Python Django:更新或更改以前保存的模型数据

Python Django:更新或更改以前保存的模型数据,python,django,database,model,updates,Python,Django,Database,Model,Updates,我在forms.py中有此代码 ref_user = User.objects.get( username=form.cleaned_data['referrer'] ) user = User.objects.create_user( username=form.cleaned_data['username'] ) # Count the referrer's direct referrals ref_recruits = D

我在forms.py中有此代码

    ref_user = User.objects.get(
    username=form.cleaned_data['referrer']
    )

    user = User.objects.create_user(
    username=form.cleaned_data['username']
    )

    # Count the referrer's direct referrals
    ref_recruits = DirectReferral.objects.filter(referrer=ref_user).count()

    # Get newly created direct referral
    get_ref = DirectReferral.objects.get(name=user)

    #Check if ref_recruits is not divisible by 2 or paired
    if ref_recruits % 2 != 0:

        # Newly created direct referral is_paired is False
        get_ref.is_paired = False
        get_ref.save()

    else:

        # Newly created direct referral is_paired is True
        get_ref.is_paired = True
        get_ref.save()

        # But I also want to update the previous is_paired to True 
        # of the same referrer
我如何更新同一推荐人的上一个is_paired to True

检查下面的图片以更好地理解我的意思。 我希望你能理解

models.py(已请求)


您需要以某种方式对
directreference
对象进行排序,以便它们以正确的顺序返回,即最后添加的对象是
directreference
对象列表中的第一个对象,然后您可以轻松更改列表中的第二个对象,在您的情况下是前一个对象

例如:

get_refs = DirectReferral.objects.filter(referrer=ref_user).ordery_by('-timestamp')

if ref_recruits % 2 != 0:
    get_refs[0].is_paired = False
    get_refs[0].save()

else:
    get_refs[0].is_paired = True
    get_refs[1].is_paired = True
    get_refs[0].save()
    get_refs[1].save()

我认为你应该重构你的代码。根据您的实现,“DirectReference”的
name
字段将是唯一的

ref_user = User.objects.get(
username=form.cleaned_data['referrer']
)

user = User.objects.create_user(
username=form.cleaned_data['username']
)

# Count the referrer's direct referrals
ref_recruits = DirectReferral.objects.filter(referrer=ref_user).count()

# Get newly created direct referral
get_ref = DirectReferral.objects.get(name=user)

#Check if ref_recruits is not divisible by 2 or paired
if ref_recruits % 2 == 0:
    get_ref.is_paired = True
    DirectReferral.objects.filter(referrer=ref_user, is_paired=False).update(is_paired=True)
else:
    get_ref.is_paired = False
get_ref.save()

你能显示DirectReferral模型代码吗?@Paul我添加了models.py。我尝试了这个,但我不知道为什么没有保存或更新is\u paired。它只在我的模型数据中返回一个问号。是的,每个refererrer只有一个唯一的DirectReference。
ref_user = User.objects.get(
username=form.cleaned_data['referrer']
)

user = User.objects.create_user(
username=form.cleaned_data['username']
)

# Count the referrer's direct referrals
ref_recruits = DirectReferral.objects.filter(referrer=ref_user).count()

# Get newly created direct referral
get_ref = DirectReferral.objects.get(name=user)

#Check if ref_recruits is not divisible by 2 or paired
if ref_recruits % 2 == 0:
    get_ref.is_paired = True
    DirectReferral.objects.filter(referrer=ref_user, is_paired=False).update(is_paired=True)
else:
    get_ref.is_paired = False
get_ref.save()