Introduce a common template for object list views

This commit is contained in:
Jeremy Stretch
2020-02-13 13:13:27 -05:00
parent 35498c17d7
commit c5f74cce80
3 changed files with 51 additions and 6 deletions

View File

@@ -1,9 +1,10 @@
import datetime
import json
import re
import yaml
import yaml
from django import template
from django.urls import NoReverseMatch, reverse
from django.utils.html import strip_tags
from django.utils.safestring import mark_safe
from markdown import markdown
@@ -11,7 +12,6 @@ from markdown import markdown
from utilities.choices import unpack_grouped_choices
from utilities.utils import foreground_color
register = template.Library()
@@ -101,6 +101,21 @@ def model_name_plural(obj):
return obj._meta.verbose_name_plural
@register.filter()
def url_name(model, action):
"""
Return the URL name for the given model and action, or None if invalid.
"""
url_name = '{}:{}_{}'.format(model._meta.app_label, model._meta.model_name, action)
try:
# Validate and return the URL name. We don't return the actual URL yet because many of the templates
# are written to pass a name to {% url %}.
reverse(url_name)
return url_name
except NoReverseMatch:
return None
@register.filter()
def contains(value, arg):
"""

View File

@@ -71,7 +71,7 @@ class ObjectListView(View):
filterset = None
filterset_form = None
table = None
template_name = None
template_name = 'utilities/obj_list.html'
def queryset_to_yaml(self):
"""
@@ -156,9 +156,11 @@ class ObjectListView(View):
# Provide a hook to tweak the queryset based on the request immediately prior to rendering the object list
self.queryset = self.alter_queryset(request)
# Compile user model permissions for access from within the template
perm_base_name = '{}.{{}}_{}'.format(model._meta.app_label, model._meta.model_name)
permissions = {p: request.user.has_perm(perm_base_name.format(p)) for p in ['add', 'change', 'delete']}
# Compile a dictionary indicating which permissions are available to the current user for this model
permissions = {}
for action in ('add', 'change', 'delete', 'view'):
perm_name = '{}.{}_{}'.format(model._meta.app_label, action, model._meta.model_name)
permissions[action] = request.user.has_perm(perm_name)
# Construct the table based on the user's permissions
table = self.table(self.queryset)