I found this in the guide, but have no idea how to implement the same
yii\db\Query::count(); returns the result of a COUNT query. Other
similar methods include sum($q), average($q), max($q), min($q), which
support the so-called aggregational data query. $q parameter is mandatory
for these methods and can be either the column name or expression.
Say for example I have a table name 'billing' with columns:
name amount
charge1 110.00
charge2 510.00
Total - 620.00
How I implement using
yii\db\Query::sum('amount');
I have also tried like
$command = Yii::$app->db->createCommand("SELECT sum(amount) FROM billing");
yii\db\Query::sum($command);
but page generates error.
Thanks.
The first part of code you tried appears to be attempting to use Query Builder. In this case, you must create an instance of a query, set the target table, and then compute the sum:
Via Query Builder
(http://www.yiiframework.com/doc-2.0/guide-db-query-builder.html):
$query = (new \yii\db\Query())->from('billing');
$sum = $query->sum('amount');
echo $sum;
The second part of code you tried appears to be attempting to use Data Access Objects. In this case, you can write raw SQL to query the database, but must use queryOne(), queryAll(), queryColumn(), or queryScalar() to execute the query. queryScalar() is appropriate for an aggregate query such as this one.
Via Data Access Objects
(http://www.yiiframework.com/doc-2.0/guide-db-dao.html):
$command = Yii::$app->db->createCommand("SELECT sum(amount) FROM billing");
$sum = $command->queryScalar();
echo $sum;
Within a model the sum could also be fetched with:
$this->find()->where(...)->sum('column');
You can directly use yii query concept in Search Model
$this->find()->from('billing')->where(['column'=>value])->sum('amount');
OR
$this->find()->where(['column'=>value])->sum('amount');
Outside the Model (Billing is model name)
Billing::find()->where(['column'=>value])->sum('amount');
i hope your model name is Billing
inside Billing model use
$this->find()->sum('amount');
in other models
Billing::find()->sum('amount');
Related
I’m using NestJS with TypeORM, database is MySQL, and I’d like to filter multiple parameters that can be passed in.
The frontend has a list of products and filters are applied as query params sent to NestJS, filtering works for a single param eg api.example.com?manufacturer=Acer but how would I filter an Array eg api.example.com?manufacturer=Acer,Toshiba,Asus.
I tried quite a few things in TypeORM, currently using the QueryBuilder to build the array with an if statement if the filter exists if so I’m doing something like a where statement.
.andWhere(manufacturer = filterOne, {filterOne: *manufacturers from the query param*})
But yeah just can’t hack something together, tried a couple of things, above is a rough example, did try methods that TypeORM had as an example on filtering arrays but it seemed like it was more for an array of integers only? Regardless, I’m open to any methods that allow for the end result of filtering the example I provided, cheers and thanks again!
You have to use IN to get all data where manufacturer equal the data came from the query, first, you have to convert the query to an array:
var manufacturerParam = filterOne.split(",");
then add it to your query:
.andWhere(manufacturer IN (:filter)", { filter: manufacturerParam })
I want to make a query that gets all the users from table "users" and i need to have only user.id.
$users = User::all();
this will get the whole model User but this is a real performance issue for my app. There is too much data going through.
I need to append some data for each user so i can calculate the working hours.
So the question is how to fetch all users without any other data except $user->id?
$name = DB::table('users')->select('id')->get();
For the certain columns, I think this is best:
$users = User::select('id')->get();
See Documentation.
Use the pluck() method:
$users = User::pluck('id');
The pluck method retrieves all of the values for a given key
Getting all User objects with id only:
$users = User::select('id')->get();
Getting all id as straight int value
According to documentation: https://laravel.com/docs/5.3/queries#selects
Specifying A Select Clause (most efficient)
$users = DB::table('users')->select('id')->get();
Retrieving A Single Column From A Row (but this will process all columns)
$name = DB::table('users')->where('name', 'John')->pluck('name');
I have models:
class Reference(models.Model):
name = models.CharField(max_length=50)
class Search(models.Model):
reference = models.ForeignKey(Reference)
update_time = models.DateTimeField(auto_now_add=True)
I have an instance of Reference and i need to get all last searches for the reference. Now i am doing it in this way:
record = Search.objects.filter(reference=reference)\
.aggregate(max_date=Max('update_time'))
if record:
update_time = record['max_date']
searches = reference.search_set.filter(update_time=self.update_time)
It is not a big deal to use 2 queries except the one but what if i need to get last searches for each reference on a page? I would have got 2x(count of references) queries and it would not be good.
I was trying to use this solution https://stackoverflow.com/a/9838438/293962 but it didn't work with filter by reference
You probably want to use the latest method.
