TypeError: view must be a callable or a list/tuple in the case of include()
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由于Django版本的原因，导致url.py文件里面的写法不同。
笔者所用的Django版本是1.10，但是按照教程编写url.py如下
from django.conf.urls import urlfrom django.contrib import adminurlpatterns = [ url(r'^admin/', admin.site.urls), url(r'^index/$',"blog.views.index"),]出现了错误信息：
TypeError: view must be a callable or a list/tuple in the case of include()这是由于上诉写法是Django旧版本的写法。
应该如下来配置url.py
from django.conf.urls import urlfrom django.contrib import adminfrom blog.views import indexurlpatterns = [ url(r'^admin/', admin.site.urls), url(r'^index/$',index),]此外笔者还碰到了修改后台支持中文的方法问题。
这里需要修改 LANGUAGE_CODE
#英文# LANGUAGE_CODE = 'en-us'#旧版本的中文写法# LANGUAGE_CODE = 'zh-cn'#Django1.10 简体中文# LANGUAGE_CODE = 'zh-Hans' #Django1.10 繁体中文# LANGUAGE_CODE = 'zh-Hant'参考资料：
1、http://blog.csdn.net/TH_NUM/article/details/52760667 2017.4.10
2、http://www.cnblogs.com/Eric-Young/p/6600461.html 2017.4.10
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- TypeError: view must be a callable or a list/tuple in the case of include()
- TypeError: view must be a callable or a list/tuple in the case of include().
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