i have (it works):
in db default connection i have
'db' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=dbcompany_0',
'username' => 'root',
'password' => 'XXXXXXXXXXXXXXXXXXXX',
'charset' => 'utf8',
],
In model I have
public function tableName()
{
$schema = '';
$user = User::find(Yii::app()->user->id);
$schema = "dbcompany_". $user->CompanyId;
return $schema . '.' . 'customer';
}
With this approach I just have a single db connection for all database companies.
How I can apply the same\similiar approach in API. I don't have sessions.
Any idea is well welcome.
Create a new component under db as:
'db1' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=dbcompany_0',
'username' => 'root',
'password' => 'XXXXXXXXXXXXXXXXXXXX',
'charset' => 'utf8',
],
'db2' => [
'class' => 'yii\db\Connection',
'dsn' => 'mysql:host=localhost;dbname=dbcompany_0',
'username' => 'root',
'password' => 'XXXXXXXXXXXXXXXXXXXX',
'charset' => 'utf8',
],
And use it as:
Yii::$app->db1
Yii::$app->db2;
All requests are authenticate using a token and this is associated to the user. When the authentication is done YII Framework assigns the userid to \Yii::$app->user->identity->id. So from anywhere I can identify the company owner in model.
public static function tableName(){
$moreinfo= account\Userextra::findOne(\Yii::$app->user->identity->id);
$schema = \Yii::$app->params['table.prefix'] . $moreinfo->CompanyId;
return $schema . '.country';
}
I need que make a select query from a table outside my user. I can make the select query from toad for example but in cakephp i get this error:
Missing Database Table Error: Table cursos for model Curso was not
found in datasource ot.
I understand the error occurs because the table doesn't exists directly in my user.
The question is: is there any way to get the data in this situation??
look what I understand the problem you have 2 databases and users want to use one for work:
Multiple Database connection issue CakePHP
in your database.php You can configure 2 connections
class DATABASE_CONFIG {
public $default = array(
'datasource' => 'Database/Mysql',
'persistent' => false,
'host' => 'localhost',
'login' => 'test',
'password' => 'test1',
'database' => 'test_portal',
'prefix' => ''
//'encoding' => 'utf8',
);
public $test = array(
'datasource' => 'Database/Mysql',
'persistent' => false,
'host' => 'localhost',
'login' => 'dfffd_23',
'password' => 'dsfsd324',
'database' => 'testdbuser',
'prefix' => ''
//'encoding' => 'utf8',
);
}
now in your model you have to specify which of the settings or connections want used for so you can select the users that you want.
class Example extends AppModel {
public $useDbConfig = 'test';
}
you can see this in the documentation
Im pretty done new to actually setting up sites as a live version, I've only done websites using local host. Im using codeigniter for this one, and i've traced my problem to the model where it loads the database. This makes me think that my database config isn't set up correctly.
The model's function is as follows:
public function register($email, $password, $first_name, $last_name, $gender, $birthday){
$salt = $this->generateSalt();
$password = $this->hash($password, $salt);
$data = array('email' => $email, 'password' => $password, 'salt' => $salt, 'first_name' => $first_name,
"last_name" => $last_name, 'gender' => $gender, "birthday" => $birthday);
$this->load->database();
if($this->db->insert('users', $data)){
$this->loginById($this->db->insert_id());
return true;
}else{
return false;
}
}
And the database config looks like this though I removed the password and username for privacy the username and pass I used are the same as I used to get into phpmyadmin though:
$db['default'] = array(
'dsn' => '',
'hostname' => 'http://130.184.99.114/',
'username' => '',
'password' => '',
'database' => 'meetings',
'dbdriver' => 'mysqli',
'dbprefix' => '',
'pconnect' => TRUE,
'db_debug' => TRUE,
'cache_on' => FALSE,
'cachedir' => '',
'char_set' => 'utf8',
'dbcollat' => 'utf8_general_ci',
'swap_pre' => '',
'autoinit' => TRUE,
'encrypt' => FALSE,
'compress' => FALSE,
'stricton' => FALSE,
'failover' => array()
);
So, how do I need to set this up? I think Im on the right track, but im really new to this, so please explain it in a way I can learn please! :D IF you need any more information please let me know.
The hostname you have listed is incorrect.
It should either be:
'hostname' => '130.184.99.114',
or
'hostname' => 'localhost',
depending on how the server has been set up.
Have you set this up in the /application/config/database.php file, or are you doing it differently? I ask as I see failover => array() in there, which isn't in the default configuration file.