I am new to zend framework,
Following is the plain mysql query which takes particular column from table,
SELECT jobs_users.id,jobs_users.first_name from jobs_users left join friends on jobs_users.id=friends.friend_id where friends.member_id=29
I tried with zend to implement the above query like below,
public function getFriendsProfileList($id){
$db = Zend_Db_Table::getDefaultAdapter();
$select = $db->select();
$select->from('jobs_users')
->joinLeft(
'friends',
'jobs_users.id=friends.friend_id',
array('jobs_users.id','jobs_users.first_name','jobs_users.last_name','jobs_users.photo')
)
->where("friends.member_id = ?", $id);
$result = $db->fetchAll($select);
return $result;
}
Here i got result with all column name , not with exact column name which i have given in query.
Kindly help me on this.
Use this instead:
$select->from('jobs_users', array('jobs_users.id','jobs_users.first_name','jobs_users.last_name','jobs_users.photo'))
->joinLeft('friends', 'jobs_users.id=friends.friend_id')
->where("friends.member_id = ?", '20');
You may also try this:
$select = $db->select();
$select->setIntegrityCheck(false);
$select->joinLeft('jobs_users','',array('jobs_users.id','jobs_users.first_name','jobs_users.last_name','jobs_users.photo'));
$select->joinLeft('friends','jobs_users.id=friends.friend_id', array());
$select->where("friends.member_id = ?", $id);
$result = $db->fetchAll($select);
return $result;
Related
I am trying to convert the following query to Laravel:
select libéllé
from application
where libéllé not in (select application_id from application_user
where user_id = $id)
Laravel whereNotIn supports closures for subqueries, so it will be as simple as this:
Using Eloquent:
// Add this to top of your file.
use App\{ Application, ApplicationUser };
// Get your entries.
$rows = Application::whereNotIn('libéllè', function($query) use ($id) {
$query->select('application_id')->from((new ApplicationUser)->getTable())->where('user_id', $id);
})->get(['libéllè']);
Using Query Builder:
$rows = DB::table('application')->whereNotIn('libéllè', function($query) use ($id) {
$query->select('application_id')->from('application_user')->where('user_id', $id);
})->get(['libéllè']);
Please Try it.
$results = DB::select(
select libéllé
from application
where (libéllé)
not in(
select application_id from application_user
where user_id = $id
)
);
Also see this answer: How to convert mysql to laravel query builder
Also see this documentation: https://laracasts.com/discuss/channels/laravel/laravel5-resolve-username-from-2-tables?page=1
I have two tables Products(id, name) and Views(id,count,time), and those two tables are not related to each other. This is my code:
$dbCommand = Yii::app()->db->createCommand("
SELECT P.`id`, P.`name`, V.`time`
FROM `products` P, `views` V
WHERE P.`type` = 2
ORDER BY V.`time` DESC
");
$data = $dbCommand->queryAll();
It is working, but I want to convert this query to CDbCriteria syntax.
$cdb = new CDbCriteria();
$cdb->select = //???
$cdb->where = //???
$cdb->order = //???
How can I do this? Can somebody help me?
You cannot use CDbCriteria, Try using query builder.
Yii::app()->db->createCommand()
->select('P.id, P.name, V.time')
->from('products P, views V')
->where('P.type = :type')
->order('V.time DESC')
->queryAll(array(
':type' => 2
));
I want to execute the below sql query using Yii framework and need help on this.
SQL query
SELECT t.*, LP.name AS lp_name FROM `user` AS `t` LEFT JOIN `level_profiles` AS `LP` ON t.prof_i = LP.id WHERE t.bld_i IN (17)
So, i tried the below steps.
$usql = 't.bld_i IN (17)';
$criteria1 = new CDbCriteria;
$criteria1->select = 't.*, LP.*';
$criteria1->join = ' LEFT JOIN `level_profiles` AS `LP` ON t.prof_i = LP.id';
$criteria1->addCondition($usql);
$criteria1->order = 't.prof_i';
$result = User::model()->findAll($criteria1);
The above step is not allowing me to access the value from 'level_profiles' table.
Then, i tried to execute:
$usql = 't.bld_i IN (17)';
$result = User::model()->with('level_profiles', array(
'level_profiles'=>array(
'select'=>'name',
'joinType'=>'LEFT JOIN',
'condition'=>'level_profiles.id="prof_i"',
),
))->findAll($usql);
This is returning an error 'Relation "level_profiles" is not defined in active record class "User". '
I know this could be executed using the below method.
