如何在django的html文件中将上下文从视图传递到动作?

时间:2016-06-14 09:11:37

标签: python django python-3.x django-templates django-views

我从view.py文件传递上下文

def change_password(request,pk=None):
    user = MyUser.objects.get(pk=pk)
    if request.method == "POST":
        form = PasswordChangeForm(user=user, data=request.POST)
        if form.is_valid():
            form.save()
            update_session_auth_hash(request, form.user)  
            return render(request, "nurse/change_password.html", {"user_id":user.id})     // passing user_id as context

我可以在我的html文件中看到这个值

<div class="form-group">
    <label class="control-label allign col-sm-3" > {{user_id}} Old Password </label>    // user_id is giving the right value
</div>

但是当我尝试在行动中获取相同的内容而不是它不起作用时:

<form class="form-horizontal login-box2" action="{% url 'booking:change_password' user_id %}" role="form" method="post">   // not getting user_id here
    <div class="form-group">  
        <label class="control-label allign col-sm-3" > {{user_id}} Old Password </label>      // getting user_id here
    </div>
    <div class="form-group"> 
        <li> <button type="submit" class="btn btn-primary f-w-b f-s-18">Change</button> </li>
    </div>
</form>

谁能告诉我我做错了什么?

由于

EDITED

添加urls.py代码:

url(r'^change_password/(?P<pk>[0-9]+)/$',views.change_password,name='change_password'),

0 个答案:

没有答案