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Django ORM查询:如何交换属性值?

[英]Django ORM query: how to swap value of a attribute?

Here is the case. 就是这种情况。 I want to swap value of user. 我想交换用户的价值。

user_with_apple = User.objects.filter(fruit=apple)
user_with_pear = User.objects.filter(fruit=pear)

user_with_apple.update(fruit=pear)
user_with_pear.update(fruit=apple)

Well, this does not work, as everything ends up with pear. 嗯,这行不通,因为一切都以梨结尾。 Because django ORM is lazy, I suppose? 因为django ORM很懒,我想?

So may I ask how can I perform such a "swap" operation? 因此,请问如何执行这种“交换”操作? Do I have to use a intermediate value? 我必须使用中间值吗?

Thank you so much! 非常感谢!

Or does it make sense if I force the evaluation of the query by slicing or calling list() on it? 还是通过切片或调用list()来强制对查询求值,这有意义吗? eg. 例如。 user_with_apple = User.objects.filter(fruit=apple)[:] user_with_pear = User.objects.filter(fruit=pear)[:] user_with_apple = User.objects.filter(fruit = apple)[:] user_with_pear = User.objects.filter(fruit = pear)[:]

Variant 1 变体1

I suppose you have CharField for fruit . 我想你有CharFieldfruit Then add unknown option to choice. 然后将unknown选项添加到选择中。

FRUIT_CHOICES = (
    ...
    (unknown, 'Unknown'),
    (apple, 'Apple'),
    (pear, 'Pear'),
    ...
)

class User(models.Model):
    ...
    fruit = models.CharField(..., choices=FRUIT_CHOICES)
    ...

Then try this. 然后试试这个。

User.objects.filter(fruit=apple).update(fruit=unknown)
User.objects.filter(fruit=pear).update(fruit=apple)
User.objects.filter(fruit=unknown).update(fruit=pear)

Variant 2 变体2

Add access null values for fruit field. fruit字段添加访问空值。

FRUIT_CHOICES = (
    ...
    (apple, 'Apple'),
    (pear, 'Pear'),
    ...
)

class User(models.Model):
    ...
    fruit = models.CharField(..., blank=True, null=True, choices=FRUIT_CHOICES)
    ...

User.objects.filter(fruit=apple).update(fruit=None)
User.objects.filter(fruit=pear).update(fruit=apple)
User.objects.filter(fruit=None).update(fruit=pear)

Here is what that comes to mind: 这是我想到的:

If you call list on your QuerySets, you force Django to give you the objects. 如果在QuerySet上调用list ,则强制Django给您对象。 The problem with this is that you can't call update on a list, so you need to iterate over the lists, and update each object. 这样做的问题是您无法在列表上调用update ,因此需要遍历列表并更新每个对象。

This may help you on your way: 这可能会帮助您:

users_with_apple = list(User.objects.filter(fruit=apple))
users_with_pear = list(User.objects.filter(fruit=pear))

for user in users_with_apple:
    user.fruit = pear
    user.save()
for user in users_with_pear:
    user.fruit = apple
    user.save()

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