Django Admin app站點名和app站點名下model排序方法


一  使用環境

  開發系統: windows

  IDE: pycharm  

  數據庫: msyql,navicat

  編程語言: python3.7  (Windows x86-64 executable installer)

  虛擬環境: virtualenvwrapper

  開發框架: Django 2.2

  Django 2.2通病===>訪問admin出現問題:

  報錯:UnicodeDecodeError: 'gbk' codec can't decode byte 0xa6 in position 9737: illegal multibyte sequence

  解決方法:https://www.cnblogs.com/djtang/p/10194811.html

二 Django Admin 左側app站點名和app站點名下model排序方法

1.在 admin.py 最上方加入如下代碼

from django.contrib import admin
from django.apps import apps
from django.utils.text import capfirst
from django.utils.datastructures import OrderedDict


# app站點名和 app站點名下model排序方法,作用於全局admin(model按注冊順序排列顯示,app站點名按 main_menu_index = 0 后數字的大小排序
def find_app_index(app_label):
    app = apps.get_app_config(app_label)
    main_menu_index = getattr(app, 'main_menu_index', 9999)
    return main_menu_index


def find_model_index(name):
    count = 0
    for model, model_admin in admin.site._registry.items():
        if capfirst(model._meta.verbose_name_plural) == name:
            return count
        else:
            count += 1
    return count


def index_decorator(func):
    def inner(*args, **kwargs):
        templateresponse = func(*args, **kwargs)
        app_list = templateresponse.context_data['app_list']
        app_list.sort(key=lambda r: find_app_index(r['app_label']))
        for app in app_list:
            app['models'].sort(key=lambda x: find_model_index(x['name']))
        return templateresponse

    return inner


registry = OrderedDict()
registry.update(admin.site._registry)
admin.site._registry = registry
admin.site.index = index_decorator(admin.site.index)
admin.site.app_index = index_decorator(admin.site.app_index)

 

2.在要排序的 apps.py 中, 加入 main_menu_index = 3 , 數字越大排的越靠后.

from django.apps import AppConfig


class Px01Config(AppConfig):
    name = 'px01'
    verbose_name = '排序001'
    main_menu_index = 3  # 數字越大排的越靠下面

3. 請各位大老多多指點! ===> 個人微信:DJtang009


免責聲明!

本站轉載的文章為個人學習借鑒使用,本站對版權不負任何法律責任。如果侵犯了您的隱私權益,請聯系本站郵箱yoyou2525@163.com刪除。



 
粵ICP備18138465號   © 2018-2025 CODEPRJ.COM