小编Ala*_*air的帖子

可能是下面的问题？看起来pip中有一个bug.我昨天使用brew安装了pip.在此之前,我已经安装了大多数python包$ python setup.py install

steves-MacBook-Pro:server steve$ pip -V
pip 1.4.1 from /Library/Python/2.7/site-packages/pip-1.4.1-py2.7.egg (python 2.7)
steves-MacBook-Pro:server steve$ pip list
altgraph (0.9)
bdist-mpkg (0.4.4)
... 
...(a bunch of python packages omitted here for brevity)
...
...
requests (2.0.0)
Exception:
Traceback (most recent call last):
  File "/Library/Python/2.7/site-packages/pip-1.4.1-py2.7.egg/pip/basecommand.py", line 134, in main
    status = self.run(options, args)
  File "/Library/Python/2.7/site-packages/pip-1.4.1-py2.7.egg/pip/commands/list.py", line 80, in run
    self.run_listing(options)
  File "/Library/Python/2.7/site-packages/pip-1.4.1-py2.7.egg/pip/commands/list.py", line 127, in run_listing
    self.output_package_listing(installed_packages)
  File "/Library/Python/2.7/site-packages/pip-1.4.1-py2.7.egg/pip/commands/list.py", line 136, in output_package_listing
    if dist_is_editable(dist):
  File "/Library/Python/2.7/site-packages/pip-1.4.1-py2.7.egg/pip/util.py", line 347, in …

我在限制formset中的可选选项时遇到了问题.我有以下模型:员工,部门,项目,项目类型,成员资格和角色.员工可以在表单集中添加/删除他们为给定部门项目播放的角色,表单应该将可选项目限制为仅属于员工所属部门的项目.

楷模:

class Department(models.Model):
    name = models.CharField(max_length=20)
    def __unicode__(self):
    return self.name

class Employee(models.Model):
    fname = models.CharField(max_length=15)
    department = models.ForeignKey(Department)
    def __unicode__(self):
        return self.fname

class Projecttype(models.Model):
    name = models.CharField(max_length=20)
    def __unicode__(self):
        return self.name

class Project(models.Model):
    projecttype = models.ForeignKey(Projecttype)
    department = models.ForeignKey(Department)
    members = models.ManyToManyField(Employee, through='Membership')
    def __unicode__(self):
       return "%s > %s" % (self.department, self.projecttype)

class Role(models.Model):
    name = models.CharField(max_length=20)
    def __unicode__(self):
       return self.name

class Membership(models.Model):
    project = models.ForeignKey(Project, null=True)
    department = models.ForeignKey(Department)
    employee = models.ForeignKey(Employee)
    role = models.ManyToManyField(Role, blank=True, null=True)
    class …

我有一个自定义表单,创建一个字段的隐藏输入:

class MPForm( forms.ModelForm ):
    def __init__( self, *args, **kwargs ):
        super(MPForm, self).__init__( *args, **kwargs )
        self.fields['mp_e'].label = "" #the trick :)

class Meta:
    model = MeasurementPoint
    widgets = { 'mp_e': forms.HiddenInput()  }
    exclude = ('mp_order') 

我必须这个小技巧来"隐藏"标签,但我想要做的是将其从表单中删除.我创建这样的表单:

forms.MPForm()

我切换到Django 1.7.当我为我的应用程序尝试makemigrations时,它会崩溃.崩溃报告是:

Migrations for 'roadmaps':
  0001_initial.py:
    - Create model DataQualityIssue
    - Create model MonthlyChange
    - Create model Product
    - Create model ProductGroup
    - Create model RecomendedStack
    - Create model RecomendedStackMembership
    - Create model RoadmapMarket
    - Create model RoadmapUser
    - Create model RoadmapVendor
    - Create model SpecialEvent
    - Create model TimelineEvent
    - Create model UserStack
    - Create model UserStackMembership
    - Add field products to userstack
    - Add field viewers to userstack
    - Add field products to recomendedstack
    - Add field product_group …

我正在尝试使用argparse创建一个Django管理命令,但是每当我运行它时,它总是不会返回有效的选项,因为此消息来自manage.py:

class Command(BaseCommand):
    def handle(self, *args, **options):
        parser = argparse.ArgumentParser('Parsing arguments')
        parser.add_argument('--max', type=float, store)
        args = parser.parse_args(sys.argv[2:])

将一些参数解析器与管理命令一起使用的正确方法是什么？

Python版本2.x.

