按照django 文档,我尝试在我的models.py 中做一个选择字段 :
class MyModel(Model):
ACTION_CHOICES = (
('d', 'delete'),
('c', 'create'),
('u', 'update'),
)
action = CharField(max_length=1, choices=ACTION_CHOICES)
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但是当我运行它时
action = MyModel(action='create')
action.save()
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并检查数据库的动作不是c,而是create。此外,即使对 max_length 的限制也不会对它产生影响。好的,我继续前进并尝试以这种方式制作它:
模型.py
class MyModel(Model):
ACTION_CHOICES = (
(1, 'delete'),
(2, 'create'),
(3, 'update'),
)
action = IntegerField(max_length=1, choices=ACTION_CHOICES)
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代码.py:
action = MyModel(action='create')
action.save()
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但是我得到一个错误,str 字段不能像 int 一样保存(数据库有 int 字段,但 MyModel.action 是 str)我在这里想念什么?选择不是应该在保存时转换价值吗?
此致,
我正在尝试将 Django 应用程序部署到 Heroku,其中一个必需的包存在于https://testpypi.python.org/pypi其中,当然 Django 位于主 PyPI 服务器上。
该requirements.txt文件如下所示:
Django==1.7.7
-i https://testpypi.python.org/pypi
foo-bar==0.4
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运行pip install -r requirements.txt失败并出现以下错误:
Could not find any downloads that satisfy the requirement Django==1.7.7 (from -r requirements.txt (line 1))
Cleaning up...
No distributions at all found for Django==1.7.7 (from -r requirements.txt (line 1))
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所以看起来pip是在试图找到 Djangotestpypi
所以我试过这个:
-i https://pypi.python.org/pypi/
Django==1.7.7
-i https://testpypi.python.org/pypi
foo-bar==0.4
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它导致相同的错误。
如果我在需求文件中只放置一个(无关紧要)包,pip 能够找到该包并安装它。
问题:index-url在单个文件中指定可由命令读取的多个不同参数的正确语法是什么pip install -r file
我认为这并不重要,但 python 是 3.4.0 版,而 pip …
我正在尝试将帖子标题转换为CamelCase来创建一个twitter标签,我正在使用strip但是它返回了一个对象而我不知道这是不是正确的方法?
# views.py
def post_create(request):
if not request.user.is_authenticated():
raise Http404
form_class = PostCreateForm
if request.method == 'POST':
form = form_class(request.POST, request.FILES)
if form.is_valid():
instance = form.save(commit=False)
instance.creator = request.user
instance.slug = slugify(instance.title)
instance.hashtag = instance.title.strip()
instance.save()
slug = slugify(instance.title)
return redirect(instance.get_absolute_url())
else:
form = form_class()
context = {
'form': form,
}
return render(request, "posts/post_create.html", context)
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哪个<built-in method strip of unicode object at 0x031ECB48>在模板var中返回,我正在寻找的结果就像这样MyPostTitle进入模板
# Template view
<h3>#{{instance.hashtag|title}}</h3>
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models.py
class Post(models.Model):
creator = models.ForeignKey(ProfileUser)
title = models.CharField(max_length=80) …Run Code Online (Sandbox Code Playgroud) 我正在关注官方的django教程https://docs.djangoproject.com/en/1.10/intro/tutorial04/
我正在使用Django 1.9.8,我现在收到以下错误/polls/views.py:
File "C:\Users\hk\Documents\mysite\polls\views.py", line 4, in <module>
from django.urls import reverse
ImportError: No module named urls
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民调/ views.py
from django.shortcuts import get_object_or_404, render
from django.http import HttpResponseRedirect, HttpResponse
from django.urls import reverse
from .models import Choice, Question
# ...
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mysite的/ urls.py
from django.conf.urls import include, url
from django.contrib import admin
admin.autodiscover()
urlpatterns=[
url(r'^admin/', admin.site.urls),
url(r'^polls/', include('polls.urls')),
]
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mysite的/ mysite的/ urls.py
from django.conf.urls import include,url
from django.contrib import admin
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^polls/', include('polls.urls')), …Run Code Online (Sandbox Code Playgroud) import factory.fuzzy
import faker
from core.strings import underscore_15
from factory import DjangoModelFactory, SubFactory
faker = faker.Factory.create()
class SubProjectFactory(DjangoModelFactory):
class Meta:
model = SubProject
django_get_or_create = ("internal_name",)
internal_name = factory.Faker('pystr')
internal_short_name = factory.Faker('pystr')
underscore_15 = factory.fuzzy.FuzzyText(length=15)
main_project = SubFactory(MainProjectFactory)
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我需要字段 underscore_15 是一个小写字符串,具体长度只有 15 个字符,并且没有空格。如果有空格,应该加下划线。我试图在周围放置一个函数factory.fuzzy.FuzzyText(length=15)。然后我意识到我假设 FuzzyText 返回一个字符串,但它是一个 FuzzyText 对象。
我应该怎么办?
