How to write SELECT query inside JOIN LEFT in a Zend model?
for eg how to convert the following mysql query to zend model query
LEFT JOIN
(SELECT count(*) as game_count,topic_id,time as g_time from games_list WHERE type < 3 GROUP BY topic_id) t3
ON chapter_list.id=t3.topic_id
I want to add converted query with the following zend query and I need game_count from the above query with the following result.
$query = $this->select()
->setIntegrityCheck(false)
->from(array('a' => 'chapter_list'), array('subtopic_id as topic','id as topic_id','grade','chapter_no as chapter','if(file_status=1,1,0) as ppt)','CONCAT("http://mysite.com?t=",subtopic_id) as link'))
->joinLeft(array('b' => 'subject_list'), 'a.subject = b.subject_id', array());
return $this->fetchAll($query)->toArray();
Finally I got it,
$subquery1= $this->select()
->setIntegrityCheck(false)
->from('games_list', array('count(*) as game_count','topic_id','time as g_time'))
->where('type<?', 3)
->group('topic_id');
//main query
$query = $this->select()
->setIntegrityCheck(false)
->from(array('a' => 'chapter_list'), array('subtopic_id as topic','id as topic_id','grade','chapter_no as chapter','if(file_status=1,1,0) as ppt)','CONCAT("http://eskool.com/learn?t=",subtopic_id) as link'))
->joinLeft(array('b' => 'subject_list'), 'a.subject = b.subject_id', array())
->joinLeft(array('c' => $subquery1), 'a.id = c.topic_id', array('IFNULL(game_count,0)','g_time'))
return $this->fetchAll($query)->toArray();
Related
I'm new in Yii2, and I have a query with right result:
SELECT DISTINCT workloadTeam.project_id, wp.project_name, workloadTeam.user_id, workloadTeam.commit_time, wp.workload_type FROM
(SELECT p.id, p.project_name, w.user_id, w.commit_time, w.comment, w.workload_type
FROM workload as w, project as p
WHERE w.user_id = 23 AND p.id = w.project_id) wp
INNER JOIN workload as workloadTeam ON wp.id = workloadTeam.project_id
But in my ModelSearch.php, I wrote:
$user_id = Yii::$app->user->id;
$subquery = Workload::find()->select('p.id', 'p.project_name', 'w.user_id', 'w.commit_time', 'w.comment', 'w.workload_type')
->from(['project as p', 'workload as w'])
->where(['user_id' => $user_id, 'p.id' => 'w.project_id']);
$query = Workload::find()
->select(['workloadTeam.project_id', 'wp.project_name', 'workloadTeam.user_id', 'workloadTeam.from_date', 'workloadTeam.to_date', 'workloadTeam.workload_type', 'workloadTeam.comment'])
->where(['', '', $subquery]);
$query->join('INNER JOIN', 'workload as workloadTeam', 'wp.id = workloadTeam.project_id');
It happended error:
SELECT COUNT(*) FROM `workload` INNER JOIN `workload` `workloadTeam` ON wp.id = workloadTeam.project_id WHERE `` (SELECT p.project_name `p`.`id` FROM `project` `p`, `workload` `w` WHERE (`user_id`=20) AND (`p`.`id`='w.project_id'))
And I can't fix it with right query above.
You have any solution about this?
Is this error shown in the Yii-debug toolbar? Then your query (which you mentioned as error) is probably only the count from the query which is listed before.
You missed to add the sub-query in from clause like you shown in your working sql. Add this in your where clause were just the wrong place. Put sub-queries in whereconditions, if you have scalar results, because you have to use this result with operands like =, >=, in...
This could work:
$user_id = Yii::$app->user->id;
$subquery = Workload::find()->select([
'p.id as id',
'p.project_name as project_name',
'w.user_id as user_id',
'w.commit_time as commit_time',
'w.comment as comment',
'w.workload_type as workload_type'
])
->from([
'project as p',
'workload as w'
])
->where([
'user_id' => $user_id,
'p.id' => 'w.project_id'
]);
$query = Workload::find()
->select([
'workloadTeam.project_id',
'wp.project_name',
'workloadTeam.user_id',
'workloadTeam.from_date',
'workloadTeam.to_date',
'workloadTeam.workload_type',
'workloadTeam.comment'
])
->from([$subquery => 'wp']); //you were missing this line
$query->join('INNER JOIN', 'workload as workloadTeam', 'wp.id = workloadTeam.project_id');
But you don't use any selects from your workload table in your main-query $query...
Since I don't know what's your goal to achieve I can't help you at this topic...
I am using Zend Framework 2.4 with TableGateway and need to execute the following query:
SELECT products.title, student.firstname, orders.order_id,
orders.profile_id, orders.status
FROM student
INNER JOIN
(products
INNER JOIN orders ON products.product_id = orders.product_id)
ON student.student_id = orders.student_id
WHERE (((orders.profile_id)='.$id.'));
Here is what I am using:
$select = $this->tableGateway->getSql()->select();
$select->columns(array('*'))
->join('products', 'products.product_id = orders.product_id', array('title'))
->join('student', 'student.student_id = orders.student_id', array('first_name'))
->where(array('orders.profile_id' => $id));
$items = $this->tableGateway->selectWith($select);
return $items;
The problem is that $items does not contain the columns title or first_name. It only contains the columns from the orders table.
