Python 在Django框架中获取ValueError作为对http请求的响应

当我访问此url时：

http://localhost:8000/ProjectOverview/proj_view/15/

Django告诉我格式完整：

Request Method:     GET
Request URL:    http://localhost:8000/ProjectOverview/proj_view/15/
Django Version:     1.3.1
Exception Type:     ValueError
Exception Value:    incomplete format

Exception Location:     /home/projects/bruens_erp/ProjectOverview/views.py in proj_view, line 11

My URL.py：

from django.conf.urls.defaults import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^ProjectOverview/$', 'ProjectOverview.views.index'),
    url(r'^ProjectOverview/login/$', 'ProjectOverview.views.login'),
    url(r'^ProjectOverview/proj_view/(?P<cust>\d+)/$', 'ProjectOverview.views.proj_view'),
    url(r'^admin/doc/', include('django.contrib.admindocs.urls')),
    url(r'^admin/', include(admin.site.urls)),
)

---

这个代码有什么问题

您需要将%s变为%s（如果是字符串）或%d（如果是整数）：

---

看。

嗯，是的，刚刚看到输入错误。。该死的阅读障碍

# Create your views here.
from django.http import HttpResponse

def index(request):
    return HttpResponse("Index.... please login")

def login(request):
    return HttpResponse("login page")

def proj_view(request, cust):
    return HttpResponse("project overview for cust: %." % cust)

def proj_view(request, cust):
    return HttpResponse("project overview for cust: %s." % cust)