例如我有一个字符串是
textstring= """
0000 Onn ch=1 n=60 v=50
0000 Onn ch=1 n=67 v=50
9600 Off ch=1 n=67 v=00
9600 Off ch=1 n=60 v=00
9600 Onn ch=1 n=62 v=50
9600 Onn ch=1 n=69 v=50
1920 Off ch=1 n=69 v=00
1920 Off ch=1 n=62 v=00
"""
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例如9600,当行的前四位数字相等时,如何将 67/60/62/69 打印在一起?(来自 之后的四行中的每一行n=)
我尝试过类似下面的方法,但我认为它没有按预期工作
for i, char in enumerate(textstring):
if char=="O" and (textstring[i+1]=="f" or textstring[i+1]=="n"):
if textstring[i-5]==textstring[i+18] and textstring[i-4]==textstring[i+19] and textstring[i-3]==textstring[i+20] and textstring[i-2]==textstring[i+21]:
if char=="n" and textstring[i+1]=="=": #we look for "n=" …Run Code Online (Sandbox Code Playgroud) 我知道已经有一个同样的问题,我也尝试过的解决方案,由于其但它力的工作对我来说因此我写我自己的问题。大家好,我正在使用apache2服务器部署django服务器,当我尝试我的主机ip地址时,它可以正常工作,因为它显示了它的“工作页面”,它也适用于IPADRESS / admin,但是在填写凭据并按Enter后,将产生此错误
***OperationalError at /admin/login/
unable to open database file***
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我的错误看起来像
Environment:
Request Method: POST
Request URL: http://192.168.1.105/admin/login/?next=/admin/
Django Version: 1.7.4
Python Version: 2.7.6
Installed Applications:
('django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
'managementdashboard')
Installed Middleware:
('django.contrib.sessions.middleware.SessionMiddleware',
'django.middleware.common.CommonMiddleware',
'django.middleware.csrf.CsrfViewMiddleware',
'django.contrib.auth.middleware.AuthenticationMiddleware',
'django.contrib.auth.middleware.SessionAuthenticationMiddleware',
'django.contrib.messages.middleware.MessageMiddleware',
'django.middleware.clickjacking.XFrameOptionsMiddleware')
Traceback:
File "/var/www/dashboardapp/env/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
111. response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "/var/www/dashboardapp/env/lib/python2.7/site-packages/django/views/decorators/cache.py" in _wrapped_view_func
52. response = view_func(request, *args, **kwargs)
File "/var/www/dashboardapp/env/lib/python2.7/site-packages/django/contrib/admin/sites.py" in login
361. return …Run Code Online (Sandbox Code Playgroud) 因此,我应该编写一个Python程序,该程序将以某个封闭的间隔[2,n](每行一个)识别并打印所有理想数。我们只需要使用嵌套的while循环/ if-else语句。我使用for循环以某种方式做到了,但使用while循环无法弄清楚。如果您能向我展示如何将我的代码转换为while循环,我将不胜感激。多谢你们。这是我所拥有的:
limit = int(input("enter upper limit for perfect number search: "))
for n in range(2, limit + 1):
sum = 0
for divisor in range(1, n):
if not n % divisor:
sum += divisor
if sum == n:
print(n, "is a perfect number")
Run Code Online (Sandbox Code Playgroud) 这是我第一次使用 Django 登录模板,但遇到了问题。它一直说 login 不是一个有效的视图函数,但我在我的用户应用程序的 urls.py 中定义了它:
不知道我做错了什么。
主要网址.py
urlpatterns = [
path('', include('blogging_logs.urls', namespace='blogging_logs')),
path('users/', include('users.urls', namespace='users')),
path('admin/', admin.site.urls),
]
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应用程序: blogging_logs:base.html
<p>
<a href="{% url 'blogging_logs:index' %}">Home</a>
<a href="{% url 'blogging_logs:categories' %}">Categories</a>
{% if user.is_authenticated %}
Hello, {{ user.username }}.
{% else %}
<a href="{% url 'users:login' %}"> login in</a>
{% endif %}
</p>
{% block content %}{% endblock content %}
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应用程序:用户:urls.py
from django.contrib import admin
from django.urls import re_path, path
from django.contrib.auth import authenticate, login
from django.contrib.auth …Run Code Online (Sandbox Code Playgroud) 我一直无法下载任何软件包,pip并且我不断收到有关确认 SSL 证书问题的相同错误。我目前在我大学的网络上,我的笔记本电脑上也安装了 Anaconda。我不确定这两个因素是否与这个问题有关。
我正在使用pythonversion 3.7.0、pipversion9.0.1和pip3version 10.0.1。
当我pip install django在终端中输入时,我得到以下结果:
Collecting django
Could not fetch URL https://pypi.python.org/simple/django/:
There was a problem confirming the ssl certificate:
[SSL: CERTIFICATE_VERIFY_FAILED] certificate verify failed (_ssl.c:749)
- skipping
Could not find a version that satisfies the requirement django (from versions: )
No matching distribution found for django
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我已经阅读了其他几个主题。我试过这样的命令没有成功:
curl https://bootstrap.pypa.io/get-pip.py | python3
pip install --upgrade pip
pip install --trusted-host pypi.org --trusted-host …Run Code Online (Sandbox Code Playgroud) python ×8
django ×7
python-3.x ×3
pip ×2
string ×2
django-admin ×1
django-views ×1
django-wsgi ×1
factory-boy ×1
faker ×1
heroku ×1
python-2.7 ×1
url ×1