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
Related
I am stuck on transforming this SQL query to an eloquent query in Laravel. The SQL query works (tested in Sequel) but I cannot write it in eloquent ...
SELECT faqs.question FROM faqs
JOIN categories c ON c.id = faqs.category_id
WHERE c.subsite = 'sport'
This is what I have tried so far, but it returns all the questions, (ignoring the subsite filter).
$cat = Faq::with(['category' => function($query) use ($subsite) {
$query->where('subsite', $subsite);
}])->get();
Thanks for the help
Try this
$cat = Faq::query();
if (isset($subsite) && !empty($subsite) {
$query->whereHas('category', function ($query) use ($subsite) {
$query->where('subsite', $subsite);
});
}
$query->with('category')->get();
as per you core sql query, here is your laravel query.
$data = DB::table('faqs')
->join('categories', 'categories.id', '=', 'faqs.category_id')
->select('faqs.question')
->where('categories.subsite', '=', 'sport')
->get();
Hope this will help you!
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);
Following query returning six values
SELECT tbl_start FROM timetable inner join route ON tbl_rte_id = id WHERE rte_origin = "UL" and rte_destination = "HW" ORDER BY(tbl_start) DESC;
And my laravel code is returning only one value
$tables = Timetable::join('route', 'tbl_rte_id', '=', 'id')
->where('rte_origin', $origin, 'AND')
->where('rte_destination', $destination)
->orderBy('tbl_start', 'desc')
->get();
foreach ($tables as $table) {
$result[$table->id] = $table->tbl_start;
}
This laravel code is not similar or similar. Can anyone help me.
Change this part:
->where('rte_origin', $origin, 'AND')
// to:
->where('rte_origin', $origin)
It will know by default that it's AND operator
And if you want to provide this operator, then do this:
->where('rte_origin', '=', $origin, 'AND')
You may try something like this:
$tables = Timetable::join('route', 'tbl_rte_id', '=', 'timetable.id')
->where('rte_origin', $origin)
->where('rte_destination', $destination)
->orderBy('tbl_start', 'desc')
->get()->lists('tbl_start', 'id');
The $tables will contain an array of id => tbl_start pairs.
Add a listener in your routes.php
Event::listen('illuminate.query', function($sql){
var_dump($sql);
});
Then execute both queries and check if you have the same result
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 $row->child.Subject_Category
as father and child are just db aliases.
I am new to zend framework,
Following is the plain mysql query which takes particular column from table,
SELECT jobs_users.id,jobs_users.first_name from jobs_users left join friends on jobs_users.id=friends.friend_id where friends.member_id=29
I tried with zend to implement the above query like below,
public function getFriendsProfileList($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;