I have such model and query
class Employer(Models.model)
name = ...
class JobTitle(Models.model)
name = ...
employer = models.ForeignKey(Employer)
and query is
Employer.objects.select_related('jobtitle')
.filter(jtt__activatedate__range=[startdate,enddate])
.annotate(jtt_count=Count('jobtitle'))
.order_by('-jtt_count')[:5]
As you see it returns 5 employer list which has maximum number of jobtitles which are related to that employer and whose activation date is in some certain range.
However, I also want to get the total number of jobtitles of each employer in that query.
Of course I may loop over each employer and make such query JobTitle.objects.filter(employer = emp) and taking length of that query but it is bad solution.
How can I achive this in that query?
Although it may not be possible to get both total number and filtered number of job titles, I may get the jobttiles of each emplyoer such that len(emp.jobtitle) however it also didn't work.
Thanks
Try the extra lookup. So, in your case it may be like this:
.extra(
select={
'jobtitle_count': 'SELECT COUNT(*) FROM YOURAPP_jobtitle WHERE YOURAPP_jobtitle.employer_id = YOURAPP_employer.id'
},
)
Related
Been searching the web for a couple hours now looking for a solution but nothing quite fits what I am looking for.
I have one model (simplified):
class SimpleModel(Model):
name = CharField('Name', unique=True)
date = DateField()
amount = FloatField()
I have two dates; date_one and date_two.
I would like a single queryset with a row for each name in the Model, with each row showing:
{'name': name, 'date_one': date_one, 'date_two': date_two, 'amount_one': amount_one, 'amount_two': amount_two, 'change': amount_two - amount_one}
Reason being I would like to be able to find the rank of amount_one, amount_two, and change, using sort or filters on that single queryset.
I know I could create a list of dictionaries from two separate querysets then sort on that and get the ranks from the index values ...
but perhaps nievely I feel like there should be a DB solution using one queryset that would be faster.
union seemed promising but you cannot perform some simple operations like filter after that
I think I could perhaps split name into its own Model and generate queryset with related fields, but I'd prefer not to change the schema at this stage. Also, I only have access to sqlite.
appreciate any help!
Your current model forces you to have ONE name associated with ONE date and ONE amount. Because name is unique=True, you literally cannot have two dates associated with the same name
So if you want to be able to have several dates/amounts associated with a name, there are several ways to proceed
Idea 1: If there will only be 2 dates and 2 amounts, simply add a second date field and a second amount field
Idea 2: If there can be an infinite number of days and amounts, you'll have to change your model to reflect it, by having :
A model for your names
A model for your days and amounts, with a foreign key to your names
Idea 3: You could keep the same model and simply remove the unique constraint, but that's a recipe for mistakes
Based on your choice, you'll then have several ways of querying what you need. It depends on your final model structure. The best way to go would be to create custom model methods that query the 2 dates/amount, format an array and return it
I have this model:
class User_Data(AbstractUser):
date_of_birth = models.DateField(null=True,blank=True)
city = models.CharField(max_length=255,default='',null=True,blank=True)
address = models.TextField(default='',null=True,blank=True)
gender = models.TextField(default='',null=True,blank=True)
And I need to run a django query to get the count of each age. Something like this:
Age || Count
10 || 100
11 || 50
and so on.....
Here is what I did with lambda:
usersAge = map(lambda x: calculate_age(x[0]), User_Data.objects.values_list('date_of_birth'))
users_age_data_source = [[x, usersAge.count(x)] for x in set(usersAge)]
users_age_data_source = sorted(users_age_data_source, key=itemgetter(0))
There's a few ways of doing this. I've had to do something very similar recently. This example works in Postgres.
Note: I've written the following code the way I have so that syntactically it works, and so that I can write between each step. But you can chain these together if you desire.
First we need to annotate the queryset to obtain the 'age' parameter. Since it's not stored as an integer, and can change daily, we can calculate it from the date of birth field by using the database's 'current_date' function:
ud = User_Data.objects.annotate(
age=RawSQL("""(DATE_PART('year', current_date) - DATE_PART('year', "app_userdata"."date_of_birth"))::integer""", []),
)
Note: you'll need to change the "app_userdata" part to match up with the table of your model. You can pick this out of the model's _meta, but this just depends if you want to make this portable or not. If you do, use a string .format() to replace it with what the model's _meta provides. If you don't care about that, just put the table name in there.
Now we pick the 'age' value out so that we get a ValuesQuerySet with just this field
ud = ud.values('age')
And then annotate THAT queryset with a count of age
ud = ud.annotate(
count=Count('age'),
)
At this point we have a ValuesQuerySet that has both 'age' and 'count' as fields. Order it so it comes out in a sensible way..
ud = ud.order_by('age')
And there you have it.
You must build up the queryset in this order otherwise you'll get some interesting results. i.e; you can't group all the annotates together, because the second one for count depends on the first, and as a kwargs dict has no notion of what order the kwargs were defined in, when the queryset does field/dependency checking, it will fail.
Hope this helps.
If you aren't using Postgres, the only thing you'll need to change is the RawSQL annotation to match whatever database engine it is that you're using. However that engine can get the year of a date, either from a field or from its built in "current date" function..providing you can get that out as an integer, it will work exactly the same way.
