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Python EmailTemplate.get_login_key_template方法代码示例

本文整理汇总了Python中evap.evaluation.models.EmailTemplate.get_login_key_template方法的典型用法代码示例。如果您正苦于以下问题:Python EmailTemplate.get_login_key_template方法的具体用法?Python EmailTemplate.get_login_key_template怎么用?Python EmailTemplate.get_login_key_template使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在evap.evaluation.models.EmailTemplate的用法示例。


在下文中一共展示了EmailTemplate.get_login_key_template方法的1个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: index

# 需要导入模块: from evap.evaluation.models import EmailTemplate [as 别名]
# 或者: from evap.evaluation.models.EmailTemplate import get_login_key_template [as 别名]
def index(request):
    """Main entry page into EvaP providing all the login options available. THe username/password
       login is thought to be used for internal users, e.g. by connecting to a LDAP directory.
       The login key mechanism is meant to be used to include external participants, e.g. visiting
       students or visiting contributors.
    """

    # parse the form data into the respective form
    submit_type = request.POST.get("submit_type", "no_submit")
    new_key_form = NewKeyForm(request.POST if submit_type == "new_key" else None)
    login_key_form = LoginKeyForm(request.POST if submit_type == "login_key" else None)
    login_username_form = LoginUsernameForm(request, request.POST if submit_type == "login_username" else None)

    # process form data
    if request.method == 'POST':
        if new_key_form.is_valid():
            # user wants a new login key
            profile = new_key_form.get_user()
            profile.generate_login_key()
            profile.save()

            EmailTemplate.get_login_key_template().send_to_user(new_key_form.get_user(), cc=False)

            messages.success(request, _(u"Successfully sent email with new login key."))
        elif login_key_form.is_valid():
            # user would like to login with a login key and passed key test
            auth_login(request, login_key_form.get_user())
        elif login_username_form.is_valid():
            # user would like to login with username and password and passed password test
            auth_login(request, login_username_form.get_user())

            # clean up our test cookie
            if request.session.test_cookie_worked():
                request.session.delete_test_cookie()

    # if not logged in by now, render form
    if not request.user.is_authenticated():
        # set test cookie to verify whether they work in the next step
        request.session.set_test_cookie()

        template_data = dict(new_key_form=new_key_form, login_key_form=login_key_form, login_username_form=login_username_form)
        return render(request, "index.html", template_data)
    else:
        user, created = UserProfile.objects.get_or_create(username=request.user.username)

        # check for redirect variable
        redirect_to = request.GET.get("next", None)
        if redirect_to is not None:
            if redirect_to.startswith("/staff/"):
                if request.user.is_staff:
                    return redirect(redirect_to)
            elif redirect_to.startswith("/contributor/"):
                if user.is_contributor:
                    return redirect(redirect_to)
            elif redirect_to.startswith("/student/"):
                if user.is_participant:
                    return redirect(redirect_to)
            else:
                return redirect(redirect_to)

        # redirect user to appropriate start page
        if request.user.is_staff:
            return redirect('evap.staff.views.index')
        elif user.is_editor_or_delegate:
            return redirect('evap.contributor.views.index')
        elif user.is_participant:
            return redirect('evap.student.views.index')
        else:
            return redirect('evap.results.views.index')
开发者ID:janno42,项目名称:EvaP,代码行数:71,代码来源:views.py


注:本文中的evap.evaluation.models.EmailTemplate.get_login_key_template方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。