How to pass a variable to YII createCommand Query? - yii

I want to do something like this
Select * from users where id = $user_id
But how can I achieve this in YII?
Here is what I am trying
$user = Yii::app->db->createCommand('SELECT * FROM users where user_id = "$user_id"')->queryAll();
which returns an empty array.

You can use bindValue
$user = Yii::app->db->createCommand('SELECT * FROM users where user_id = :user_id')
->bindValue(':user_id', $user_id)
->queryAll();

Related

Get user data from database codeignter

I have two tables in one database, table one called: ci_admin, and table two called active_ingredient.
I want to get user data from the active_ingredient table where user_id equals admin_id from ci_admin table using the current user session.
this code, get data for all user, I need the data inserted by the user only:
$wh =array();
$SQL ='SELECT * FROM active_ingredient';
$wh[] = "SELECT 1
FROM ci_admin
WHERE ci_admin.admin_id = active_ingredient.user_id";
if(count($wh)>0)
{ $WHERE = implode(' and ',$wh);
return $this->datatable->LoadJson($SQL,$WHERE);
}
else
{
return $this->datatable->LoadJson($SQL);
}
Try this, it will help you.
$this->db->select('ai.*,ca.*');
$this->db->from('active_ingredient ai');
$this->db->join('ci_admin ca', 'ca.admin_id = ai.user_id', 'left');
$this->db->where($data); //pass the session data for exact filter
$query = $this->db->get();
return $query->result();

SQL query to Laravel

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

LastInserID from other table insert to table [duplicate]

I have a query, and I want to get the last ID inserted. The field ID is the primary key and auto incrementing.
I know that I have to use this statement:
LAST_INSERT_ID()
That statement works with a query like this:
$query = "INSERT INTO `cell-place` (ID) VALUES (LAST_INSERT_ID())";
But if I want to get the ID using this statement:
$ID = LAST_INSERT_ID();
I get this error:
Fatal error: Call to undefined function LAST_INSERT_ID()
What am I doing wrong?
That's because that's an SQL function, not PHP. You can use PDO::lastInsertId().
Like:
$stmt = $db->prepare("...");
$stmt->execute();
$id = $db->lastInsertId();
If you want to do it with SQL instead of the PDO API, you would do it like a normal select query:
$stmt = $db->query("SELECT LAST_INSERT_ID()");
$lastId = $stmt->fetchColumn();
lastInsertId() only work after the INSERT query.
Correct:
$stmt = $this->conn->prepare("INSERT INTO users(userName,userEmail,userPass)
VALUES(?,?,?);");
$sonuc = $stmt->execute([$username,$email,$pass]);
$LAST_ID = $this->conn->lastInsertId();
Incorrect:
$stmt = $this->conn->prepare("SELECT * FROM users");
$sonuc = $stmt->execute();
$LAST_ID = $this->conn->lastInsertId(); //always return string(1)=0
You can get the id of the last transaction by running lastInsertId() method on the connection object($conn).
Like this $lid = $conn->lastInsertId();
Please check out the docs https://www.php.net/manual/en/language.oop5.basic.php

How to get a single value in findbyPk() method in yii?

In my controller
$agent = University::model()->findByPK($university_id);
I hope it will return value of a row of value.
I want a single attribute(field3) value say university_name, (with out using findByPK), how to get it
SELECT field3 FROM table [WHERE Clause]
Try this
$usercriteria = new CDbCriteria();
$usercriteria->select = "university_name";
$usercriteria->condition = "university_id=$university_id";
$university = University::model()->findAll($usercriteria);
echo $university->university_name;
Or simply do like u did first
$agent = University::model()->findByPK($university_id);
echo $agent-> university_name;
$agent = University::model()->findByPK($university_id);
echo $agent->university_name;
It should be like this:
$agent = University::model()->findByPK($university_id)->university_name;
Try this:
$university_name = University::model()->findByPK($university_id, array('select'=>'univeersity_name'))->university_name;
#Query: SElECT university_name FROM table_name where id=x;
Instead of
$university_name = University::model()->findByPK($university_id)->university_name;
#Query: SElECT * FROM table_name where id=x;
Second query returns all the fields. So better to avoid those fields are not necessary.
in case if you need to view on the yii _views, i have implemented this on my project
i put this inside my '_view.php' at /protected/views/myTable/
$agent = University::model()->findByPK($data->id_university/*this is the PK field name*/);
echo $agent->university_name /*university field name*/;
sorry again for my bad english :o
It can be done like this:
$agent = University::model()->findAllByAttributes(array('field3'),"WHERE `id` = :id", array(':id' => $university_id));
First argument of findByAttributes is an array of attributes you wish it to return. If left empty it returns all (*).

How to get particular column in zend using Left join

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;