[Django 07] API 연습

대화식 Python 쉘을 통해 Django API를 사용해볼 시간

$ python manage.py shell

Python 3.8.3 (default, May 18 2020, 11:25:01) 
[GCC 7.5.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
(InteractiveConsole)
>>> from polls.models import Choice, Question
>>> Question.objects.all()
<QuerySet []>
>>> from django.utils import timezone
>>> q = Question(question_text="What's new?", pub_date=timezone.now())
>>> q
<Question: Question object (None)>
>>> q.save()
>>> q.id
1
>>> q.question_text
"What's new?"
>>> q.pub_date
datetime.datetime(2021, 4, 13, 12, 48, 15, 294472, tzinfo=<UTC>)
>>> q.question_text = "What's up?"
>>> q.save()
>>> Question.objects.all()
<QuerySet [<Question: Question object (1)>]>
>>> 

Question object (1)은 객체 표현에 도움이 되지 않는다.
polls/models.py 파일의 Question 모델을 수정하여, __str__() 메소드를 Question과 Choice에 추가해보자

class Question(models.Model):    
    question_text = models.CharField(max_length=200)
    pub_date = models.DateTimeField('date published')
    
    def __str__(self):
        return self.question_text

__str__() 메소드를 추가하는 것은 객체의 표현을 대화식 프롬프트에서 편하게 보려는 이유 말고도, Django가 자동으로 생성하는 관리 사이트에서도 객체의 표현이 사용하기 때문이다.

>>> from polls.models import Choice, Question

# Make sure our __str__() addition worked.
>>> Question.objects.all()
<QuerySet [<Question: What's up?>]>

# Django provides a rich database lookup API that's entirely driven by
# keyword arguments.
>>> Question.objects.filter(id=1)
<QuerySet [<Question: What's up?>]>
>>> Question.objects.filter(question_text__startswith='What')
<QuerySet [<Question: What's up?>]>

# Get the question that was published this year.
>>> from django.utils import timezone
>>> current_year = timezone.now().year
>>> Question.objects.get(pub_date__year=current_year)
<Question: What's up?>

# Request an ID that doesn't exist, this will raise an exception.
>>> Question.objects.get(id=2)
Traceback (most recent call last):
    ...
DoesNotExist: Question matching query does not exist.

# Lookup by a primary key is the most common case, so Django provides a
# shortcut for primary-key exact lookups.
# The following is identical to Question.objects.get(id=1).
>>> Question.objects.get(pk=1)
<Question: What's up?>

# Make sure our custom method worked.
>>> q = Question.objects.get(pk=1)
>>> q.was_published_recently()
True

# Give the Question a couple of Choices. The create call constructs a new
# Choice object, does the INSERT statement, adds the choice to the set
# of available choices and returns the new Choice object. Django creates
# a set to hold the "other side" of a ForeignKey relation
# (e.g. a question's choice) which can be accessed via the API.
>>> q = Question.objects.get(pk=1)

# Display any choices from the related object set -- none so far.
>>> q.choice_set.all()
<QuerySet []>

# Create three choices.
>>> q.choice_set.create(choice_text='Not much', votes=0)
<Choice: Not much>
>>> q.choice_set.create(choice_text='The sky', votes=0)
<Choice: The sky>
>>> c = q.choice_set.create(choice_text='Just hacking again', votes=0)

# Choice objects have API access to their related Question objects.
>>> c.question
<Question: What's up?>

# And vice versa: Question objects get access to Choice objects.
>>> q.choice_set.all()
<QuerySet [<Choice: Not much>, <Choice: The sky>, <Choice: Just hacking again>]>
>>> q.choice_set.count()
3

# The API automatically follows relationships as far as you need.
# Use double underscores to separate relationships.
# This works as many levels deep as you want; there's no limit.
# Find all Choices for any question whose pub_date is in this year
# (reusing the 'current_year' variable we created above).
>>> Choice.objects.filter(question__pub_date__year=current_year)
<QuerySet [<Choice: Not much>, <Choice: The sky>, <Choice: Just hacking again>]>

# Let's delete one of the choices. Use delete() for that.
>>> c = q.choice_set.filter(choice_text__startswith='Just hacking')
>>> c.delete()

모델의 관계에 대한 더 많은 정보는 관련 객체에 접근하기를 참조하세요. API에서 이중 밑줄(__) 을 이용해서 어떻게 필드를 조회할 수 있는지는 필드 조회를 읽어보세요.데이터베이스 API에 대한 자세한 내용을 보려면, 데이터베이스 API 레퍼런스를 읽어보세요.