最多Django / pythonic的创建或覆盖记录的方式是什么？ [英] Whats the most Django/pythonic way to create or overwrite a record?

问题描述

使用Django 1.2我正在做一个葡萄酒评论网站。用户应该只能检查一次葡萄酒，但应该可以回去并重新检查葡萄酒，而不会出现错误。

Working with Django 1.2 I am making a wine review site. A user should only be able to review each wine once, but should be able to go back and re-review a wine without raising an error.

使用get_or_create方法似乎最合理的解决方案，但我一直在执行它的各种问题。搜索我发现这篇文章看起来很有希望：
正确的方法来使用get_or_create ？

Using the get_or_create method seems the most rational solution but I have been running into various problems implementing it. Searching I found this article which looked promising: Correct way to use get_or_create?

当然还有django文档：
http://docs.djangoproject.com/en/1.2/ref/models/querysets/#get-or-create

and of course the django documentation on it: http://docs.djangoproject.com/en/1.2/ref/models/querysets/#get-or-create

但似乎没有回答我的问题。这是我的代码：

But didn't seem to answer my question. Here is my code:

Views.py

@login_required
def wine_review_page(request, wine_id):
wine = get_object_or_404(Wine, pk=wine_id)

if request.method == 'POST':
form = WineReviewForm(request.POST)
if form.is_valid():
  review, created = Review.objects.get_or_create(
    user = request.user,
    wine = wine,
    like_dislike = form.cleaned_data['like_dislike'],
    ...
    )
variables = RequestContext(request, {
 'wine': wine
  })   
  review.save()
  return HttpResponseRedirect(
    '/detail/%s/' % wine_id
  )
else:
  form = WineReviewForm()
  variables = RequestContext(request, {
  'form': form,
  'wine': wine
 })
return render_to_response('wine_review_page.html', variables)

Models.py p>

Models.py

class Review(models.Model):
  wine = models.ForeignKey(Wine, unique=True)
  user = models.ForeignKey(User, unique=True)
  like_dislike = models.CharField(max_length=7, unique=True)
  ...

如果我明白如何正确使用get_or_create ，因为我不匹配所有的值，如__dislike等等，那么django会觉得它是独一无二的。我尝试删除其他表单参数，但是后来没有提交发布请求。

If I understand how to use get_or_create correctly, since I am not matching on all the values like_dislike, etc... then django perceives it to be unique. I tried removing the other form parameters, but then they are not submitted with the post request.

建议将不胜感激。

推荐答案

在制作基于CRUD的应用程序时，我也遇到了这个问题。我不知道是否有更好的方法，但是我最终的做法是使用 exists（） 来检查是否存在条目。

I came across this too when making a CRUD based app. I'm not sure if there's a better way but the way I ended up getting doing was using a exists() to check if an entry ... exists.

您可以在is_valid（）范围内使用get_or_create，但是在显示表单之前，您需要检查是否存在审阅，以便在审核已经存在的情况下将实例数据加载到表单中。

You can use get_or_create within the is_valid() scope, however, you need to check if the review exists before displaying your form in order to load instance data into the form in the case that the review already exists.

您的models.py可能如下所示：

Your models.py might look like this:

from django.db import models
from django.contrib.auth.models import User

class Wine(models.Model):
    name = models.CharField()

class Review(models.Model):
    wine = models.ForeignKey(Wine)
    user = models.ForeignKey(User)
    like = models.BooleanField(null=True, blank=True) # if null, unrated

您的forms.py可能如下所示：

Your forms.py might look like this:

from django import forms

class WineReviewForm(forms.ModelForm):
    class Meta:
        model = Review
        fields = ['like',] # excludes the user and wine field from the form

使用get_or_create可以让您如此使用：

Using get_or_create will let you do this if used like so:

@login_required
def wine_review_page(request, wine_id):
    wine = get_object_or_404(Wine, pk=wine_id)

    review, created = Review.objects.get_or_create(user=request.user, wine=wine)

    if request.method == 'POST':
        form = WineReviewForm(request.POST, instance=review)
        if form.is_valid():
            form.save()   
            return HttpResponseRedirect('/detail/%s/' % wine_id )
    else:
        form = WineReviewForm(instance=review)

    variables = RequestContext(request, {'form': form, 'wine': wine })
    return render_to_response('wine_review_page.html', variables) 

只需访问该页面即可创建评论，并要求其他信息默认或允许在模型级别为空。

Doing creates a review just by visiting the page and requires that the other information either have a default or are allowed to be blank at the model level.

使用 exists（），如果审核存在，您将得到两个db命中，但是除非用户提交有效的表单，否则您不创建对象：

With exists(), you get two db hits if the review exists, however you don't create an object unless the user submits a valid form:

@login_required
def wine_review_page(request, wine_id):
    wine = get_object_or_404(Wine, pk=wine_id)

    review = None
    if Review.objects.filter(user=request.user, wine=wine).exists():
        review = Review.objects.get(user=request.user, wine=wine)

    if request.method == 'POST':
        form = WineReviewForm(request.POST, instance=review)
        if form.is_valid():
            form.save()   
            return HttpResponseRedirect('/detail/%s/' % wine_id )
    else:
        form = WineReviewForm(instance=review)

    variables = RequestContext(request, {'form': form, 'wine': wine })
    return render_to_response('wine_review_page.html', variables)

我使用 exists（）我认为这可能会更好？

I used exists() but I think that this might be better?

try:
    review = Review.objects.get(user=request.user, wine=wine)
except Review.DoesNotExist:
    review = None

充满希望有更多经验的人将会进入。

Hopefully someone with more experience will chime in.

编辑：

以下是Daniel Roseman的 键博客 。我不知道它是否仍然适用，但可能与您的问题有关。

Here is a fairly old post from Daniel Roseman's Blog. I don't know if it is still applicable, but might be relevant to your question.

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