From the docs, "Returns the latest object in the table, by date, using the field_name provided as the date field."
https://docs.djangoproject.com/en/1.8/ref/models/querysets/#latest
so your query would be
Search.objects.filter(reference=reference).latest('update_time')
I implemented a snippet from someone in gist but I don't remember the user neither have the link.
A bit of context:
I have a model named Medicion that contains the register of mensuration of a machine, machines are created in a model instance of Equipo, Medicion instances have besides of a Foreign key to Equipo, a foreign key to Odometro, this model serves as a kind of clock or metre, that's why when I want to retrieve data (measurements aka instances of Medicion model) for a certain machine, I need to indicate the clock as well, otherwise it would retrieve me a lot of messy and unreadable data.
Here is my implementation:
First I retrieve the last dates:
ult_fechas_reg = Medicion.objects.values('odometro').annotate(max_fecha=Max('fecha')).order_by()
Then I instance an Q object:
mega_statement = Q() # This works as 'AND' Sql Statement
Then looping in every date retrieved in the queryset(annotation) and establishing the Q statement:
for r in ult_fechas_reg:
mega_statement |= (Q(odometro__exact=r['odometro']) & Q(fecha=r['max_fecha']))
Finally passed this mega statement to the queryset that pursues to retrieve the last record of a model filtered by two fields:
resultados = Medicion.objects.filter(mega_query).filter(
equipo=equipo,
odometro__in=lista_odometros).order_by('odometro', 'fecha') # lista_odometros is a python list containing pks of another model, don't worry about it.
I am struggling with a basic problem. i am using cake php 2.5. i try to apply the find query in the company model and receiving all the data from companies and with its associations, but i only want to receive the data from company table and want to exclude the data from rest of relationships, can anyone help me with this. below are my queries.
$this->loadModel('Company');
$fields=array('id','name','logo','status');
$conditions=array('status'=>1);
$search_companies = $this->Company->find('first',
compact(array('conditions'=>$conditions,'fields'=>$fields)));
print_r($search_companies);die();
echo json_encode($search_companies);die();
With out seeing your data output, I am just going to take a stab at the problem.
Inside your $search_companies variable you are getting a multidimensional array probably with the other values of the other tables.
Why not just select the one array:
$wantedData = $search_companies['Company'];
// The key Company (which is the model) should be the data you are wanting.
Try setting model's recursive value to -1
$this->Company->recursive = -1;
$search_companies = $this->Company->find('first',
compact(array('conditions'=>$conditions,'fields'=>$fields)));
With this you will not fire the joins queries and therefore you only retrieve model's information.
Cakephp provide this functionality that we can unblind few/all associations on a any model. the keyword unbindModel is used for this purpose. inside the unblindModel you can define the association type and model(s) name that you want to unblind for that specific association.
$this->CurrentModelName->unbindModel(array('AssociationName' => array('ModelName_Youwwant_unblind')));
I have a datamodel, let's say: invoices (m:n) invoice_items
and currently I store the invoice total, calculated in PHP by totalling invoice_items, in a column in invoices. I don't like storing derived data as it paves the way for errors later.
How can I create a logical column in the invoices table in MySql? Is this something I would be better handling in the PHP (in this case CakePHP)?
There's something called Virtual Fields in CakePHP which allows you to achieve the same result from within your Model instead of relying on support from MySQL. Virtual Fields allow you to "mashup" various data within your model and provide that as an additional column in your record. It's cleaner than the other approaches here...(no afterFind() hacking).
Read more here: http://book.cakephp.org/view/1608/Virtual-fields
Leo,
One thing you could do is to modify the afterFind() method in your model. This would recalculate the total any time you retrieve an invoice (costing runtime processing), but would mean you're not storing it in the invoices table, which is apparently what you want to avoid (correct if I'm wrong).
Try this:
class Invoice extends AppModel {
// .. other stuff
function afterFind() {
parent::afterFind();
$total = 0;
foreach( $this->data['Invoice']['InvoiceItems'] as $item )
$total += ($item['cost'] * $item['quantity']);
$this->data['Invoice']['total'] = $total;
}
}
I may have messed up the arrays on the hasMany relationship (the foreach line), but I hope you get the jist of it. HTH,
Travis
Either you can return the derived one when you want it via
SELECT COUNT(1) as total FROM invoice_items
Or if invoices can be multiple,
//assuming that invoice_items.num is how many there are per row
SELECT SUM(num) as total FROM invoice_items
Or you can use a VIEW, if you have a certain way you want it represented all the time.
http://forge.mysql.com/wiki/MySQL_virtual_columns_preview
It's not implemented yet, but it should be implemented in mysql 6.0
Currently you could create a view.