查詢設定了太多物件，無法在djangotemplateView中解壓預期(2)

我寫了一個視圖，在點擊時在儀表板上顯示打開、完成、接受和關閉的票證，進入特定的 URL 以相應地顯示票證，并根據他們的狀態切換模板，我正在查詢票證狀態，我得到以下資訊錯誤太多要解包的物件（預期為 2）

模型.py

class Ticket(models.Model):
    ticket_title = models.CharField(max_length=200)
    ticket_description = models.TextField()
    created_by = models.ForeignKey(User,related_name = 'created_by',blank=True,null=True,on_delete=models.CASCADE)

    STATUS_CHOICES = (
        ('Opened','Opened'),
        ('Accepted','Accepted'),
        ('Completed','Completed'),
        ('Closed','Closed')
    )

    status = models.CharField('Status',choices=STATUS_CHOICES,max_length = 100,default = 'Opened')

    closed_date = models.DateTimeField(blank=True,null=True)
    completed_date = models.DateTimeField(blank=True,null=True)
    accepted_date = models.DateTimeField(blank=True,null=True)
    opened_date = models.DateTimeField(blank=True,null=True)
    

    accepted_by = models.ForeignKey(User,related_name='assigned_to',on_delete=models.CASCADE,blank=True,null=True)

    created = models.DateTimeField(auto_now_add=True)
    updated = models.DateTimeField(auto_now=True)


    def __str__(self):
        return self.ticket_title

視圖.py

class DeveloperTicketView(TemplateView):
  

def get_template_names(self):
    ticket_type = Ticket.objects.filter('status')
    if ticket_type == "Opened":
        template_name = 'app/open_tickets.html'
    elif ticket_type == 'Accepted':
        template_name = 'app/dev_accepted_ticket.html'
    elif ticket_type == 'Completed':
        template_name = 'app/dev_completed_tickets.html'
    else:
        template_name = 'app/dev_closed_tickets.html'
    return template_name


def get_context_data(self, **kwargs):
    context =  super(TemplateView,self).get_context_data(**kwargs)
    context['open_tickets'] = Ticket.objects.filter(status = 'Opened')
    context['accepted_tickets'] = Ticket.objects.filter(status = 'Accepted',accepted_by = self.request.user)
    context['completed_tickets'] = Ticket.objects.filter(status = 'Completed',accepted_by = self.request.user)
    context['closed_tickets'] = Ticket.objects.filter(status = 'Closed',accepted_by = self.request.user)
    return context

uj5u.com熱心網友回復：

錯誤是由于此查詢...

Ticket.objects.filter('status')

您應該指定如下內容：

if Ticket.objects.filter(status='Opened'):
    template_name = 'app/open_tickets.html')

另外請檢查邏輯。上面的查詢在 get_template_name 中沒有意義...您應該根據請求獲取一個模板。因此，如果請求是針對“已打開”的，那么您應該僅獲取“已打開”的模板。

假設如果您想獲取“已打開”，您需要在查詢中傳遞它并執行以下操作：

if self.request.GET.get('status', '') == 'Opened':
    template_name = 'app/open_tickets.html')

uj5u.com熱心網友回復：

def show_ticket(request,status):
    if status.title() in ['Opened','Accepted','Completed','Closed']:

        if status.title() == 'Opened':
            tickets = Ticket.objects.filter(status='Opened')
        else:
            tickets = Ticket.objects.filter(status=status.title(),accepted_by=request.user)
        templates = {'Opened':'app/open_tickets.html','Accepted':'app/dev_accepted_ticket.html','Completed':'app/dev_completed_tickets.html','Closed':'app/dev_closed_tickets.html'}
        return render(request,templates[status.title()],{'tickets':tickets})

在你的 urls.py 里面

path('tickets/<str:status>/',views.show_ticket,name='show_ticket')

標籤：姜戈 django 模型 django-views django-queryset 基于 Django 类的视图

上一篇：Django-Rest-Framework無法覆寫序列化程式唯一的error_messages

下一篇：我如何在django模型上制作一個對我有用的計數器