无法将表单数据发布到操作网址
[英]Not able to post form data to action url
问题描述
I have a login form. 
我有一个登录表格。 After pressing the login button the the post data is sent to the view login_auth that authenticates the user data and redirects accordingly. 按下登录按钮后，帖子数据将发送到视图login_auth，该视图对用户数据进行身份验证并相应地重定向。 However,after pressing the login button, I am not being redirected to the appropriate page. 但是，按登录按钮后，我没有被重定向到相应的页面。
views.py views.py
def login_successful(request):
return render(request,"login_successful.html")
def login_invalid(request):
return render(request,"login_invalid.html")
def login(request):
return render(request,'login.html',c)
def loginauth(request):
username=request.POST.get("username",'')
password=request.POST.get("password",'')
user=auth.authenticate(username=username,password=password)
if user is not none:
user.login(request.user)
return redirect(login_successful)
else:
return redirect(login_invalid)
urls.py urls.py
urlpatterns = [
url(r'^registration/',views.registration),
url(r'^registration_successful/',views.registration_successful),
url(r'^home/',views.home),
url(r'^login/',views.login),
url(r'^login_successful/',views.login_successful),
url(r'^login_invalid/',views.login_invalid),
url(r'^login/auth',views.loginauth)
]
login.html login.html
<html>
<form action="/login/auth" method="POST">{% csrf_token %}
Username :<input type="textbox" name="username" >
Password :<input type="password" name="password">
<input type="submit" value="Login">
</form>
</html>
2 个解决方案
解决方案1
0 已采纳 2016-06-02 09:01:26
Your login url pattern is missing a trailing $ . 您的登录网址格式缺少尾随的$ 。 It should be: 它应该是：
url(r'^login/$', views.login),
Without the dollar, the /login/auth is matched by r'^login/ , so the request is handled by your login view. 如果没有美元，则/login/auth与r'^login/匹配，因此请求由您的登录视图处理。
It's a bit unusual to process the form on a different url. 在不同的URL上处理表单有点不寻常。 Django comes with authentication views , including a login view. Django带有身份验证视图 ，包括登录视图。 I would recommend using this rather than writing your own. 我建议使用此方法，而不要自己编写。
解决方案2
0 2016-06-02 09:17:52
Use name for url 使用名称作为网址
views.py views.py
def login_successful(request):
return render(request,"login_successful.html")
def login_invalid(request):
return render(request,"login_invalid.html")
def login(request):
return render(request,'login.html',c)
def loginauth(request):
username=request.POST.get("username",'')
password=request.POST.get("password",'')
user=auth.authenticate(username=username,password=password)
if user is not none:
user.login(request.user)
return redirect('login_successful')
else:
return redirect('login_invalid')
urls.py urls.py
urlpatterns = [
url(r'^registration/',views.registration),
url(r'^registration_successful/',views.registration_successful),
url(r'^home/',views.home),
url(r'^login/$',views.login),
url(r'^login_successful/',views.login_successful, name='login_successful'),
url(r'^login_invalid/',views.login_invalid, name='login_invalid'),
url(r'^login/auth',views.loginauth)
]
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