小编BoJ*_*man的帖子

我刚刚加入了一个拥有相当大的现有代码库的项目.我们在linux中开发并且不使用和IDE.我们通过命令行运行.我正在试图弄清楚如何在运行项目模块时让python搜索正确的路径.例如,当我运行类似的东西:

python someprojectfile.py

我明白了

ImportError: no module named core.'somemodule'

我得到了所有我的导入,我认为这是路径的问题.

TLDR:

如何~/codez/project/在导入语句期间让Python搜索*.py文件的所有文件和文件夹.

当我尝试将属性与另一个具有M到M关系的属性关联时,我收到此错误:

get()返回了多个主题 - 它返回2!

你能告诉我这意味着什么,也许可以提前告诉我如何避免这个错误？

楷模

class LearningObjective(models.Model):
    learning_objective=models.TextField()

class Topic(models.Model):
    learning_objective_topic=models.ManyToManyField(LearningObjective)
    topic=models.TextField()

输出 LearningObjective.objects.all()

[<LearningObjective: lO1>, <LearningObjective: lO2>, <LearningObjective: lO3>]

输出 Topic.objects.all()

[<Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>, <Topic: Topic object>]

意见

 def create_themen(request):
     new_topic=Topic(topic=request.POST['topic'])
     new_topic.save()
     return render(request, 'topic.html', {'topic': topic.objects.all()})

 def create_learning_objective(request):
     new_learning_objective=LearningObjective(learning_objective=request.POST['learning_objective']) …

尝试使用非常简单的表单将文件上载到新的类实例中.我希望有两个文件,request.FILES但它是空的.我在捆绑的开发服务器上.

被困在这里并经历了所有相关问题.

wayfinder_map.media_file = request.FILES['media_file'] 

生成

MultiValueDictKeyError:"在MultiValueDict中找不到键'media_file':{}>"

模型

class WayfinderMap(models.Model):
    """ Way-finding Map Config"""


    media_file = models.FileField(upload_to="maps", null=True, blank=True) 
    wall_file = models.FileField(upload_to="maps_data", null=True, blank=True) 

视图

@login_required
def create_map(request, form_class=WayfinderMapForm, template_name="wayfinder/map/create.html"):
wayfinder_map_form = form_class(request.user, request.POST or None, request.FILES)

    if wayfinder_map_form.is_valid():
        wayfinder_map = wayfinder_map_form.save(commit=False)
        wayfinder_map.media_file = request.FILES['media_file']
        wayfinder_map.data_file = request.FILES['data_file']
        wayfinder_map.creator = request.user
        wayfinder_map.save()
    return HttpResponseRedirect(wayfinder_map.get_absolute_url())

return render_to_response(template_name, {
    "wayfinder_map_form": wayfinder_map_form,
}, context_instance=RequestContext(request))

模板

<form enctype="multipart/form-data" class="uniForm" id="wayfinder_map_form" method="POST" action="">
        <fieldset class="inlineLabels">
            {{ wayfinder_map_form|as_uni_form }}
            <div class="form_block">
                <input type="hidden" name="action" …

我想在select小部件中将字段限制为值0-10.

field=models.IntegerField(max_length=10, choices=CHOICES)

我可以写出从(0,0),(1,1)开始的所有选择元组,但必须有一个明显的方法来处理它.

非常感谢帮助.

我正在我的本地机器上使用Django(1.5.1)构建一个库.在我的相册模型中,我有一个ImageField.可以显示相册的所有图像.它运作良好,但最终图像不会显示.您可以看到图像的边框,但图像无法加载.

截图

models.py

class Category(models.Model):
 ###  
class Album(models.Model):   
    category = models.ForeignKey(Category, related_name='albums')
 ###
class Image(models.Model):
    album = models.ForeignKey(Album)
    image = models.ImageField(upload_to = 'images/albums/')

views.py

def detail(request, album_id):
    album = get_object_or_404(Album, pk=album_id)
    return render(request, 'gallery/detail.html', {'album': album})

detail.html

<h1>{{ album.title }}</h1>
{% for image in album.image_set.all %}
  <a>  <img src="{{ image.image.url }}" height="420"></a>
{% endfor %}

如果这是我的专辑地址: http://localhost:8000/gallery/1/

然后图像URL是:http://localhost:8000/media/images/albums/photo_4.JPG (I get 404 when enter it in browser)

这个媒体根和网址:

MEDIA_ROOT = '/media/'    
MEDIA_URL = '/localhost:8000/media/'  

我的媒体root有777权限.

我现在应该怎么做？问题出在哪儿？

尝试使用pip从cmd安装Django.

Python 2.7.3, Windows 7

Env var(...C:\Python27;C:\Python27\Lib\site-packages\;C:\Python27\Scripts\;)

    help("pip")
    .
    .
    .
    .
    VERSION
        1.3.1

    pip install Django
    File "<stdin>", line 1
    pip install Django
          ^
    SyntaxError: invalid syntax

我使用tinymce作为动态生成至少5个文本的网页.
我使用的配置仅适用于第一个textarea.

tinyMCE.init({
    height : "300",
    mode : "exact",
    elements : "content",
    theme : "simple",
    editor_selector : "mceEditor",
    ...

<textarea class="mceEditor" name="content" rows="15" cols="40">content</textarea>

什么是在所有人中 启用tinymce编辑的配置textarea.

我有一个名为baseusers的用户组,其中包含我通过管理员添加的一组权限.如果baseusers不存在,我想创建它并设置它现在拥有的相同权限.我可以创建组但​​是如何从shell中找到权限然后将其添加到我的代码中？例如,两个权限名为can_fm_list和can_fm_add.应用程序称为fileman.但我需要使用id:s吗？这是我想要设置权限的代码.

userprofile/views.py

from fileman.models import Setting (Has the permissions)
from fileman.models import Newuserpath
from django.contrib.auth.models import Group

def register(request):

...

            if Group.objects.count() > 0:
            newuser.groups.add(1)
            else:
            newgroup = Group(name='baseusers')
            newgroup.save()
            newuser.groups.add(1)

fileman/models.py

class Setting(models.Model):
    owner = AutoForeignKey(User, unique=True, related_name='fileman_Setting')
    root = models.CharField(max_length=250, null=True)
    home = models.CharField(max_length=250, null=True)
    buffer = models.TextField(blank=True)
    class Meta:
        permissions = (
            ("can_fm_list", _("Can look files list")),
            ("can_fm_add", _("Can upload files")),
            ("can_fm_rename", _("Can rename …

import mysql.connector

config = {
    'user' : 'root',
    'passwd' : ' ',
    'host' : 'localhost',
    'raise_on_warnings' : True,
    'use_pure' : False,
    }
cnx = mysql.connector.connect(**config)

cnx.close()

我用这段代码来检查我使用mysql提供的安装程序安装的mysql包

我在终端运行文件,结果是,

Traceback (most recent call last):
 File "/Users/Krishna/Documents/check.py", line 1, in <module>
   import mysql.connector
ImportError: No module named 'mysql'

非常感谢帮助.

我正在运行一个基于python flask的web服务,我想执行一个小的MySQL查询.当我获得SQL查询的有效输入时,一切都按预期工作,我得到了正确的值.但是,如果该值未存储在数据库中,则会收到一个TypeError

    Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1478, in full_dispatch_request
    response = self.make_response(rv)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1566, in make_response
    raise ValueError('View function did not return a response')
ValueError: View function did not return a response

我试图自己处理错误处理并将此代码用于我的项目,但似乎这不能正常工作.

#!/usr/bin/python

from flask import …