What do I need to do to get $items to contain the columns from the joined tables?
try to use an associative array
$select = $this->tableGateway->getSql()->select();
$select->columns(array('*'))
->join('products', 'products.product_id = orders.product_id', array('title' => 'product_title'))
->join('student', 'student.student_id = orders.student_id', array('first_name' => 'student_first_name'))
->where(array('orders.profile_id' => $id));
$items = $this->tableGateway->selectWith($select);
return $items;
How would i convert below query to sub query?
I don't want to use JOINS I want to do same through sub query. i-e I need three subqueries out of the three joins. How it would be possible for below query ?
protected $_name = 'sale_package_features';
public function getAllSalePackageFeatures(){
$sql = $this->select()->setIntegrityCheck(false)
->from(array('spf' => $this->_name))
->joinLeft(array('sd' => 'sale_devices'),'sd.sale_device_id = spf.sale_device_id',array('sd.sale_device_name AS deviceName'))
->joinLeft(array('sp' => 'sale_packages'),'sp.sale_package_id = spf.sale_package_id',array('sp.sale_package_name AS packageName'))
->joinLeft(array('sf' => 'sale_features'),'sf.sale_feature_id = spf.sale_feature_id',array('sf.sale_feature_name AS featureName'))
->where('sf.parent_id != ?',0)
->order('spf.sale_package_feature_id ASC');
return $sql->query()->fetchAll();
}
Edited :
SELECT
`spf`.*, `sd`.`sale_device_name` AS `deviceName`,
`sp`.`sale_package_name` AS `packageName`,
`sf`.`sale_feature_name` AS `featureName`
FROM `sale_package_features` AS `spf`
LEFT JOIN `sale_devices` AS `sd`
ON sd.sale_device_id = spf.sale_device_id
LEFT JOIN `sale_packages` AS `sp`
ON sp.sale_package_id = spf.sale_package_id
LEFT JOIN `sale_features` AS `sf`
ON sf.sale_feature_id = spf.sale_feature_id
WHERE (sf.parent_id != 0)
ORDER BY `spf`.`sale_package_feature_id` ASC
You can use
$dbAdapter = Zend_Db_Table::getDefaultAdapter();
$subSelect = $dbAdapter
->select()
->from( 'tablename',
array(
'col1_alias' => 'column1_name',
'col2_alias' => 'column2_name',
))
->where('somefield = ?', $value_to_be_quoted);
$select = $dbAdapter
->select()
->from(array('sub_alias' => $subSelect ))
->where('otherfield = ?', $other_value_to_be_quoted);
$sql = $select->assemble();
$sql will contain
SELECT
sub_alias . *
FROM
(SELECT
tablename.column1_name AS col1_alias,
tablename.column2_name AS col2_alias
FROM
tablename
WHERE
(somefield = 'quoted_value')) AS sub_alias
WHERE
(otherfield = 'quoted_other_value')
If you need to use SELECT .. WHERE x IN (subselect) than go with
$select->where( 'column IN (?)', new Zend_Db_Expr($subselect->assemble()) );
I have defined the following criteria in Yii, and tries to use it to fetch an array of customers.
$criteria = new CDbCriteria(array(
"condition"=>"hidden = 0".(Yii::app()->user->GetState('is_admin') ? "" : " AND franchisesMunicipalities.franchise_id=".Yii::app()->user->getState('fid')),
"with" => array('municipality','municipality.franchisesMunicipalities')
));
$customers = Customers::model()->findAll($criteria);
This (i thought) should result in that Yii joined the table Customers with the table Municipalities, and then in the same query join the table Municipalities with the table Franchises_Municipalities. However it doesn't work since it doesn't join franchisesMunicipalities.
The resulting query is this
SELECT `t`.`id` AS `t0_c0`,
`t`.`municipality_id` AS `t0_c1`,
`t`.`personal_code_number` AS `t0_c2`,
`t`.`name` AS `t0_c3`,
`t`.`adress` AS `t0_c4`,
`t`.`zip` AS `t0_c5`,
`t`.`phone` AS `t0_c6`,
`t`.`mobile` AS `t0_c7`,
`t`.`email` AS `t0_c8`,
`t`.`hidden` AS `t0_c9`,
`municipality`.`id` AS `t1_c0`,
`municipality`.`county_id` AS `t1_c1`,
`municipality`.`name` AS `t1_c2`
FROM `customers` `t`
LEFT OUTER JOIN `municipalities` `municipality`
ON ( `t`.`municipality_id` = `municipality`.`id` )
WHERE ( hidden = 0
AND municipality.franchisesmunicipalities.franchise_id = 7 )
LIMIT 30
As you can see it only joins on one relation. The relations in the model should be correctly defined, since i am able to use them in other contexts.