Yii::app()->db->createCommand('SELECT query')->queryAll();
But i dont want to use the above.
I am a beginner with Yii and tried to look into the forums. But, i am getting confused how to execute the query using "User::model()" approach .
class User extends CActiveRecord
{
......
public function relations()
{
return array(
'level_porfile_relation'=>array(self::BELONGS_TO, 'Level_Profiles_Modelname', 'prof_i'),
);
}
and your query will be:
$result = User::model()->with('level_porfile_relation')->findAll($usql);
I'm attempting to join two tables while using codeigniter. I've done this same SQL query writing regular SQL. When I attempt to do the same in codeigniter, I keep getting errors. I'm not quite sure what I'm doing wrong. What do you guys think I'm doing wrong?
My function in model_data.php
function getJoinInformation($year,$make,$model)
{
//$this->db->distinct();
// here is where I need to conduct averages but lets just work on extracting info.
// from the database and join tables.
$this->db->select('*');
$this->db->from('tbl_car_description');
$this->db->join('tbl_car_description', 'd.id = p.cardescription_id');
$this->db->where('d.year', $year);
$this->db->where('d.make', $make);
$this->db->where('d.model', $model);
$result = $this->db->get();
/*
$query = $this->db->get_where('tbl_car_description',
array(
'year' => $year,
'make' => $make,
'model' => $model
)
);
if($query->num_rows()) return $query->result();
return null;
*/
}
My error message
A Database Error Occurred
Error Number: 1066
Not unique table/alias: 'tbl_car_description'
SELECT * FROM (`tbl_car_description`) JOIN `tbl_car_description` ON `d`.`id` = `p`.`cardescription_id` WHERE `d`.`year` = '2006' AND `d`.`make` = 'Subaru' AND `d`.`model` = 'Baja'
Filename: C:\wamp\www\_states\system\database\DB_driver.php
Line Number: 330
Here is the code written in SQL and it's working great. I want to do the something in codeigniter but I'm confused as to how. Any help would be much appreciated. Thanks everyone.
$sql_2 = mysql_query("SELECT ROUND(AVG(p.value),1) AS AvgPrice, ROUND(AVG(p.mileage),1) AS AvgMileage
FROM tbl_car_description d, tbl_car_prices p
WHERE (d.id = p.cardescription_id)
AND (d.year = '".$year."')
AND (d.make = '".$make."')
AND (d.model = '".$model."')
AND (p.approve = '1')");
Your CI query references the same table twice, which I assume is a typo. However, you only need include your table alias with the table name in your Active Record call:
//$this->db->distinct();
$this->db->select('*');
$this->db->from('tbl_car_description d');
$this->db->join('tbl_car_prices p', 'd.id = p.cardescription_id');
$this->db->where('d.year', $year);
$this->db->where('d.make', $make);
$this->db->where('d.model', $model);
$result = $this->db->get();
Couple issues: You need to include your table aliases and your join should have the name of the second table ...
$this->db->from('tbl_car_description AS d');
$this->db->join('tbl_car_prices AS p', 'd.id = p.cardescription_id');
Whenever I tried to execute this sql query in a function module in drupal I am not able to get the results but when I try to execute this in MySQL I can view the result. My code looks like this :
function _get_subject_sub_category() {
$options = array();
$sql = "SELECT father.Subject_Code, child.Subject_Category
FROM {subjects} as child
INNER JOIN {subjects} as father ON (child.Parent_Category = father.Subject_Code
AND child.Level =2 )";
$result = db_query($sql);
foreach ($result as $row) {
$options[$row->father.Subject_Code] = $row->child.Subject_Category;
}
return $options;
}
The error I encountered is in the line $options[$row->father.Subject_Code] = $row->child.Subject_Category;`.
Any help will be highly appreciated.
Try changing this line:
$options[$row->father.Subject_Code] = $row->child.Subject_Category;
To this:
$options[$row->Subject_Code] = $row->Subject_Category;
The name of the table is not in the result. If you need to avoid confusion, you can use alias in your SQL query.
I don't know drupal and don't use php, but I would say :
remove
father. in $row->father.Subject_Code
and
child. in $row->child.Subject_Category
as father and child are just db aliases.