我试图在允许用户查看特定用户设置页面之前检查某些条件.我试图使用user_passes_test装饰器来实现这一点.该函数位于基于类的视图中,如下所示.我正在使用方法装饰器来装饰视图中的get_initial函数.

class UserSettingsView(LoginRequiredMixin, FormView):
    success_url = '.'
    template_name = 'accts/usersettings.html'

    def get_form_class(self):
        if self.request.user.profile.is_student:
            return form1
        if self.request.user.profile.is_teacher:
            return form2
        if self.request.user.profile.is_parent:
            return form3

    @method_decorator(user_passes_test(test_settings, login_url='/accounts/usertype/'))
    def get_initial(self):
        if self.request.user.is_authenticated():
            user_obj = get_user_model().objects.get(email=self.request.user.email)
            if user_obj.profile.is_student:
                return { ..........
       ...... .... 

以下是test_settings函数:

def test_settings(user):
    print "I am in test settings"
    if not (user.profile.is_student or user.profile.is_parent or user.profile.is_teacher):
        return False
    else:
        return True

我与装饰器得到以下错误.

File "../django/core/handlers/base.py", line 111, in get_response
  response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "../django/views/generic/base.py", line 69, in view
  return …

我将Django从1.8升级到1.9.之后,在Django管理员登录后,我在本地主机上收到此错误:

Referer checking failed - Referer is insecure while host is secure.

生产中的一切都很好.以下是我的settings.py文件的片段:

SECURE_PROXY_SSL_HEADER = ('HTTP_X_FORWARDED_PROTO', 'https')
SESSION_COOKIE_SECURE = True
CSRF_COOKIE_SECURE = True

我最近更新到Django 1.9并尝试更新我的一些模型字段以使用内置的JSONField(我正在使用PostgreSQL 9.4.5).当我试图创建和更新我的对象的字段时,我遇到了一些特殊的东西.这是我的模型:

class Activity(models.Model):
    activity_id = models.CharField(max_length=MAX_URL_LENGTH, db_index=True, unique=True)
    my_data = JSONField(default=dict())

这是我正在做的一个例子:

>>> from proj import models
>>> test, created = models.Activity.objects.get_or_create(activity_id="foo")
>>> created
True
>>> test.my_data['id'] = "foo"
>>> test.save()
>>> test
<Activity: {"id": "foo"}>
>>> test2, created2 = models.Activity.objects.get_or_create(activity_id="bar")
>>> created2
True
>>> test2
<Activity: {"id": "foo"}>
>>> test2.activity_id
'bar'
>>> test.activity_id
'foo'

似乎每当我更新任何字段时my_data,我创建的下一个对象都会预先填充my_data来自前一个对象的数据.出现这种情况我是否使用get_or_create或只create.有人可以向我解释发生了什么吗？

我在网站上使用Django联系表格,允许访问者发送电子邮件.

目前,它的转义字符,所以单引号和双引号转换为'和"分别.如果引号显示为'和,则电子邮件将更易读".

我理解为什么我永远不应该将访问者的非转义输入放入我的网页,因为xss存在风险.电子邮件是否存在相同的风险,或者发送访问者未转义的输入是否可以？

请问我为什么在模板中显示“fav”时收到此错误消息

QuerySet object has no attribute game_id\n

我尝试用 替换game_id，game但id_game什么也没有......

视图.py

from django.contrib import messages\nfrom django.conf import settings\nfrom django.contrib.auth.models import User\nfrom django.contrib.auth import authenticate, login, logout\nfrom django.shortcuts import render\nfrom start.models import Games, FavoriteGames\nimport urllib, json\n\ndef view_recap(request):\n    if request.user.is_authenticated():\n        username = request.user.username\n        id_user = request.user.id\n        fav = FavoriteGames.objects.filter(user_id=id_user).game_id\n        return render(request, \'recap.html\', locals())\n    else:\n        from start.views import view_logoff\n        from start.views import view_logon\n        messages.add_message(request, messages.INFO, \'Vous devez \xc3\xaatre connect\xc3\xa9 pour acc\xc3\xa9der \xc3\xa0 cette page.\')\n …