Why doesn't this work?
According to http://www.yiiframework.com/doc/api/1.1/CActiveRecord#with-detail I believe you should be doing it this way:
$criteria = new CDbCriteria(array(
"condition"=>"hidden = 0".(Yii::app()->user->GetState('is_admin') ? "" : " AND franchisesMunicipalities.franchise_id=".Yii::app()->user->getState('fid'))
));
$customers = Customers::model()->with('municipality','municipality.franchisesMunicipalities')->findAll($criteria);
Hopefully that works for you.
Can someone Tell me how to join 3 table with php?
Example
SELECT FROM table1, table2,table on INNERJOIN -------------------
let I have 3 table.(question table ,answer table and category table)
Here is example form my webpage.
Time remaining 30 minutes(I will get "30 minutes" form Category table)
1. Question (from question table)
2. answer (from answer table)
I don't know how to join 3 table.
it should be like that,
$this->db->select('*');
$this->db->from('table1');
$this->db->join('table2', 'table1.id = table2.id');
$this->db->join('table3', 'table1.id = table3.id');
$query = $this->db->get();
as per CodeIgniters active record framework
I believe that using CodeIgniters active record framework that you would just use two join statements one after the other.
eg:
$this->db->select('*');
$this->db->from('table1');
$this->db->join('table1', 'table1.id = table2.id');
$this->db->join('table1', 'table1.id = table3.id');
$query = $this->db->get();
Give that a try and see how it goes.
I created a function to get an array with the values for the fields and to join. This goes in the model:
public function GetDataWhereExtenseJoin($table,$fields,$data) {
//pega os campos passados para o select
foreach($fields as $coll => $value){
$this->db->select($value);
}
//pega a tabela
$this->db->from($table);
//pega os campos do join
foreach($data as $coll => $value){
$this->db->join($coll, $value);
}
//obtem os valores
$query = $this->db->get();
//retorna o resultado
return $query->result();
}
This goes in the controller:
$data_field = array(
'NameProduct' => 'product.IdProduct',
'IdProduct' => 'product.NameProduct',
'NameCategory' => 'category.NameCategory',
'IdCategory' => 'category.IdCategory'
);
$data_join = array
( 'product' => 'product_category.IdProduct = product.IdProduct',
'category' => 'product_category.IdCategory = category.IdCategory',
'product' => 'product_category.IdProduct = product.IdProduct'
);
$product_category = $this->mmain->GetDataWhereExtenseJoin('product_category', $data_field, $data_join);
result:
echo '<pre>';
print_r($product_category);
die;
I think in CodeIgniter the best to use ActiveRecord as wrote above.
One more thing: you can use method chaining in AR:
$this->db->select('*')->from('table1')->join('table2','table1.id=table2.id')->...
$this->db->select('*');
$this->db->from('table1');
$this->db->join('table2', 'table1.id = table2.id','JOIN Type');
$this->db->join('table3', 'table1.id = table3.id');
$query = $this->db->get();
this will give you result from table1,table2,table3 and you can use any type of join in the third variable of $this->db->join() function such as inner,left, right etc.
For executing pure SQL statements (I Don't Know About the FRAMEWORK- CodeIGNITER!!!)
you can use SUB QUERY!
The Syntax Would be as follows
SELECT t1.id
FROM example t1 INNER JOIN
(select id from (example2 t1 join example3 t2 on t1.id = t2.id)) as t2 ON t1.id = t2.id;
Hope you Get My Point!
$this->db->select('*');
$this->db->from('table1');
$this->db->join('table2', 'table1.id = table2.id', 'inner');
$this->db->join('table3', 'table1.id = table3.id', 'inner');
$this->db->where("table1", $id );
$query = $this->db->get();
Where you can specify which id should be viewed or select in specific table. You can also select which join portion either left, right, outer, inner, left outer, and right outer on the third parameter of join method.
you can modiv your coding like this
$this->db->select('a.nik,b.nama,a.inv,c.cekin,c.cekout,a.tunai,a.nontunai,a.id');
$this->db->select('DATEDIFF (c.cekout, c.cekin) as lama');
$this->db->select('(DATEDIFF (c.cekout, c.cekin)*c.total) as tagihan');
$this->db->from('bayar as a');
$this->db->join('pelanggan as b', 'a.nik = b.nik');
$this->db->join('pesankamar_h as c', 'a.inv = c.id');
$this->db->where('a.user_id',$id);
$query = $this->db->get();
return $query->result();
i hope can be resolve your SQL
function fetch_comments($ticket_id){
$this->db->select('tbl_tickets_replies.comments,
tbl_users.username,tbl_roles.role_name');
$this->db->where('tbl_tickets_replies.ticket_id',$ticket_id);
$this->db->join('tbl_users','tbl_users.id = tbl_tickets_replies.user_id');
$this->db->join('tbl_roles','tbl_roles.role_id=tbl_tickets_replies.role_id');
return $this->db->get('tbl_tickets_replies');
}