小编tde*_*lam的帖子

我有一个运行我们几年前在内部构建的迷你CMS的Django网站,它使用的是postgresql.保存简单标题和文本段落时,我收到以下错误:

value too long for type character varying(100)

奇怪的是,没有一个列是变化的(100)它们都是200或250,即使是默认的Django已经从100改为200由于这里提到的重新打开的票

有谁知道这个问题的解决方案？

我想知道使用Django/Python通过Web应用程序上传大小约为4GB的文件是否有任何后果？我记得在过去使用Java的流式上传是首选的方法但是今天仍然这样做或者使用Django/Python这样做是否完全安全？

我config.action_mailer.default_url_options = { :host => 'localhost:3000' }按照建议安装了devise并添加到我的development.rb文件中.当我运行黄瓜时,我收到一个错误:

缺少主机链接!请提供:host参数或设置default_url_options [:host](ActionView :: Template :: Error)

有谁知道这与什么有关？谷歌没有太多关于此的信息

有没有办法着色Django测试输出？基本上是红色/绿色因素导致通过或失败的结果？

我在Mac OS X上使用Terminal.app.

我一直收到这个错误......

[2012-06-14 11:54:50,072: ERROR/MainProcess] Hard time limit (300s) exceeded for movies.tasks.encode_media[14cad954-26e2-4511-94ec-b17b9a4149bb]

[2012-06-14 11:54:50,111: ERROR/MainProcess] Task movies.tasks.encode_media[bc173429-77ae-4c96-b987-75337f915ec5] raised exception: TimeLimitExceeded(300,)

Traceback (most recent call last): File "/srv/virtualenvs/filmlib/local/lib/python2.7/site-packages/celery/concurrency/processes/pool.py", line 370, in _on_hard_timeout raise TimeLimitExceeded(hard_timeout)
TimeLimitExceeded: 300

即使我有CELERYD_TASK_TIME_LIMIT = 86400和CELERYD_TASK_SOFT_TIME_LIMIT = 86400我的settings.py项目.有谁知道为什么它仍然默认为300s？

谢谢

我目前正在从周一开始到周日结束,但我怎样才能从周一开始到周日结束？这是我现在的本周:

>>> import datetime
>>> today = datetime.date.today()
>>> weekday = today.weekday()
>>> start_delta = datetime.timedelta(days=weekday)
>>> start_of_week = today - start_delta
>>> week_dates = []
>>> for day in range(7):
...     week_dates.append(start_of_week + datetime.timedelta(days=day))
...
>>> week_dates
[datetime.date(2013, 10, 28), datetime.date(2013, 10, 29), datetime.date(2013, 10, 30),     datetime.date(2013, 10, 31), datetime.date(2013, 11, 1), datetime.date(2013, 11, 2), datetime.date(2013, 11, 3)]
>>> week_dates[0], week_dates[-1]
(datetime.date(2013, 10, 28), datetime.date(2013, 11, 3)) <--- Monday, Sunday

我在我的生产服务器上运行了迁移,并且我正在使用MySQL,我收到此错误:

Mysql2 ::错误:'admin'的默认值无效:ALTER TABLE usersADD admintinyint(1)DEFAULT'false'`

我的迁移看起来像这样:

class AddAdminToUsers < ActiveRecord::Migration
  def change
    add_column :users, :admin, :boolean, default: :false
  end
end

我理解错误是因为"false"不是tinyint的正确值,在这种情况下应该为0.我认为默认:: false是将布尔值默认为false的正确方法.

我如何解决这个问题,以便MySQL不会抱怨坏的价值？

我的所有用户在获得管理员批准之前都将被取消批准,管理员将登录该网站以将用户标记为已批准.我正在关注这里的Devise文档,这些文档效果很好但是如果新用户注册后如何向管理员发送电子邮件以便管理员知道并且可以批准注册？

我一直收到错误 Could not import movies.views. Error was: No module named models

这是我的追溯:

    Environment:


    Request Method: GET
    Request URL: http://localhost:8000/movies/awesome-movie/

    Django Version: 1.3.1
    Python Version: 2.7.3
    Installed Applications:
    ['django.contrib.auth',
     'username_patch',
     'django.contrib.contenttypes',
     'django.contrib.sessions',
     'django.contrib.sites',
     'django.contrib.messages',
     'django.contrib.staticfiles',
     'django.contrib.admin',
     'django.contrib.flatpages',
     'south',
     'tagging',
     'tagging_autocomplete',
     'accounts',
     'movies',
     'tracking',
     'djcelery',
     'pagination']
    Installed Middleware:
    ('django.middleware.common.CommonMiddleware',
     'django.contrib.sessions.middleware.SessionMiddleware',
     'django.middleware.csrf.CsrfViewMiddleware',
     'django.contrib.auth.middleware.AuthenticationMiddleware',
     'django.middleware.locale.LocaleMiddleware',
     'django.contrib.messages.middleware.MessageMiddleware',
     'django.contrib.flatpages.middleware.FlatpageFallbackMiddleware',
     'pagination.middleware.PaginationMiddleware')