I have a table which has a list of invoices and their details:
class Invoice(models.Model):
invoiceNum = models.CharField(etc...)
invoiceDate = models.DateField(etc...)
customerID = models.ForeignKey(etc...)
isPaid = models.CharField(etc...)
The Invoice records do not hold the actual invoice total. Instead, an invoice's total is made up of several Invoice_Line_Item records, held in another table:
class Invoice_Line_Item(models.Model):
invNum = models.ForeignKey(Invoice)
itemName = models.CharField(etc...)
itemPrice = models.DecimalField(etc...)
I have a webapp that shows all the invoices in a big HTML table, along with all the details of that invoice on the table's tr row. Details such as, Invoice Date, Invoice Number, Customer ID, all come from that Invoice table. There are hundreds of invoices to display in this HTML table.
What I would like to do is also show each invoice's total value - which is the sum of all the line items. However, I can't think of a simple way to acomplish this since the invoice details and the line items that make up the invoice's total are in two different tables.
One way I thought is to pass the entire Invoice_Line_Item querySet into the HTML template, then for each invoice displayed in a table tr, I could iterate over the entire Invoice_Line_Item querySet, adding up all the line items that match the current invoice. This, however, seems hugely inefficient.
Any better ideas on how to do this?
Thanks!
One word: Aggregation
Invoice_Line_Item.objects.filter(invNum=invoice).aggregate(Sum('itemPrice'))
https://docs.djangoproject.com/en/dev/topics/db/aggregation/
Another way is to store the total in Invoice and update it whenever you change a related Invoice_Line_Item
One more word: annotate.
from django.models import Sum
Invoice.objects.filter( .. ).annotate( InvTotal=Sum( 'invoice_line_number__itemPrice' ) )
InvTolal becomes a new attribute of Invoice object, you can use it in template the same way as invoiceNum or invoiceDate.
With this approach you do not have to pass any additional data structures to your template, only a QuerySet of Invoices.
Please note:
Argument of Sum is a string, which is a concatenation of the name of related model converted to lowercase, than double '_', and than the name of a field in related model.
I have two models such that
class Employer(models.Model):
code = models.CharField(null=False,blank=False,default="")
class JobTitle(models.Model):
employer = models.ForeignKey(Employer,unique=False,null=False,default=0)
name = models.CharField(max_length=1000,null=False,default="")
and I would like to get all employers whose jobtitle name is less than X chracters. How can I achive this in Django ?
Thanks
The correct code for this is
Employer.objects.filter(jobtitle__name__regex="^.{0,20}$")
This will select all the employers who have a job title name up to and including 20 characters long. Just replace the 20 with whatever number you need.
Note that if an Employer has multiple JobTitles whose name are less than 20 characters long, it will return that Employer in the list multiple times. If you don't want this to happen, you should add distinct() to the query as follows:
Employer.objects.filter(jobtitle__name__regex="^.{0,20}$").distinct()
You'll now only get the Employer back once, even if they have multiple short JobTitles.
try
Employer.objects.filter(jobtitle__name__regex="^\W{0, X}$")
When you use regex, you can filter records from database with provided regular expression. In this case all records with name which contains 0 to X signs will be returned
I have two models such that
class Employer(models.Model):
name = models.CharField(max_length=1000,null=False,blank=False)
eminence = models.IntegerField(null=False,default=4)
class JobTitle(models.Model):
name = models.CharField(max_length=1000,null=False,blank=False)
employer= models.ForeignKey(JobTitle,unique=False,null=False)
class People(models.Model):
name = models.CharField(max_length=1000,null=False,blank=False)
jobtitle = models.ForeignKey(JobTitle,unique=False,null=False)
I would like to list random 5 employers and one job title for each employer. However, job title should be picked up from first 10 jobtitles of the employer whose number of people is maximum.
One approach could be
employers = Employer.objects.filter(isActive=True).filter(eminence__lt=4 ).order_by('?')[:5]
for emp in employers:
jobtitle = JobTitle.objects.filter(employer=emp)... and so on.
However, loop through selected employers may be ineffiecent. Is there any way to do it in one query ?
Thanks
There is! Check out: https://docs.djangoproject.com/en/dev/ref/models/querysets/#select-related
select_related() tells Django to follow all the foreign key relationships using JOINs. This will result in one large query as opposed to many small queries, which in most cases is what you want. The QuerySet you get will be pre-populated and Django won't have to lazy-load anything from the database.
I've used select_related() in the past to solve almost this exact problem.
I have written such code block and it works. Although I loop over employers because I have used select_related('jobtitle'), I consider it doesn't hit database and works faster.
employers = random.sample(Employer.objects.select_related('jobtitle').filter(eminence__lt=4,status=EmployerStatus.ACTIVE).annotate(jtt_count=Count('jobtitle')).filter(jtt_count__gt=0),3)
jtList = []
for emp in employers:
jt = random.choice(emp.jobtitle_set.filter(isActive=True).annotate(people_count=Count('people')).filter(people_count__gt=0)[:10])
jtList.append(jt)