DjangoCMS插件可从挂接模型中列出特定人员

如何解决DjangoCMS插件可从挂接模型中列出特定人员

我是DjangoCMS的新手。因此，如果这个问题太琐碎，请忍受。

我制作了一个应用钩子，用于为 model.py

from django.db import models
from filer.fields.image import FilerImageField
from djangocms_text_ckeditor.fields import HTMLField
from django.urls import reverse
from cms.models.fields import PlaceholderField

# Create your models here.

class Designations(models.Model):
    class Meta:
        app_label = 'voxstaff'
        verbose_name_plural = 'designations'
    desingation = models.CharField(
        blank=False,help_text="Please provide a label for the Designation",unique=True,max_length=100
    )

    def __str__(self):
        return self.desingation

class Staffs(models.Model):

    class Meta:
        app_label = 'voxstaff'

    full_name = models.CharField(
        blank=False,help_text="Please enter the full name of the staff",max_length=100
    )

    slug = models.SlugField(
        blank=False,default='',help_text='Provide a unique slug for this staff member',max_length=100,)

    desingation = models.ForeignKey(
        Designations,on_delete=models.SET_NULL,blank=True,null=True
    )

    photo = FilerImageField(
        blank=True,null=True,)
    staff_intro = HTMLField(blank=True)
    
    

    bio = PlaceholderField("staff_bio")

    is_featured = models.BooleanField()

    def get_absolute_url(self):
        return reverse("voxstaff:staffs_detail",kwargs={"slug": self.slug})
        

    def __str__(self):
        return self.full_name

class LinkTypes(models.Model):
    link_type = models.CharField(max_length=100)
    
    def __str__(self):
        return self.link_type


class Links(models.Model):
    staff = models.ForeignKey(Staffs,on_delete=models.CASCADE)
    link_type = models.ForeignKey(LinkTypes,blank=True)
    link_url = models.CharField(max_length=200)

    def __str__(self):
        return self.link_type.link_type

和 cms_apps.py 如下

from cms.app_base import CMSApp
from cms.apphook_pool import apphook_pool
from django.utils.translation import ugettext_lazy as _

from .menu import StaffSubMenu

@apphook_pool.register
class StaffApp(CMSApp):
    name = _('VOXStaff')
    # urls = ['voxstaff.urls',]
    app_name = 'voxstaff'
    menus = [StaffSubMenu,]
    def get_urls(self,page=None,language=None,**kwargs):
        return ["voxstaff.urls"]

现在，我需要在主页上显示人员的插件。在职员模型中，这些职员应具有 is_featured 为是。

如何进行相同的操作。请帮助，我被困住了。

解决方法

我可以在下面的帖子中满足我的要求

django cms plugin that display model with specific value checked

这个问题可以解决。