    Traceback:
    File "/Users/jeff/Code/filmlibrary/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
      101.                             request.path_info)
    File "/Users/jeff/Code/filmlibrary/lib/python2.7/site-packages/django/core/urlresolvers.py" in resolve
      252.                     sub_match = pattern.resolve(new_path)
    File "/Users/jeff/Code/filmlibrary/lib/python2.7/site-packages/django/core/urlresolvers.py" in resolve
      252.                     sub_match = pattern.resolve(new_path)
    File "/Users/jeff/Code/filmlibrary/lib/python2.7/site-packages/django/core/urlresolvers.py" in …

我正在尝试将超过1个文件资源上传到电影中.我一直收到错误Asset model missing required attr_accessor for 'asset_file_name',我不知道为什么,这是我的代码:

asset.rb模型

class Asset < ActiveRecord::Base
  belongs_to :movie

  has_attached_file :asset
end

movie.rb模型

class Movie < ActiveRecord::Base
  belongs_to :user

  has_many :assets
  accepts_nested_attributes_for :assets

  validates :title, presence: true

  default_scope order: 'movies.created_at DESC'
end

我还运行rails g paperclip movie asset了生成以下迁移文件

class AddAttachmentAssetToMovie < ActiveRecord::Migration
  def self.up
    add_column :movies, :asset_file_name, :string
    add_column :movies, :asset_content_type, :string
    add_column :movies, :asset_file_size, :integer
    add_column :movies, :asset_updated_at, :datetime
  end

  def self.down
    remove_column :movies, :asset_file_name
    remove_column :movies, :asset_content_type
    remove_column :movies, :asset_file_size …

我正在使用邪恶的宝石来构建表单向导.在文档中,它使用的是current_user对象,但我想使用其他对象.我正在努力为redirect_to添加一个参数,所以我可以使用这个对象.

products_controller.rb

def create
  @product = current_user.product.build(params[:product])
  @product.ip_address = request.remote_ip

  if @product.save
    redirect_to product_steps_path # This is where I think I need to pass product object but not sure how
  else
    render 'new'
  end
end

product_steps_controller.rb

class ProductStepsController < ApplicationController
  include Wicked::Wizard
  steps :apps, :templates

  def show
    @product = Product.find(params[:id])
    render_wizard
  end
end

路线:

resources :products
resources :product_steps

我用上面的代码得到的错误是:

ActiveRecord::RecordNotFound in ProductStepsController#show

Couldn't find Product with id=apps

如何在这样的特定对象上使用邪恶的gem而不是文档中的current_user示例？

我有一种情况,用户将被要求输入他们的邮政编码,一旦他们这样做,他们将被重定向到与他们输入的邮政编码相关的内容的网站.将有类别等.简而言之,用户将看到仅与其区域相关的内容.

计划 - 我想将他们的邮政编码存储在一个cookie中,并在每次回到网站时重复使用它,当然如果没有邮政编码,他们应该被指示在表格中输入.表格将在任何其他内容出现之前出示.

我的问题是我应该使用装饰器并用自定义装饰器装饰视图吗？还是应该写中间件？如果我为此编写一个自定义中间件,我应该使用process_request process_view,在我看来,每次请求都会调用process_request,这可能会造成问题.

谢谢!

JeffC

我有一个这样的表格:

from django import forms
from django.contrib.auth.models import User

from django_countries.countries import COUNTRIES

from statuses.models import Status

class StatusForm(forms.Form):
    country = forms.ChoiceField(choices=COUNTRIES)
    mood = forms.IntegerField()
    sleep_quality = forms.IntegerField()

此表单仅显示给登录的用户,如何设置request.user以便在用户提交此表单时,我可以将表单条目与它们关联？我的模型与用户FK类似:

from django.db import models
from django.contrib.auth.models import User

from django_countries import CountryField

class Status(models.Model):
    user = models.ForeignKey(User)
    country = CountryField()
    mood = models.SmallIntegerField(default=4)
    sleep_quality = models.SmallIntegerField(default=4)

以下是我对此表单的看法:

@login_required
def index(request, template_name="status/index.html"):
    if request.method == 'POST':
        postdata = request.POST
        form = StatusForm(postdata)
        if form.is_valid():
            messages.success(request, 'Something happened, good!')
            return redirect(urlresolvers.reverse('profile')) …