in code below i get a user_name columm not found if i remove my "workaround", how i can set the alias before the get command? have a trick?
//Select
$qb = DB::table('log as l');
//Joins
$qb->join('user as u', 'l.user_id', '=', 'u.id');
//Orders
if($sort[0]["field"]=="user_name")// This is a workaround for my problem
$sort[0]["field"] = "u.name";
$qb->orderBy($sort[0]["field"],$sort[0]["dir"]);
//Wheres
Data::applyFilter($qb,$filters);
$total = $qb->count("l.id");
$qb->skip($skip)
->take($take);
$result = $qb->get(array("l.id", "l.action", "l.entity_type", "l.entity_id", "u.name as user_name", "l.datetime"));
In documentation there is
$users = DB::table('users')->select('name as user_name')->get();
If you set a key-value variable instead of "user_name" it should do what you want ?
Bye
Related
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
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 (*).
SocialEngine4: How can I get the gender of the user using sql query?
$subject = Engine_Api::_()->core()->getSubject(); //Get user
$fields = $subject->fields()->getFieldsValuesByAlias();
$gender = $fields['gender'];
$this->view->gender = $gender;
Use this listed below code:
<?php
$currentUserID=Engine_Api::_()->user()->getViewer()->getIdentity();
$sSqlInviter="select gender from engine4_user_fields_search where item_id=".$currentUserID;
$dbInviterSqli=Zend_Db_Table_Abstract::getDefaultAdapter();
$stmtInviter=$dbInviterSqli->query($sSqlInviter);
$rowInviter=$stmtInviter->fetch();
$gender=$rowInviter['gender'];
if($gender==2)
echo "<b> Male</b>";
else
echo "<b> Female</b>";
?>
$viewer = Engine_Api::_()->user()->getViewer();
$fields = Engine_Api::_()->fields()->getFieldsValuesByAlias($viewer);
echo $fields['gender'];
Are you using default profile question? Then:
From engine4_ser_fields_option
There is and option id for the particular field.
I am new to zend framework,
Following is the plain mysql query which takes particular column from table,
SELECT jobs_users.id,jobs_users.first_name from jobs_users left join friends on jobs_users.id=friends.friend_id where friends.member_id=29
I tried with zend to implement the above query like below,
public function getFriendsProfileList($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;
How to write SQL LIKE Query in drupal ,
SELECT title FROM { node } WHERE type='%s'
i want to add the LIKE CONDITION IN THAT
SELECT title FROM { node } WHERE type='%s' AND LIKE '%S%'
i think i writtern wrong like query formnat, can rewrite and tell me,
Just use % to escape.
$result = db_query('SELECT title FROM {node} WHERE type = "%s" AND title LIKE "%%%s%%"', 'type', 'title');
while ($row = db_fetch_object($result)) {
// do stuff with the data
}
Node type does not need escaping.
And here is an example with how to use LIKE in a dynamic query (Drupal 7 Only):
$query = db_select('node', 'n')
->fields('n', array('title'))
->condition('type', 'my_type')
->condition('title', '%' . db_like(search_string) . '%', 'LIKE');
$result = $query->execute()->fetchCol();
db_like() is used to escapes characters that work as wildcard characters in a LIKE pattern.
drupal_query replace %% to % and %s to value string
so your code will be
$sql = "SELECT title FROM node WHERE type='%%%s' AND title LIKE '%%%S%%'";
$type = "type to use in query";
$title = "title to use in query";
$result = db_result(db_query($sql, $type, $title));
OK, so you want the LIKE operator to refer to the title column. Use this query:
$sql = "SELECT title FROM node WHERE type='%s' AND title LIKE '%S%'";
$type = "type to use in query";
$title = "title to use in query";
$result = db_result(db_query($sql, $type, $title));
This is because the LIKE operator requires a column name to be specified. Otherwise, your database doesn't have any idea what value you want to perform the comparison on. See here.